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Interesting problems from the AMATYC Student Math League Exams

(November 2003, #1) If L has equation , M is its reflection across the y-axis, and N is its reflection across the x-axis, which of the following must be true about M and N for all nonzero choices of a, b, and c?

Suppose that L has an x-intercept of A and a y-intercept of B. Then its slope is . M has

an x-intercept of -A and a y-intercept of B. So its slope is . N has an x-intercept

of A and a y-intercept of -B,. So its slope is .

So the correct answer is C) the slopes are equal.

Or just notice that the addition of the dashed line must result in a parallelogram which forces M and N to have the same slope.

Or notice that the equation of M must be so its slope is , and the equation

of N must be so its slope must be .[See the section on Graph Properties]

LM

N

(November 2003, #2) A collection is made up of an equal number of pennies, nickels, dimes, and quarters. What is the largest possible value of the collection which is less than $2?

. So the largest possible value is .

So the correct answer is A) $1.64.

(November 2003, #3) When the polynomial is divided by , the remainder is

. What is the remainder when is divided by ?

, so . But

this means that the remainder when is divided by is

.

So the correct answer is B) . [See the section on Polynomial Properties]

(November 2003, #4) If , find .

.

So the correct answer is B) 55.

(November 2003, #5) What is the remainder when is divided by ?

. So the answer is .

So the correct answer is A) . [See the section on Polynomial Properties]

(November 2003, #6) Let p be a prime number and k an integer such that has two positive integer solutions. What is the value of ?

must factor into where a and b are integers. But the only way this can happen is if one of them is p and the other is 1. This means that

, so . This rearranges into .

So the correct answer is B) . [See the section on Polynomial Properties]

(November 2003, #7) What is the least number of prime numbers (not necessarily different) that 3185 must be multiplied by so that the product is a perfect cube?

. So to turn it into a perfect cube with the least number of multiplications by primes, we would need to multiply it by .

So the correct answer is E) 5. [See The Fundamental Theorem of Arithmetic]

(November 2003, #8) Two adjacent faces of a three-dimensional rectangular box have areas 24 and 36. If the length, width and height of the box are all integers, how many different volumes are possible for the box?

and , and , where H is a common factor of

36 and 24. So the number of different volumes is the same as the number of common factors of 36 and 24. The common factors are 1, 2, 3, 4, 6, and 12.

So the correct answer is E) 6.

(November 2003, #9)

.

So the correct answer is C) . [See the section on Trigonometric Formulas]

L W

H

(November 2003, #10) The counting numbers are written in the pattern at the right. Find the middle number of the 40th row.

1

2 3 4

5 6 7 8 9

10 11 12 13 14 15 16

The middle numbers are generated by adding consecutive even numbers to 1.

.

Or another pattern is that the middle number of the nth row is given by , so

.

So the correct answer is A) 1561. [See the section on Algebraic Formulas]

(November 2003, #11) The solution set of is a subset of the solution set of which of the following inequalities?

No

No

6-3

S SSS

5-4

S SSS

4-3

S SSS

Yes

So the correct answer is C) .

(November 2003, #12) If and , find .

.

So the correct answer is A) . [See the section on Algebraic Formulas]

(November 2003, #13) Square ABCD is inscribed in circle O, and its area is a. Square EFGH is inscribed in a semicircle of O. What is the area of square EFGH?

The area of an inscribed square in a circle of radius r, is .

D C

BA

E F

GH

S SSS

The area of an inscribed square in a semicircle of radius r, is .

So if , then the area of the square inscribed inside the semicircle is .

So the correct answer is B) . [See the section on Geometric Formulas]

(November 2003, #14) Consider all arrangements of the letters AMATYC with either the A’s together or the A’s on the ends. What fraction of all possible such arrangements satisfy these conditions?

A’s together:

A’s on the ends:

Total number of different arrangements:

.

So the correct answer is D) . [See the section on Sets and Counting]

r

r

(November 2003, #15) The year 2003 is prime, but its reversal, 3002, in not. In fact, 3002 is the product of exactly three different primes. Let N be the sum of these three primes. How many other positive integers are the products of exactly three different primes with this sum N?

, so . , since the sum of three distinct primes not equal to 2 must be an odd number, one of the primes must be 2.

Now let’s check the possible prime values for , and see which ones make a prime number as well.

So there are two other numbers: and

So the correct answer is C) 2.

(November 2003, #16) In a group of 30 students, 25 are taking math, 22 English, and 19 history. If the largest and smallest number who could be taking all three courses are M and m respectively, find .

So .

So the correct answer is E) 25. [See the section on Sets and Counting]

(November 2003, #17) A boat with an ill passenger is 7½ miles north of a straight coastline which runs east and west. A hospital on the coast is 60 miles from the point on shore south of the boat. If the boat starts toward shore at 15 mph at the same time an ambulance leaves the hospital at 60 mph and meets the ambulance, what is the total distance(to the nearest .5 mile) traveled by the boat and the ambulance?

boat

hospital

15t

60t60 60t

152

.

So the total distance traveled is .

So the correct answer is E) 62.5.

(November 2003, #18) If each letter in the equation represents a different decimal digit, find T’s value.

and to produce a 6-digit number, MYM must be at least 323. So let’s start checking the squares of MYM values:

, so .


(November 2003, #19) If a, b, c, and d are nonzero numbers such that c and d are solutions of and a and b are solutions of , find .

If c and d are solutions of , then

, so and .

If a and b are solutions of , then

, so and .

So we get the system

.

Solving the linear part

Leads to

With solutions of . Plugging this into the other two

equations leads to and . So the solution of the system is

, so .

So the correct answer is A) .

(November 2003, #20) Al and Bob are at opposite ends of a diameter of a silo in the shape of a tall right circular cylinder with radius 150 feet. Al is due west of Bob. Al begins walking along the edge of the silo at 6 feet per second at the same moment that Bob begins to walk due east at the same speed. The value closest to the time in seconds when Al first can see Bob is

At a point on the circle where a tangent line will intersect the x-axis, , it will

intersect the x-axis at the point .

Here’s why:

The slope of the tangent line at the point on the circle is , so

an equation of the tangent line is . Setting y equal

to zero and solving for x to find the x-intercept results in .

Al’s path can be parametrized by . So we need to solve the

equation . An approximate solution of this equation is

48.00747736.

Here’s the graph of both sides of the equation:


(February 2004, #1) A stock loses 60% of its value. What must the percent of increase be to recover all of its lost value?

.

So the correct answer is C) 150%.

(February 2004, #2) Which of the following is NOT a factor of ?

, so is not a factor of

.

So the correct answer is D) . [See the section on Polynomial Properties]

(February 2004, #3) The library in Johnson City has between 1000 and 2000 books. Of these,

25% are fiction, are biographies, and are atlases. How many books are either

biographies or atlases?

The total number of books must be a common multiple of 4, 13, and 17. The LCM of these numbers is 884. The total number of books must be a multiple of the LCM that’s between 100

0 and 2000, so it must be 1768. The number of biographies is , and the number of

atlases is , so the number of books that are either biographies or atlases is 240.

So the correct answer is A) 240. [See the section on LCM and GCF]

(February 2004, #4) A tricimal is like a decimal, except the digits represent fractions with

powers of 3 instead of 10. For example, as a tricimal. How is

expressed as a tricimal?

So the correct answer is E) .2212.

(February 2004, #5) The function can be written as sums and differences of

powers of . When is written that way, what is the coefficient of ?

Since , we get that

. Expanding the

right side will only yield even powers of , so the coefficient of is zero.

So the correct answer is A) 0. [See the section on Trigonometric Formulas]

(February 2004, #6) If , find .

.

So the correct answer is D) 96. [See the section on Logarithmic Properties]

(February 2004, #7) The number 877530p765q6 is divisible by both 8 and 11, with p and q both digits from 0 to 9. The number is also divisible by

In order for the number to be divisible by 8, 5q6, the number consisting of the ones, tens, and hundreds digits must be divisible by 8. This means that the only possible values for q are 3 and 7. In order for the number to be divisible by 11, the difference in the sum of the even digit positions and the odd digit positions must be divisible by 11. For 877530p765q6 the difference of these sums is . So the only

possibility is , which leads to the number 877530376536. Since the sum of its digits is which is divisible by 3, the number must be also be divisible by 3, 6, 12, 24, and 33.

