rate/speed · web viewto find the average speed of 45 miles at 55 mph and 15 miles at 30, the...

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Work/Rate/Speed

Whenever a problem mentions rate, they are referring to

amount of work time spent .

Example 1: Suppose on a highway, the speed limit is 50 miles per hour. That means your car should not travel more than 50 miles in one hour of time.

Speed = distance/ time

In this case, it is 50 miles 1 hour

If you continue to drive on this highway, how long would it take to reach Grandmother's house 60 miles away?

speed * time = distance traveled

In this formula, you know all the elements except for the time. So make time = x (or any other variable) and solve for x. In harder problems, it’s easy to lost track of what you are looking for and assigning a variable can make it easier to solve.

50 miles 1 hour

Simple algebra says that x = 1.2, meaning the time needed is 1.2 hours, 1 hour and 12 minutes.

Example 2: Now say that you traveled for 45 miles on the rural highway and the rest of the trip (15 miles) in a residential area with a speed limit of 30 mph. What was your average speed?

First instinct may tell you to take the mean of 50 and 30, which is what an average of several numbers usually implies. But if speed was only a matter of numbers, you could have traveled for 30 miles on the rural highway and 30 in a residential area or 5 minutes on the highway and the remaining time in residential and the average of 50 and 30 will always be 40. Obviously, that is not the case.

Think of an average speed as the total distance covered in a certain time. If a car had begun the journey alongside you, traveled at one speed throughout the entire trip, and arrived at the same time as you did, you both would have the same average speed and the speed that car traveled would be the average speed. This is because you covered the same distance in the same amount of time.

IMSA JHMC 1 2000 Edition

=x 60 miles

To calculate this speed, find the total distance time

Since average speed is what you are looking for, make x = average speed. To find the average speed of 45 miles at 55 mph and 15 miles at 30, the distance you already know is 60 miles.

Now, rate = amount of work time

but if you manipulate the formula, time = amount of work rate

As you know, dividing by a fraction is the same as multiplying by a reciprocal, so we will do just that.

45 miles * 1 hour + 15 miles * 1 hour = x 50 miles 30 miles 45 miles * 1 hour + 15 miles * 1 hour = x 50 miles 30 miles

0.9 hours + 0.5 hours = 1.4 hours

60 miles = 42 and 6/7 miles per hour = average speed of trip1.4 hours

Example 3: Now suppose you come across one of those problems that says Susy takes this long to do this task and Tom takes this long, how long does it take them to do it together?

Always use the basic principle that rate = amount of work time

Then don’t forget to identify your variable.

So let’s pretend that Susy can make a pizza in 2 hours and Tom can make 2 pizzas in 3 hours. How long does it take them to make one pizza together?

Suzy’s rate = 1 pizza 2 hours

Tom’s rate = 2 pizzas 3 hours

Just like the problem about average speed, you are trying to figure out how long it will take to make 1 pizza using 2 different speeds. If you manipulate the original formula


rate = amount of work time

It is also:

rate * time = amount of work

You want to find out the amount of time for Suzy and Tom to make one pizza, so x = total time. You also know the two rates, so:

x = 6/7

Because x = total time to make one pizza, it is measured in hours.It will take Susy and Tom 6/7 of an hour to make one pizza together.

Remember:

Use and manipulate the formula rate = amount of work time

Identify the unknown variable.


Interest

Pretend your friend had loaned you $10 and demanded that you pay her back with 6% interest. You would give her the $10 and add 6%, which is:

What you just calculated is simple interest plus the original amount.

However, banks do not use simple interest. An average bank today may give you 3% interest. This does not mean that you deposit money into an account and when you withdraw it, you will have 3% more.

The Opulent Bank provides 3% interest compounded yearly, and Dave deposited $1000 into his account. After the first year, he will have

The next year, he will have 3% interest from $1030 = $1060.90, and so on. If Dave kept his account unchanged for five years, his sum will be

In other words,

where: A0 = initial amountr = rate as a decimalt = number of years


Dave also has an account in The Superior Bank, which provides 3% annual interest compounded monthly. He will earn more than 3% each year. Compounding monthly means the rate will be divided by 12 and then compounded each month. After one month, he will have

In one year, he will have

In 5 years, he will have

In other words to calculate annual interest that is compounded more than once a year (quarterly, monthly, daily):

where : A0 = initial amountr = annual raten = times compounded per yeart = number of years

This formula assumes discrete growth (at a fixed point in time, a certain amount is added in). However, it may be used to approximate growth and decay.


Triangle Tips

Special Right Triangles

30°-60°-90° Triangles

45°-45°-90° Triangles

When a problem does not provide all the necessary elements of the Pythagorean theorem (refer to geometry formula sheet if you don’t know what the Pythagorean Theorem is), these two rules will become very useful.

Angle-side Relationships

Triangle Inequality

Triangle inequality, the postulate states that the length of each side of a triangle

must be less than the sum of the lengths of the two remaining sides. So, if we think of a

triangle as the roads connecting three cities A, B, and C to each other, we know that

traveling directly from A to B will be shorter than the indirect route through C.

City A


60°

30°

x

2xx

As you can see, in a 30°-60°-90° triangle, the ratio of the sides are 1: :2. This is true in any triangle with angle measures 30°, 60°, 90°.

45°

45°

2 The ratio of the sides of a 45°-45°-90° triangle, then is : :2.

80°

40°60°

A

BC

The longest side of any triangle is opposite the largest interior angle; the shortest side is opposite the smallest angle. In the following triangle, side A is longer than side B, which is longer than side C, because 80>60>40.