So the correct answer is B) 12. [See the section on Divisibility Rules]

(February 2004, #8) Teams A and B play a series of games; whoever wins two games first wins the series. If Team A has a 70% chance of winning any single game, what is the probability that Team A wins the series?

Team A will be the winner only if the following results occur:

AA, BAA, ABA.

These occur with the following probabilities:

.

So the probability that Team A wins the series is .

So the correct answer is E) .784. [See the section on Probability Formulas]

(February 2004, #9) The Venn diagram at the right represents sets A, B, and C(not necessarily in that order). Depending on how the diagram is labeled, how many different answers are possible for the number of elements in the set ? (Note: is all elements which are in A but not in B.)

The possible designations of the sets A, B, and C are

These lead to the following values for the number of elements in :

4, 8, 4, 5, 5, 2, so there are 4 different values.


4 4 1

20

2

3A B

CA C

BB A

CC A

B

B CA

C BA

(February 2004, #10) A fixed point for a function is a real number r such that

. How many of the following functions must have a fixed point?

A polynomial function of even degree > 0 without a fixed point would be .

A rational function without a fixed point would be .

For a polynomial function of odd degree > 1, . The equation leads to the

polynomial equation . Since the left side is an odd degree polynomial, it must have at least one real solution, and hence the original function must have a fixed point.

For a trigonometric function , the graph must intersect the graph of the identity function, and hence it must have at least one fixed point.


(February 2004, #11) Which of the following is the identity function for all real numbers?

only for .

for all x.

only for .

only for

only for

So the correct answer is B) .

(February 2004, #12) A circular table is pushed into the corner of a rectangular room so that it touches both walls. A point on the edge of the table between the two points of contact is 2 inches from one wall and 9 inches from the other wall. What is the radius of the table?

So .

So the correct answer is D) 17 inches. [See the section on Geometric Formulas]

(February 2004, #13) In , and . If D is the midpoint of side AC,

find .

2

2

5

3

So .


(February 2004, #14) Enrique walks along a level road and then up a hill. At the top of the hill he immediately turns and walks back to his starting point. He walks 4 mph on level ground, 3 mph uphill, and 6 mph downhill. If the entire walk takes 6 hours, how far does he walk?

His walk consists of two equal level ground portions of length L, one uphill portion of length U, and one equal downhill portion of length D. So the total distance he walks is equal to

.

We also know that .

So the correct answer is C) 24 mi..

(February 2004, #15) If , then

.


(February 2004, #16) A bag holds 5 cards identical except for color. Two are red on both sides, two are black on both sides, and one is red on one side and black on the other. If you pick a card at random and see that the only side you can see is red, what is the probability that the other side is red?

An equally likely sample space for the experiment is

, where the first part of the ordered pair is the color you see, and the second part is the color you don’t see.

If the color you see is red, then the sample space has been reduced to

and of these, the only outcomes with the other side also

red are , so the probability is .

So the correct answer is D) . [See the section on Probability Formulas]

(February 2004, #17)The set S contains the number 2, and if it contains the number n, it also contains and (assume S contains only numbers produced by these rules). Which of the following is NOT in S?

For a number to be in S, then some sequence of subtractions of 5 and divisions by 3 should take you back to the number 2.

This leaves 2000.

So the correct answer is A) 2000.

(February 2004, #18) Let , with both positive integers. If for positive

integers p and q, and , what is the value of b?

and , so subtracting leads to . From this we can conclude that . Since and both are positive integers, we can disregard 1. If , then the possible values of b are 1 and 2. leads to , which doesn’t yield an integer solution. leads to , which doesn’t yield an integer solution. If , then the possible values of b are 1, 2, 3, 4, 5, and 6. leads to

and which yield integer solutions. So b must be 4.


(February 2004, #20) Ed has four children, Al, Bo, Cy, and Di(in order oldest to youngest). Bo is 4 years older than Cy and 12 years older than Di. This year Ed notices that he is twice as old as Bo, and the sum of the squares of the children’s ages equals the square of Ed’s age. If Di just became a math teacher, what is the sum of the children’s ages?

. So we get that

. Let’s check values of B starting with 30 to see if the right side is a square and the other conditions are satisfied. The

first value to work is . It produces the values , and all the other conditions are satisfied. The sum of the children’s ages is

.


(October 2004, #2) In square ABCD, point E is between A and B, and point F is between B and C. Find the sum of the measures of and .

Consider the pentagon AEFCD. It can be dissected into three triangles with angle sum of , and so the angle sum of the pentagon is also . To get the sum of the angles and

, we just have to subtract the three angles at A, D and C. So the sum of the angles and is .

So the correct answer is C) . [See the section on Geometric Formulas]

(October 2004, #4) A newspaper advertises that it sells the Sunday paper for one-third the price of the rest of the week’s papers. If a weekly subscription costs between $2.20 and $2.30, what is the cost of one Sunday paper and one daily paper?

. . So a daily paper costs 28 cents and a Sunday paper costs 56 cents, giving a total of 84 cents.

So the correct answer is C) $.84.

(October 2004, #6) A date is called weird if the number of its month and the number of its day have greatest common factor 1. What are the fewest number of weird days in any month?

We want to look at months with numbers that have a lot of different factors: 6(June) and 12(December). It might also be nice to look at months with a fewer number of days.

6(June): 1, 5, 7, 11, 13, 17, 19, 23, 25, 29 for a total of 10 weird days.

12(December): 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31 for a total of 11 weird days.


A B

CD

E

F

(October 2004, #7) Lucia is not yet 80 years old. Each of her sons has as many sons as brothers. The combined number of Lucia’s sons and grandsons equals her age, and her oldest grandson is 29. How old is Lucia? Place your numerical answer in the corresponding answer blank.

Let B be the number of Lucia’s sons, S be the number of Lucia’s grandsons, and L be Lucia’s age. Then we get the following equations: . So we know that Lucia’s age must be a square number less than 80 and greater than 59. Lucia must be 64.

So the correct answer is 64.

(October 2004, #8) What is ?

Let .

From the pictures, you can see that

.


(October 2004, #9) George bought groceries with a $10 bill. The cost of the groceries had three different digits, and the amount of his change had the same 3 digits in a different order. What was the sum of the digits in the cost?

Suppose the cost was $a.bc and the change was $c.ab, then it must be that . So

, and hence . So the sum of the digits in the cost is 14.


4 5

3

5

4

(October 2004, #10) Let N be the smallest number divisible by 33 which is greater than 1,000,000 and whose digits are all 0’s and 1’s. What are N’s leading four digits?

Assuming that N is a seven-digit number whose digits are all 0’s and 1’s that is divisible by 33, then it must start with a 1 and be divisible by 3 and 11. To be divisible by 3, the sum of its digits must be divisible by 3, which means that it has three 1’s or six 1’s. To be divisible by 11 means that the difference in alternate digit sums must be divisible by 11, which means that the difference in alternate digit sums must be 0, so the number of 1’s must be even. Now we know that it starts with a 1 and has 5 additional 1’s and one 0. So the 6 possibilities are

1011111, 1101111, 1110111, 1111011, 1111101, 1111110.

But the only ones divisible by 11 are 1101111, 1111011, and 1111110. The smallest one is 1101111.

So the correct answer is D) 1101. [See the section on Divisibility Rules]

(October 2004, #12) The song “What a Beautiful Life” has the lyric, “Day 18,253, well, honey, that’s fifty years.” If the lyric was supposed to be exactly correct, by how many days is it wrong?

Every fourth year is a leap year of 366 days, unless the year is divisible by 100 and doesn’t leave a remainder of 200 or 600 when divided by 900. This gives the average year approximately 365.24 days. , so the lyric is off by about 9 days.

So the correct answer is D) 8 to 10.

(October 2004, #13) Chris traveled 1 hour longer and 2 miles farther than Calvin, but averaged 3 mph slower. If the sum of their times was 4 hours, what was the sum in miles of the distances they traveled?

, , , and . From the first and

last equations we get that and . Combining this with the second and third

equations leads to . So the sum of the distances is

.

So the correct answer is D) 30.5.

(October 2004, #15) Find the sum of the x and y intercepts of the line with slope which is

the hypotenuse of a right triangular region in Quadrant I with legs the x and y axes and area

.

The slope of implies that . Area of implies that . Combining the

two equations leads to . So the x-intercept is 28

and the y-intercept is , giving a sum of .

So the correct answer is D) . [See the section on Equations of Lines]

[See the section on Geometric Formulas]

(October 2004, #16) Let . How many three-element subsets of A contain at least two consecutive integers?

Let’s count how many sets of 3 don’t have any consecutive integers:

Smallest element Other two elements

0 2,4

0 2,5

0 2,6

0 2,7

0 2,8

x

y

0 2,9

0 3,5

0 3,6

0 3,7

0 3,8

0 3,9

0 4,6

0 4,7

0 4,8

0 4,9

0 5,7

0 5,8

0 5,9

0 6,8

0 6,9

0 7,9

The number of three element sets with smallest element 0 without any consecutive integers is . Similarly, the number of three element sets with smallest element 1 is

. Similarly, the number of three element sets with smallest element 2 is . Similarly, the number of three element sets with smallest element 3 is

. Similarly, the number of three element sets with smallest element 4 is . Similarly, the number of three element sets with smallest element 5 is . So the number of subsets of size 3 without any consecutive integers is . Since the

total number of subsets of size 3 is , there must be

subsets of size 3 that contain at least two consecutive integers.


(October 2004, #17) If x, y and z are positive integers with and , find the smallest possible value of .

Consider the system . Subtracting twice the first equation from the second

equation leads to . Adding the second equation to the first equation leads to

. Dividing the second equation by leads to . So the

solutions are given by , where t is an even integer with .

This means that , where t is an even integer with . So the

smallest possible value of is 1003.


(October 2004, #18) A store has four open checkout stands. In how many ways could six customers line up at the checkout stands?

Here are the possibilities sorted by largest number at a single checkout stand:

# of people 6 0 0 0 ways to line up # of arrangements Total

Checkout stand 1 2 3 4 2880









# of people 3 2 1 0 ways to line up # of arrangements TotalCheckout stand 1 2 3 4 17280







So the total number of ways the customers can line up is

.

So the correct answer is D) 60480. [See the section on Sets and Counting]

(October 2004, #20) Suppose , (a, b integers). If and

, then find the product of a and b.

.

.

So , and this means that and . Therefore, .

So the correct answer is B) 12. [See the section on Algebraic Formulas]

(February 2005, #5) A palindrome is a word or a number (like RADAR or 1221) which reads the same forwards and backwards. If dates are written in the format MMDDYY, how many dates in the 21st century are palindromes?

Let’s deal with the months that are one digit:

1 9 2 1 1 1 18

M(0) M(1-9) D(1,2)D(same as first day

digit)

Y(same as second month

digit)

Y(same as first month

digit)Total

Now for the months that are two digit:

1 3 2 1 1 1 6

M(1) M(0,1,2) D(1,2)D(same as first day

digit)

Y(same as second month

digit)

Y(same as first month

digit)Total

So the number of dates in the 21st century that are palindromes is 24.

So the correct answer is C) 24. [See the section on Sets and Counting]

(February 2005, #8) Mrs. Abbott finds that the number of possible groups of 3 students in her class is exactly five times the number of possible groups of 2 students. How many students are in her class?

The number of possible groups of 3 students is .

The number of possible groups of 2 students is . So

.

So the correct answer is B) 17. [See the section on Sets and Counting]

(February 2005, #9) In how many ways can slashes be placed among the letters AMATYCSML to separate them into four groups with each group including at least one letter?

Three slashes must be placed in the spaces between the letters. Since there are 8 spaces, there

are ways to do this.


(February 2005, #10) Two motorists set out at the same time to go from Danbury to Norwich, 100 miles apart. They follow the same route and travel at different but constant speeds of an integral number of miles per hour. The difference in their speeds is a prime number of miles per hour, and after driving for two hours, the distance of the slower car from Danbury is five times that of the faster car from Norwich. What is the faster car’s speed?

. So we need to find a prime number, p,

so that is divisible by 6. does the job, and we get and .

So the correct answer is B) 42 mph.

(February 2005, #11) The sum is equal to

Rearranging the sum leads to

So the correct answer is E) . [See the section on Trigonometric Formulas]

(February 2005, #12) If and , find .

. .

So the correct answer is A) . [See the section on Matrix Multiplication]

(February 2005, #13) A basketball team scores 78 points on 41 baskets(field goals count 2 points, free throws 1 point, and 3-point shots 3 points). If the number of each type of basket is different, and the number of baskets of any of the three types differs by no more than 4, how many field goals are scored?

Let B be the number of field goals, F the number of free throws, and T the number of 3-point

shots. Then we have the system . Subtracting twice the first equation from

the second equation leads to . Adding the second equation to the first equation

leads to . So we get that . The number of free throws

is always 4 more than the number of 3-point shots. We also need to have

. So the possible values of T are 12 and 13, but only 12

works. This leads to .


(February 2005, #14) Which of the following is a factor of

?

Let .

Then

.

So the factors are and .

So the correct answer is D) .

(February 2005, #15) The volume of cylinder A is , which is twice the volume of cylinder B. If the radius and height of A are the height and radius respectively of B, find the height of cylinder B.

.


(February 2005, #17) Two triangular regions are formed in the first quadrant, one with vertices , , and , the other with vertices , , and . Find the area to the

nearest integer of the region they have in common.

The common area can be calculated as the sum of the areas of a trapezoid and a triangle.

. So the nearest integer is 19.

So the correct answer is C) 19. [See the section on Geometric Formulas]

12

5

6

8

(February 2005, #18) A triangle has sides of length a, b, and c which are consecutive integers

in increasing order, and . Find .

From the Law of Cosines, we get

.

Again, from the Law of Cosines, we get

.


(February 2005, #19) If is a prime number, what is the largest integer which must be a factor of ?

120 150 180 240 4002400 Y Y N Y Y14640 Y N Y N

So the answer is either 120 and 240. If is always divisible by 240, then it will always be

divisible by 120 as well. If we can show that always has at least four factors of 2, then

the answer must be 240. . Both and are

consecutive even integers and hence there product must be divisible by 8, and must be

even, so must have 16 as a factor, so the answer is 240.

So the correct answer is D) 240.

(October 2005, #6) Let M and L be two perpendicular lines tangent to a circle with radius 6. Find the area bounded by the two lines and the circle.

The area is the difference between the area of the square of side 6 and the quarter circle of radius 6. So it’s .



(October 2005, #7) When I am as old as my father is now, I will be five times as old as my son is now. By then, my son will be eight years older than I am now. The sum of my father’s age and my age is 100 years. How much older am I than my son?

. These lead to . So .

So the correct answer is D) 22 yrs..

(October 2005, #9) If and have integer solutions with , find the value of .

.

which means that would have to be a factor

of 817. Since 3 and 11 are not factors of 817, would have to be 1.

So the correct answer is A) 1. [See the section on Algebraic Formulas]

(October 2005, #10) has sides of length 6, 7, and 8. Find the exact value of .

From the Law of Cosines:

So .

So the correct answer is B) . [See the section on Trigonometric Formulas]

6

6

3

h

(October 2005, #11) Find the sum of all the solutions of for which .

. So the

solutions are , and the sum of the solutions is .


(October 2005, #13) For , let , where a, b, and c

represent every possible different arrangement of 2, 4, and 8. The product can be expressed in the form . Find N.

. The arrangements of

2, 4, and 8 are the following: 248,284,428,482,824,842

These lead to the following values:

, so .

So the correct answer is E) 50. [See the section on Logarithmic Properties]

(October 2005, #14) A triangle has vertices , , and . If the triangle is rotated counterclockwise around the origin until C lies on the positive y-axis, find the area of the intersection of the region bounded by the original triangle and the region bounded by the rotated triangle.

The intersection is a right triangle whose

area is .

,

so .

This means that the area of the intersection is

So the correct answer is A) . [See the section on Geometric Formulas]

[See the section on Trigonometric Formulas]

(October 2005, #15) When written as a decimal number, has D digits and leading digit L. Find .

In general, the number of decimal digits of a number N is , where is the greatest

integer less than or equal to x. Since , the number of

digits in the number is . We also have that , so the leading

digit is 5. So .


(October 2005, #16) if , , and , how many of the following are

true?

and

, so all four are true.

So the correct answer is E) 4. [See the section on Trigonometric Formulas]

(October 2005, #17) Let and for all . Find the value that

approaches as n increases without bound.

If approaches a number, let’s call it L. Then it must be that and

. Since the process won’t produce negative

numbers, the answer is .


(October 2005, #19) If and , which of the following is a possible value for ?

Adding the two equations leads to and to

. For , you get . So

the possible values for are .


(February 2006, #2) How many different four-digit numbers can be formed by arranging the digits 2, 0, 0, and 6?

2 3 2 1# of choices for 1st digit

# of choices for 2nd digit

# of choices for 3rd digit

# of choices for 4th digit

So it looks like there are 12 different four-digit numbers, but since two of the digits are zeros, we have to divide by 2, giving us 6 of them.

So the correct answer is A) 6. [See the section on Sets and Counting]

(February 2006, #6) Let , , , and . Then

P equals

, ,

.

So the correct answer is A) . [See the section on Logarithmic Properties]

(February 2006, #7) Which of the following imply that the real number x must be rational?

I. are both rational

II. are both rational

III. are both rational

If are both rational, then so is and , and so is . If are both

rational, then consider . and . If are both rational, then so

is and , and so is .

So the correct answer is B) I and III, only.

(February 2006, #8) A positive integer less than 1000 is chosen at random. What is the probability it is a multiple of 3, but a multiple of neither 2 nor 9?

The 333 multiples of 3 are . The multiples of 2 in T would have to be

multiples of 6, and the 166 multiples of 6 in T is . The 111 multiples of 9

in T is . The numbers in T that are multiples of 2 and 9 would have to be multiples of 18, and the 55 multiples of 18 in T is . The quantity of numbers less 1000 that are multiples of 3, but not multiples of 2 or 9 is . So the probability of selecting a positive integer less than 1000 that is a multiple of 3, but not a

multiple of 2 or 9 is .


(February 2006, #9) Let r and s be the solutions to the equation . If , then find the value of c.

The solutions of are and . Squaring and adding

them together leads to . So we get that .

Or

We know that .

So the correct answer is B) . [See the section on Algebraic Formulas]

[See the section on Polynomial Properties]

(February 2006, #12) The midrange of a set of numbers is the average of the greatest and least values in the set. For a set of six increasing nonnegative integers, the mean, the median, and the midrange are all 5. How many such sets are there?

Call the six nonnegative integers . We know that ,

, and , so we get that , , and

. Here are the possibilities:

1 0 1 2 8 9 102 0 1 3 7 9 103 0 1 4 6 9 104 0 2 3 7 8 105 0 2 4 6 8 106 0 3 4 6 7 107 1 2 3 7 8 98 1 2 4 6 8 99 1 3 4 6 7 910 2 3 4 6 7 8

So the correct answer is A) 10. [See the section on Statistics Formulas]

(February 2006, #13) The sum of the absolute values of all solutions of the equation

can be written in the form , c a prime. Find

.

leads to two equations:

with solutions

of .

with

solutions of . So the sum of the absolute values of all the solutions is .


(February 2006, #14) Find the number of three-digit numbers containing no even digits which are divisible by 9.

So using only the digits 1, 3, 5, 7, and 9, we must make a three-digit number that is divisible by 9. In order for a number to be divisible by 9, the sum of its digits must be divisible by 9. The three-digit numbers whose digit sum is 9 are arrangements of 1,1,7 and 1,3,5 and 3,3,3. The three-digit numbers whose digit sum is 27 are arrangements of 9,9,9. So there are

.

So the correct answer is D) 11. [See the section on Divisibility Rules]

(February 2006, #15) If is the acute angle formed by the lines with equations and , find .

2 5y x 1 3y x

, so .

So the correct answer is C) 1. [See the section on Trigonometric Formulas]

(February 2006, #16) Find the number of points of intersection of the unit circle and the graph of the equation .

For , the equation becomes . So for , the graph looks like

For , the equation becomes . So for , the graph looks like

So putting all the pieces together along with the graph of the unit circle leads to

So the number of intersection points is 3.


(February 2006, #17) Suppose that for a function , for all x. Let A be the

point with x-coordinate a on the function and B be the point on the graph of the line for which is perpendicular to the line. Find an expression for the distance from A to

B.

A has coordinates and B has coordinates . The slope of the line through points

A and B is , so . Now apply the distance formula to get

.

So the correct answer is C) . [See the section on Equations of Lines]


(October 2006, #2) A fraction is chosen at random from all positive unreduced proper fractions with denominators less than 6. Find the probability that the fraction’s decimal representation terminates.

Here are the positive unreduced proper fractions with denominator less than 6:

.

The only ones that don’t terminate are and , so the probability is .

So the correct answer is E) . [See the section on Probability Formulas]

(October 2006, #3) Two adjacent faces of a rectangular box have areas 36 and 63. If all three dimensions are positive integers, find the ratio of the largest possible volume of the box to the smallest possible volume.

The adjacent faces must share a side whose measurement is a common factor of 36 and 63. These common factors are 1, 2, 3, and 9. The volume of the box is the product of 63 and this common factor, so the largest possible volume is 63 times 9 and the smallest possible volume is 63 times 1. This means that the ratio of the largest volume to the smallest volume is 9.

36

63


(October 2006, #5) In the expression , each different letter is replaced by a different digit 0 to 9 to form three two-digit numbers. If the product is to be as large as possible, what are the last two digits of the product?

To get the largest product it must be , with C, T, and M replaced by 5, 6, and 7. Here are the possibilities:


(October 2006, #6) A basketball player has a constant probability of 80% of making a free throw. Find the probability that her next successful free throw is the third or fourth one she attempts.

Probability that the third free throw is successful: .

Probability that the fourth free throw is successful: .

The probability that the third or fourth free throw is successful is

.

So the correct answer is B) .0384. [See the section on Probability Formulas]

(October 2006, #7) If , find the smallest possible value of

, if a, b, c, and d are all positive integers.

Multiplying and equating the matrices leads to the system

.

This reduces into the system

.

So we get that

.

So . Since this must be a positive integer, has to be a positive multiple of 5. So the smallest possible value of is 8.

So the correct answer is B) 8. [See the section on Matrix Multiplication]

(October 2006, #8) Sue works weekdays for $10 an hour, Saturdays for $15 an hour, and Sundays for $20 an hour. If she worked 180 hours last month and earned $2315, how many more weekday hours than Sunday hours did she work last month?

dividing the second equation by 5 leads to

Subtracting twice the first equation from the second equation leads to

Subtracting the second equation from the first equation leads to . So she worked

77 more weekday hours than Sunday hours.


(October 2006, #10) The year 2006 is the product of exactly three distinct primes p, q, and r. How many other years are also the product of three distinct primes with sum equal to ?

and . Let’s make adjustments to the prime factors to get new prime factors which also sum to 78. To make 2 into a different prime factor means we have to add an odd number to it, but adjusting the other prime factors by an odd number would make them even and hence not prime, so the 2 must be left alone. Here are the allowable modifications:

.


(October 2006, #11) How many positive integers less than 1000 are relatively prime to 105? Two integers are relatively prime if their greatest common divisor is 1.

, so we have to count the number of positive integers less than 1000 that are not multiples of 3, 5, or 7.


(October 2006, #13) The equation has a solution in common with which of the following?

The solutions of A) are .

The solutions of B) are .

The solutions of C) are .

The solutions of D) are .

The solutions of E) are .

.


[See the section on Logarithmic Properties]

(October 2006, #19) Find the tens digit of .

Let’s find the pattern in the tens digits:

Power of 3 Tens digit000

2842868444260268600028428

Starting with , the tens digit repeats the same 20 numbers. From to is 2008 tens digits. , and the eighth number in the repeating 20 number sequence is 8.


(October 2006, #20) In the sequence , , , , and for all ,

. Find .

. Consider the new

sequence for . We get that , , , for . This means

that for . .


(February 2007, #2) The teachers at Oak Tech have cars with average mileage 39,000 miles. George buys a brand-new car, keeping his old car, and the average mileage drops to 36,400. How many cars do the teachers now own?

Let T be the current total mileage of the cars and N be the current number of cars. So

. We also have that , so . This implies

that , or .


(February 2007, #3) The sequence is best described as which of the following?

So the correct answer is C) arithmetic with common difference .

[See the section on Algebraic Formulas]

(February 2007, #4) A set of seven different positive integers has mean and median both equal to 20. What is the largest possible value this set can contain?

Let’s call the seven positive integers . The median being 20 implies that , and the mean being 20 implies that the sum of the other values must be 120.

20

We want the value of to be as large as possible, so we need to make the values of as small as possible. This leads to

1 2 3 20 21 22 71

So the correct answer is C) 71. [See the section on Statistics Formulas]


and , so . .

So the correct answer is A) 3.6. [See the section on Logarithmic Properties]

(February 2007, #8) A function is symmetric to the origin and periodic with period 8. If , what is the value of ?

A particular function that has these properties is .

.


(February 2007, #9) For how many integer values of k do the graphs of and NOT intersect?

Intersections correspond to real solutions of the system . Using substitution, you get

. In order for this quadratic equation to not have real solutions, its

discriminant must be negative. So we want or . This leads to .

So the correct answer is D) 3. [See the section on Algebraic Formulas]

(February 2007, #10) A point is chosen at random from the interior of a square of side 16. Find the probability that the point is at least units from both diagonals.

The point must be chosen from the four triangles outside of the central dashed cross. Each has an area of 36 square units for a total of 144 square units. The area of the entire square is 256

square units, so the probability is .

So the correct answer is A) . [See the section on Geometric Formulas]

[See the section on Probability Formulas]

(February 2007, #12) If (x in radians), then can be expressed in the form

. Find .

.

So the correct answer is A) 4. [See the section on Trigonometric Formulas]

(February 2007, #17) If , the inverse of can be written as

. Find .

Notice that , so . This means that

.

2

6

6


(February 2007, #18) Choose k so that the system is dependent. For which

pair below does there exist a z such that will satisfy the resulting dependent system?

subtract the first equation from the second equation, and subtract k times the

first equation from the third equation. add the second equation to

the third equation. simplify into . For , the

system has no solution, but for , you get the dependent system , which has

solutions of . The only pair that works is , and it corresponds to

.


(February 2007, #19) A pentagon is circumscribed about a circle of diameter 6 cm. If the pentagon has area 42 cm2 , find it perimeter in centimeters.

The area of the pentagon is , so we get that . So the perimeter of the

pentagon is .

3s

So the correct answer is D) 28. [See the section on Geometric Formulas]

(February 2007, #20) The sum of the solutions of

is

But , so the sum of the solutions is 3.

So the correct answer is E) prime. [See the section on Trigonometric Formulas]

(February 2008, #3) The equation has a solution in which a, b, and c are distinct even positive integers. Find .

. The cubic numbers we

need to consider are 1, 8, 27, 64, 125, 216, and the three that add up to 251 are 8, 27, and 216. These yield , and this implies that .


(February 2008, #4) For how many different integers b is the polynomial factorable over the integers?

The integer factor pairs of 16 are 1 and 16, -1 and -16, 2 and 8, -2 and -8, 4 and 4, -4 and -4. These lead to values of b of 17, -17, 10, -10, 8, -8. So there are six possible values of b.

So the correct answer is E) 6. [See the section on Polynomial Properties]

(February 2008, #5) Let . Which of the following is a factor of

?

So the correct answer is D) . [See the section on Algebraic Formulas]

(February 2008, #7) A fair coin is labeled A on one side and M on the other; a fair die has two sides labeled T, two sides labeled Y, and two labeled C. The coin and die are each tossed three times. Find the probability that the six letters can be arranged to spell AMATYC.

The outcome of the coin tosses must be AMA in any order. This has probability of . The

outcome of the die tosses must be TYC in any order. This has probability of . So the

probability of being able to spell AMATYC is .


(February 2008, #8) What is the value of ?

Continuing in this manner leads to

.

So the correct answer is C) 4. [See the section on Logarithmic Properties]

(February 2008, #9) The letters AMATYC are written in order, one letter to a square of graph paper, to fill 100 squares. If three squares are chosen at random, without replacement, find the probability to the nearest 1/10 of percent of getting three A’s.

Writing the six letters in order will fill 96 squares, leaving room for an additional AMAT. So the total number of A’s is 34, the total number of M’s is 17, the total number of T’s is 17, the total number of Y’s is 16, and the total number of C’s is 16. So the probability of selecting

three A’s without replacement is .

So the correct answer is B) 3.7%. [See the section on Probability Formulas]

(February 2008, #10) A student committee must consist of two seniors and three juniors. Five seniors are able to serve on the committee. What is the least number of junior volunteers needed if the selectors want at least 600 different possible ways to pick the committee?

The number of different possible ways to pick the committee is

. So we want or . So the least value of J is 9.

So the correct answer is D) 9. [See the section on Sets and Counting]

(February 2008, #12) Each bag to be loaded onto a plane weighs either 12, 18, or 22 pounds. If the plane is carrying exactly 1000 lbs. of luggage, what is the largest number of bags it could be carrying?

or , so , and

. So to make as large as possible, you need to have

divisible by 6 and as small as possible. Multiples of 6 less than 500

are 498, 492, 486, 480, 474, …. Choosing makes , so the

largest number of bags is .


(February 2008, #17) Let r, s, and t be nonnegative integers. For how many such triples

satisfying the system is it true that ?

Adding the equations together leads to

. Here are the factor pairs of 48: 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8. 1 and 48 produces no solutions. For 2 and 24, if , then we get 26 values of r and t that add up to 24. 3 and 16 produces no solutions. 4 and 12 produces no solutions. 6 and 8 produces no solutions.


(October 2008, #3) If is one solution of , what is the other solution?

If is one solution, then . So the other solution is .

So the correct answer is D) . [See the section on Polynomial Properties]

(October 2008, #4) Ryan told Sam that he had 9 coins worth 45 cents. Sam said, “There is more than one possibility. How many are pennies?” After Ryan answered truthfully, Sam said, “Now I know what coins you have.” How many nickels did Ryan have?

Here are the possibilities:

# of pennies # of nickels # of dimes # of quarters Total value5 0 4 0 455 3 0 1 450 9 0 0 45

So Ryan has 9 nickels.


(October 2008, #5) A point is a lattice point if both a and b are integers. It is called

visible if the line segment from to does not pass through any other lattice points. Which of the following lattice points is visible?

The slope of the line segment from to is . So if the fraction can be reduced,

then the point is not visible. Otherwise, it is visible. The only point in the list for which

can’t be reduced is .


(October 2008, #6) A flea jumps clockwise around a clock starting at 12. The flea first jumps one number to 1, then two numbers to 3, then three numbers to 6, then two to 8, then one to 9, then two, then three, etc. What number does the flea land on at his 2008th jump?

The sequence of jumps is 1, 2, 3, 2, 1, 2, 3,2 1, 2, 3, 2, …. If you break them into groups of four, you get 8,8,8,8, …. Four goes into 2008 502 times, so the flea will have traveled a total of

spaces on the 2008th jump. The position of the flea will be the remainder when 4016 is divided by 12: .


(October 2008, #8) All nonempty subsets of are selected. How many different sums do the

elements of each of these subsets add up to?

Subset Sum2457679911121113141618

There are 12 different sums.

So the correct answer is C) 12. [See the section on Sets and Counting]

(October 2008, #9) Luis solves the equation , and Ahn solves . If they get the same correct answer for x, and a, b, and c are distinct and nonzero, what must be true?

and , so

This rearranges into .


(October 2008, #12) The equation has a solution in positive integers a, b, and c in which exactly two of a, b, and c are powers of 2. Find .

Since the power on a is the largest, let’s try to eliminate its term first. The largest value possible for a is 3. With this value, we get . To find the other values as powers of 2, let’s see how many factors of 2 are in 1280.

. So .


(October 2008, #20) For all integers ,

is always the product of two

integers and . Find the smallest value of k for which .

First let’s generalize it into .

For , you get .

For , you get .

For , you get .

In general,

. Notice that

. So we want . This

means that , so must be at least a 1,001 digit decimal number. Let’s just force to be a 1,001 digit decimal number, since and always have the same number of digits.

.


(February 2009, #3) The perimeter of a rectangle is 36 ft and a diagonal is ft. Its area in

is

.

So the correct answer is D) 77. [See the section on Algebraic Formulas]

(February 2009, #5) For what values of k will the equation have exactly two real solutions?

The equation can be rearranged into . The discriminant of the quadratic formula can be used to determine when there will be exactly two real solutions, for . For this quadratic equation for , is . In order to have exactly two real solutions,

. But this includes the value , where the equation is

not quadratic, and in fact has only one real solution.

So the correct answer , which is not one of the possible answers.


(February 2009, #6) If x and n are positive integers with and , find .

. So for and

, you get that .


(February 2009, #7) In a tournament, of the women are matched against half of the men.

What fraction of all the players is matched against someone of the other gender?

This fraction is , and we know that , so plugging into the first expression

yields .

So the correct answer is D) .

(February 2009, #11) At one point as Elena climbs a ladder, she finds that the number of rungs above her is twice the number below her(not counting the rung she is on). After climbing 5 more rungs, she finds that the number of rungs above and below are equal. How many more rungs must she climb to have the number below her be four times the number above her?

, so



.

So the correct answer is D) .296. [See the section on Trigonometric Formulas]


(February 2009, #13) How many asymptotes does the function have?

It has vertical asymptotes of and . Since

and ,

it has horizontal asymptotes of and . So it has four asymptotes.


(February 2009, #14) For how many solutions of the equation are both x and y integers?

.

Assuming x and y are integers and since 2 is prime There are only four possibilities:

, , ,

None of these lead to integer solutions.


(February 2009, #17) How many different ordered pairs of integers with are solutions for the system of equations and ?

Adding the equations yields . Since , we get that

. Since y must be an integer and

, the only possible values for y are . both lead to

which won’t yield an integer solution. both lead to , and hence to the

ordered pair solutions of and .


(February 2009, #18) The graph of the equation is the union of a

.

So , which is a line, or , which is a hyperbola.

So the correct answer is E) line and a hyperbola. [See the section on Algebraic Formulas]

(February 2009, #19) A four-digit number each of whose digits is 1, 5, or 9 is divisible by 37. If the digits add up to 16, find the sum of the last two digits.

The possible four-digit numbers that meet the conditions are

the arrangements of three 5’s and one 1: 1555, 5155, 5515, 5551, none of which are divisible by 37

and the arrangements of two 1’s, one 9, and one 5: 1195,1159, 9115, 5119, 5911, 9511, 1591, 1951, 1519, 1915, 9151, 5191, of which only 1591 is divisible by 37.


(October 2009, #1) Find the sum of the solutions to the equations and which DO NOT satisfy both equations at once.

The solutions of are 6 and from . The solutions of

are and from . So the sum of the solutions which don’t satisfy both is .


(October 2009, #2) Four consecutive integers are substituted in every possible order for a, b, c, and d. Find the difference between the maximum and minimum values of .

For the four consecutive integers , the maximum value of is , and the minimum value of is

. So the difference between the maximum and minimum values of is .


(October 2009, #3) The product of a number and b more than its reciprocal is y(b>0). Express the number in terms of b and y.

If the number is x, then , so , and . So the answer is A.


(October 2009, #4) If , find the sum of all x values satisfying .

If , then must be a solution of . The solutions come from

or . So and . So the sum of the x values is .


(October 2009, #5) Sue bikes 2.5 times as fast as Joe runs, and in 1 hr they cover a total of 42 miles. What is their combined distance if Sue bikes for .5 hr and Joe runs for 1.5 hr?

, and , from which we can deduce that

and . So the combined distance is .


(October 2009, #6) The equation (a, b, c positive integers) has a solution in which a and b are both perfect squares. Find .

, , ,

, , . Of the numbers 2007, 1944, 1279, 1752, 1689, and 1024, the only perfect square is 1024. So the numbers are , and the sum .


(October 2009, #7) How many 3-digit numbers have one digit equal to the average of the other 2?

1 14 256

1 14 649 729

The total is .


(October 2009, #8) A rectangular solid has integer dimensions with length width height and volume 60. How many such distinct solids are there?

Average digit Other two digits # of 3-digit numbers with these digits

1 0 and 2 or 1 and 1 5

20 and 4 or 1 and 3

or 2 and 211

3

0 and 6 or 1 and 5

or 2 and 4 or

3 and 3

17

4

0 and 8 or 1 and 7

or 2 and 6 or

3 and 5 or 4 and 4

23

5

1 and 9 or 2 and 8

or 3 and 7 or

4 and 6 or 5 and 5

25

6

3 and 9 or 4 and 8

or 5 and 7 or

6 and 6

19

75 and 9 or 6 and 8

or 7 and 713

8 7 and 9 or 8 and 8 7

9 9 and 9 1

So there are distinct rectangular solids.


(October 2009, #9)

.

So the correct answer is A) . [See the section on Trigonometric Formulas]

(October 2009, #10) If and , find the largest value of .

Multiplying the two equations together leads to . Let

to get . Solving for z leads to , . So the largest

value is 2.


(October 2009, #12) The sum of the squares of the three roots of is

. The roots of

are the same as the roots of . So it must be that the roots a,

Height Width Length1 1 601 2 301 3 201 4 151 5 121 6 102 2 152 3 102 5 63 4 5

b, and c satisfy the equations .

So the correct answer is B) 6. [See the section on Polynomial Properties]

(October 2009, #13) The value of is

.

So the correct answer is C) 2. [See the section on Algebraic Formulas]

(October 2009, #14) The figure shows a circle of radius 4 inscribed in a trapezoid whose longer base is three times the radius of the circle. Find the area of the trapezoid.

We need to find the value of m that will make the line tangent to the circle, so we want the

system to have just one solution. This means that

must have a double root, so its discriminant must be zero. This implies that , and the

upper base measurement must be 6. So the area of the trapezoid is .

So the correct answer is A) 72. [See the section on Geometric Formulas]

(October 2009, #16) The integer is both the common ratio of an integer geometric sequence and the common difference of an integer arithmetic sequence. Summing the corresponding terms of the sequences yields 7, 26, 90, … . The value of r is

Summing leads to , so we get the system

Subtracting the first equation from the other two yields

Subtracting twice the first equation from the second equation yields

.

So .


(October 2009, #17) A hallway has 8 offices on one side and 5 offices on the other side. A worker randomly starts in one office and randomly goes to a second and then a third office(all different). Find the probability that the worker crosses the hallway at least once.

The probability of crossing the hallway at least once is equal to one minus the probability of not crossing the hallway. The probability of not crossing the hallway is

. So the probability that we want is .

So the correct answer is D) . [See the section on Probability Formulas]

(October 2009, #19) In square ABCD, AB=10. The square is rotated around point P, the intersection of and . Find the area of the union of ABCD and the rotated square to the nearest square inch.

So the area of the union is the area of square ABCD along with the area of the four small

triangles. This gives .


(October 2009, #20) The sum of the 100 consecutive perfect squares starting with equals the sum of the next 99 consecutive perfect squares. Find a.

Subtracting the left side from the right side leads to

This leads to

. This leads to

. This leads to

. This leads to

. So we get the equation

. The quadratic formula

yields .

A B

CD

P

5 2 5

A B

CD

E2

4

So the correct answer is . [See the section on Algebraic Formulas]

(February 2010, #1) Let . Find the largest prime factor of

.

. So the

largest prime factor of is 19.


(February 2010, #2) A circle of radius 2 and center E is inscribed inside square ABCD. Find the area that is inside but outside the circle.

We want the area of the triangle minus the area of the quarter circle: .



(February 2010, #3) The unique solution to the equation is , and the unique solution to the equation is . Find .

subtract twice the second equation from the first equation to get now add

times the first equation to the second equation to get and divide the first equation

by to get . So .


(February 2010, #5) Let be an arithmetic sequence with , , and . Find .

So it must be that . This leads to and . Since

, it must be that , so .

So the correct answer is C) 8. [See the section on Algebraic Formulas]

(February 2010, #6) All solutions to the equation (a, b, c positive integers) have the same value for . Find this value of .

, so let’s check for factors of

11, 12, 13, 14, and 15:

c 11 12 13 14 1544 74 N N N N N43 161 N N N N N42 246 N N N N N41 329 N N N N N40 410 N N N N N39 489 N N N N N38 566 N N N N N37 641 N N N N N36 714 N N N Y N35 785 N N N N N34 854 N N N Y N

, so we need to find values of a and b so that and , but it’s not possible.

, so we need to find values of a and b so that and , and and both work. So .


(February 2010, #7) If (a and b real) and ,

. So and .

This leads to and . So and .

So .


(February 2010, #8) A point C is chosen on the line segment AB such that . Find

.

or . This leads to

, and a solution of .


(February 2010, #9) Let represent the greatest integer . Find .

So .

So the correct answer is E) 7264. [See the section on Logarithmic Properties]

(February 2010, #10) If you roll three fair dice, what is the probability that the product of the three numbers rolled is a prime?

The possible outcomes are 1,1,2 in any order, 1,1,3 in any order, and 1,1,5 in any order. There

are 9 outcomes with the product a prime, so the probability is .

So the correct answer is B) . [See the section on Probability Formulas]

(February 2010, #12) Three faces of a rectangular box that share a common vertex have areas of 48, 50, and 54. Find the volume of the box.

If the three dimensions of the box are a, b, and c, then we have that , , and . The volume of the box is given by .


(February 2010, #14) For a function , let , , and

so on. For the function on the domain ,

, so the

sequence is . So .


(February 2010, #16) A 100 m long railroad rail lies flat along level ground, fastened at both ends. Heat causes the rail to expand by 1% and rise into a circular arc. To the nearest meter, how far above the ground is the midpoint of the rail? rr

h100

101

So we get that , , and

. From the quadratic formula, we get that ,

so we just need to find the value of r. If we can solve the equation , then we

can plug it into the formula for h and get the result. An approximate solution is 206.8852258,

and plugging it into the formula yields

.

So the correct answer is E) 7. [See the section on Geometric Formulas]


(October 2010, #1) A square is cut into two equal rectangles, each with a perimeter of 36. Find the area of the square.

x x

x x

y y y

rr h

50 50.5

So , and the area of the square is 144.


(October 2010, #2) Last year, the cost of milk was 150% of the cost of bread. If the cost of milk has risen by 20%, and the cost of bread has risen by 25%, what percentage of the current cost of bread is the current cost of milk?

So


(October 2010, #3) Angles are complements if they add to . Let be nine times , and the complement of be nine times the complement of . Find .

So , which means that . So .


(October 2010, #4) Find the product of all values of x for which is

undefined.

, so the product is .


(October 2010, #5) If you roll three fair dice, what is the probability that the product of the three numbers rolled is even?

In order for the product not to be even, all three numbers would have to be odd. The probability

that all three numbers are odd is . So the probability of rolling three numbers

whose product is even is .


(October 2010, #6) If , , , , find .

So

So , , and . Therefore .


(October 2010, #7) A lattice point is a point with both coordinates integers. How many lattice points lie on or inside the triangle with vertices , , and ?

So there are lattice points on or inside the triangle.

Or you could use Pick’s Theorem which says that the area of the triangle is equal to the number of interior lattice points plus half the number of boundary lattice points minus 1. So you get that . This means that the number of interior lattice points is 31. These 31 combined with the 20 boundary lattice points also give you a total of 51 lattice points.


(October 2010, #8) The perimeter of a rectangle is 52, and its diagonal is 20. Find its area.

Squaring the first equation leads to

Now subtract the second equation from the first to get

9 8 7 6 5 5 4 3 2 1 1

x

y


(October 2010, #9) The consecutive even numbers are written side-by-side to form an infinite decimal . Find the digit in the 2010th decimal place.

The one-digit evens(2,4,6,8) occupy positions 1 through 4.

The two-digit even numbers(10-98) occupy positions 5 through 94.

The three-digit even numbers(100-998) occupy positions 95 through 1444.

The four-digit even numbers(1000-9998) occupy positions 1445 through 19444.

To get from the 1444 position to the 2010 position requires 566 digits. This amounts to the first 141.5 four-digit even numbers. The 141st four-digit even number is 1280, and the 142nd four-digit even number is 1282. So the digit in the 2010th place is a 2.


(October 2010, #13) The equation (a, b, and c positive integers) has a solution in which b and c have a common factor . Find d.

The only possible values of a are 1, 2, 3, and 4. So we want to find d with

So .


(October 2010, #17) The integer is both the common ratio of an integer geometric sequence and the common difference of an integer arithmetic sequence. Summing the corresponding terms of the sequences yields . The value of r is

So

Eliminating b leads to

So .


(October 2010, #19) Let be a polynomial with nonnegative integer coefficients. If

, and , then find the sum of its coefficients.

So must be divisible by 2, and must be divisible by 77. This leads to . Now we have




So must all be zero.

So , and the sum of its coefficients is 19.


(October 2010, #20) If for all real numbers x, find the maximum possible value for m.

For an odd number of values , has its

minimum value at the median of . For an even number of values ,

has its minimum value at any value at or between the

middle two values of .

In this case, the minimum of occurs for .

So

So the correct answer is B) . [See the section on Statistics Formulas]

(February 2011, #1) After Ed eats 20% of a pie and Ahn eats 40% of a pie, Ed has twice as much left as Ahn. Find Ed’s original amount of pie as a percentage of Ahn’s original amount.

Let E be Ed’s original amount of pie, and A be Ahn’s original amount of pie.

.


(February 2011, #2) The expression for integers . If , find .

.

The factor pairs of 25 are 1, 25 and 5, 5. This means that and .


(February 2011, #3) Alicia always climbs steps 1, 2, or 4 at a time. For example, she climbs 4 steps by 1-1-1-1, 1-1-2, 1-2-1,2-1-1,2-2, or 4. In how many ways can she climb 10 steps?

Only 1’s

Ten 1’s

Only 2’s

Five 2’s

1’s and 2’s:

eight 1’s and one 2 six 1’s and two 2’s Four 1’s and three 2’s Two 1’s and four 2’s

1’s and 4’s:

six 1’s and one 4 two 1’s and two 4’s

2’s and 4’s:

three 2’s and one 4 one 2 and two 4’s

1’s, 2’s, and 4’s:

four 1’s, one 2, and one 4 two 1’s, two 2’s, and one 4

This gives .


(February 2011, #4) The sum of six consecutive positive integers beginning at n is a perfect cube. The smallest such n is 2. Find the sum of the next two smallest such n’s.

.

Since this must be an odd number, we’ll only consider odd cubic numbers:

27, 125, 343, …. 27 gives the value 2, . 729 gives the value 119.

From the list of answer choices, we can check to see if

they generate a cubic number. .


(February 2011, #5) The sum of the infinite geometric series S is 6, and the sum of the series whose terms are the squares of the terms of S is 15. Find the sum of the infinite geometric series with the same first term and opposite common ratio as S.

.

So the correct answer is B) 2.5. [See the section on Algebraic Formulas]

(February 2011, #11) Multiplying the corresponding terms of a geometric and an arithmetic sequence yields 96, 180, 324, 567, …. Find the next term of the new sequence.

, so we get

. Assuming that a and b are whole numbers, , let’s try and . This leads to the system

, which has as a solution and

. These values lead to the sequences and the product sequence

.



Assuming that x and y are whole numbers, then since , the possible values for x and y are , , , . For the pair , you get

.


(February 2011, #13) The equation (a, b, c positive integers) has a solution in which two of the three numbers are prime. Find the value of the nonprime number.

The possible values of a are 1, 2, 3, and 4. For , we get . For , we get . For , we get . For , we get . We can

eliminate and . So now we need to check and .

For , , which is not a square, , which is not a square, , which is not a square, , which is not a square,

.

For , , which is , but 18 is not a prime. , which is not a square, , which is , so we get 3, 2, and 42.


(February 2011, #14) A palindrome is a number like 121 or 1551 which reads the same from right to left and from left to right. How many 4-digit palindromes are divisible by 17?

4-digit palindromes are of the form abba, where a is 1,2,3,4,5,6,7,8,or 9 and b is 0,1,2,3,4,5,6,7,8,9. Now , so we can just examine the 4-digit numbers which are multiples of both 17 and 11, and hence just multiples of 187. We can skip multiples of 10.

1122 3553 5797 7854

1309 3927 5984 8041

1496 4114 6171 8228

1683 4301 6358 8415

2057 4488 6545 8602

2244 4675 6732 8789

2431 4862 6919 8976

2618 5049 7106 9163

2805 5236 7293 9537

3179 5423 7667 9724

And .


(February 2011, #16) The increasing sequence of positive integers satisfies the equation for all . If , find .

The sequence is . We know that , which means that must be a multiple of 5. If we go with , then

, but it doesn’t work. If we go with , then , it works. .


Trigonometric Formulas:

1.

2.

3.

4.

5.

6.

7. Law of Cosines:

8. Law of Sines:

9. If , then

Algebraic Formulas:

1.

2.

3.

4.

5.

6.

7.

8. Geometric Sequences and Series:

9. Arithmetic Sequences and Series:

10. Quadratic Formula:

For the equation with a, b, and c real numbers and , the solution(s) are

given by . If the discriminant, , is positive, then there are two real

solutions; if it’s negative, then there are two imaginary solutions; and if it’s zero, then there is one real solution.

Logarithmic Properties:

1.

2.

3.

4.

5.

6.

Geometric Formulas:

1. Area of a triangle:

or Heron’s formula, , where s is the

semiperimeter, .

2. Area of a parallelogram:

3. Area of a trapezoid:

4. Area of a circle:

5. Circumference of a circle:

6. Distance between the points and :

7. Midpoint between the points and :

8. Angle sum of a triangle:

The sum of the angles in a triangle is .

9. Pick’s Theorem:

In a square lattice, the area contained by a closed figure is equal to the number of interior lattice points plus half the number of boundary lattice points minus 1.

Equations of Lines:

1. Point-Slope:

2. Two-Point:

3. Slope-Intercept:

4. Intercept-Intercept:

5. Slope:

Polynomial Properties:

1. Factor Theorem:

if and only if is a factor of .

2. Remainder Theorem:

When is divided by , the remainder is .

3. Division Theorem:

For polynomials and , there are unique polynomials and (

) with . is called the quotient, and

is called the remainder.

4. Rational Zero Theorem:

If is an nth degree polynomial with integer coefficients,

then the rational zeros of can be expressed in the form , where p is a factor of and q

is a factor of .

5. Properties of Zeros:

The n zeros of the nth degree polynomial , must satisfy the following:

, , …

Fundamental Theorem of Arithmetic:

Every positive integer greater than one can be written uniquely as a product of prime factors, i.e. , where are distinct prime numbers.

LCM and GCF:

For two whole numbers A and B with and ,

The formulas generalize to more than two numbers.

Sets and Counting:

For the universal set U

1.

2.

3.

4. The maximum possible value of is the minimum of and , and this generalizes to any finite number of sets.

5. The minimum possible value of is if this quantity is positive, and zero otherwise.

6. The number of different subsets of a set with n elements, including the empty set, is .

7. Fundamental Counting Principle:

If a decision process consists of k stages with the number of options equal to , respectively, then the number of different ways of completing the decision process is

.

8. Combinations:

The number of different subsets with k elements from a set with n elements is

.

9. Permutations:

The number of different arrangements in a line of k elements from a set with n

elements(permutations) is .

The number of different arrangements in a line of n objects in which of the objects are identical, of the objects are identical,…, of the objects are identical, with

is .

Graph Properties:

1. Replacing with in an equation in the two variables x and y has the effect of reflecting the graph of the solutions of the equation across the y-axis.

2. Replacing with in an equation in the two variables x and y has the effect of reflecting the graph of the solutions of the equation across the x-axis.

Divisibility Rules:

1. A positive integer is divisible by 2 if and only if its one’s digit is even.

2. A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. This process may be repeated.

3. A positive integer is divisible by 4 if and only if the ten’s and one’s digits form a two-digit integer divisible by 4.

4. A positive integer is divisible by 5 if and only if its one’s digit is either a 5 or a 0.

5. A positive integer is divisible by 6 if and only if it’s both divisible by 2 and divisible by 3.

6. A positive integer is divisible by 7 if and only if when you remove the one’s digit from the integer and then subtract twice the one’s digit from the new integer, you get an integer divisible by 7. This process may be repeated.

7. A positive integer is divisible by 7 if and only if when you remove the one’s digit from the integer and then subtract nine times the one’s digit from the new integer, you get an integer divisible by 7. This process may be repeated.

8. A positive integer with more than three digits is divisible by 7 if and only if when you split the digits into groups of three starting from the right and alternately add and subtract these three digit numbers you get a result which is divisible by 7.

9. A positive integer is divisible by 8 if and only if the hundred’s, ten’s, and one’s digits form a three-digit integer divisible by 8.

10. A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. This process may be repeated.

11. A positive integer is divisible by 11 if and only if when you remove the one’s digit from the integer and then subtract the one’s digit from the new integer, you get an integer divisible by 11. This process may be repeated.

12. A positive integer is divisible by 11 if and only if when you subtract the sum of the ten’s digit and every other digit to the left from the sum of the one’s digit and every other digit to the left you get a number divisible by 11. This process may be repeated.

13. A positive integer with more than three digits is divisible by 11 if and only if when you split the digits into groups of three starting from the right and alternately add and subtract these three digit numbers you get a result which is divisible by 11.

Statistics Formulas:

1. Mean or average:

For the group of numbers , the mean is .

2. Median:

For the ordered group of numbers , if n is odd, then the median is the middle number, and if n is even, then the median is the average of the middle two numbers.

3. Mode:

The mode is the most frequently occurring number, if there is one.

Probability Formulas:

1. Equally-likely probabilities: If all the outcomes in the sample space, S, are equally-likely to

occur, then the probability that an event E occurs is given by .

2. Complementary probabilities: .

3. Intersection probability: If the events E and F are independent, then . This formula generalizes for any finite number of independent

events.

4. Union probability: .

5. Conditional probability: .

6. Geometric probability: If a point is to be chosen at random from a region S, then the

probability that the point is in the region E is given by .

Matrix Multiplication:

If the matrix A has n rows and k columns and the matrix B has k rows and m columns, then the matrix product AB is defined and will have n rows and m columns. The entry in the ith row and jth column of AB is the product of the ith row of A with the jth column of B. For example:

and , the product AB is defined, has 2 rows and 2 columns, and

.

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