AB AC

City B BC City C

AB < AC + BC, AC < AB + BC, BC < AB + AC

To understand the concept of triangle inequality, one must consider the fact that

the shortest distance between two points on a plane is the length of the straight line

segment joining them. Since no two sides of a triangle can be collinear because all angles

in a triangle must be between 0o and 180o, exclusive, it is easy to see that why triangle

inequality must be true.

This concept is useful in that it allows us to determine whether a set of three

numbers can represent the lengths of the sides of a triangle at a glance. All one has to do

is make sure that each number is less than the sum of the other two. This can be useful in

eliminating extraneous solutions in some problems.


A

F

Similar Triangles

What do you notice about the two triangles?

ABC and DEF are identical except that DEF is a larger version of ABC, right?

Their angles are the same and the lengths of their sides are proportional to each other. Triangles such as these are called similar triangles.

In other words, if triangle ABC and triangle DEF are similar, then:

and

What if you only knew that A = D and B = E? Would ABC and DEF still be similar?

Yes! Because all triangles have 180°, if you know two angles, you also know the third.

One more way to identify similar triangles is called SAS (side-angle-side).

If two triangles have one common angle and the sides that make up that angle are proportional, they are similar.

(Note: Unless the problem states that figures are drawn to scale, assume they’re not.)


BC E

D

A = DB = EC = F

P

ET55°

F

OG55°

3

4

6

8

PET and FOG are similar by SAS.

By knowing that triangles are similar, you can figure out information about one triangle based on information about the other.

For example:

Here is the information provided:

T = 55°

= 7= 4= 3

G = 55°

O = 50°

= 6= 8

From this, you should be able to solve for:

a. Pb. Ec. F

d.


P

ET55

3

F

OG55

6

84

7

50

Shoelace

The Shoelace Algorithm is a procedure used to find the area enclosed by any set of coordinates.

For example:

Find the area of the triangle determined by the lines: y-x=4, -x=10+2y, and y=-2x+10.

In this problem, the standard approach is to graph the lines, find the points of intersection, and use a box plot to solve the area. Instead of using box plot, shoelace may be applied instead.

In this case, the coordinates are: (-6,-2), (2,6), and (10,-10). (This can be solved using a system of equations.) The area of the triangle, K, can be found by following the procedure:

1)

2)

3)

4)Determine the absolute value of the difference of the “\”

products and “/” products:

==192


Select a vertex and travel around the triangle ending with the starting point. Write a matrix of the coordinates of the path including both the starting and ending coordinates.

Determine the sum of all “\” products: (-6)(6)+(2)(10)+(10)(2)=-36+-20+-20=-76

Determine the sum of all “/” products:(-2)(2)+(6)(10)+(-10)(-6)=-4+60+60=116

5)Take ½ of (4)

½(192)=96= area of triangle

The formula for the area is displayed:

K=1/2 abs =1/2 |(-76)-(96)|=1/2(192)=96

Note: Do you see the “shoelaces”?


Circles

1. A central angle is equal in degrees to its intercepted arc. A central angle is one in which the point at which the rays meet is at the center of the circle (shown below).

AIf arc AB = 80º, then 80ºAngle AOB = 80º

B

2. An inscribed angle is equal in degree to one-half its intercepted arc. An inscribed angle is one in which the point at which the rays meet is on the circumference of the circle (shown below).

If arc AC = 120º, then AAngle ABC = 60º B

120º C

3. An interior angle, formed by two chords intersecting in a circle, is equal in degrees to one half the sum of its intercepted arcs.

If arc AD = 40º and arc CB = 80º, CThen angle CEB = 60º = angle AED

Angle AED = ½(AD + BC) =1/2(40 + 80) = 60

4. An exterior angle, formed outside of a circle by two secants, a secant and a tangent, or two tangents, is equal in degrees to one-half the difference if its intercepted arcs.

CIf arc AD = 120º and Barc BD = 70º, then angle ACD = 25º A

Angle ACD = ½(AD-BD) =1/2(120°-70°)=25°


O

EA

D B

8040

D120

70

5. Two tangent segments drawn to a circle from the same external point are congruent (same length).

If AC and CE are tangent B Ato circle O at B and D,Then CB = CD.

C

D

6. Angle O and angle C are supplementary angles when OB is tangent to AC and OD is tangent to CE.


O

E

90

90

O

E

90

90D

C

BA

Interior Angles

Polygons are closed plane figures bounded by three or more line segments. Polygons cannot be composed of any curves since true curves cannot be constructed using straight-line segments. Circles and ellipses, therefore, are not polygons.

Polygons are named according to their number of line segments, or sides. Examples of polygons:

name number of sidestriangle 3

quadrilateral 4pentagon 5hexagon 6heptagonoctagon 8nonagon 9decagon 1011-gon 11

dodecagon 12… …

n-gon n

When we refer to the angles of a polygon, we are in fact referring to interior angles. An important property of polygons is that all polygons with the same number of sides have the same sum of interior angles. For example, adding the three interior angles of any triangle (three-sided polygon) always produces 180°, adding the four angles of any quadrilateral always produces 360°, and adding the five angles of any pentagon always produces 540°. In general, adding the interior angles of any n-sided polygon always produces 180(n-2) degrees.

Why does this pattern occur?Every time we add another side to a polygon, a triangle is added on to the

polygon. Since the angles of a triangle total 180 degrees, 180 degrees are added to the interior angle sum of the polygon.

In summary, every three-sided polygon has an interior angle sum of 180°. For every side more than 3, 180 degrees is added to the sum. Thus, the interior angle sum of a polygon with n sides is 180°(n-2).


7

Shown here is an example. Each triangle has 180 degrees and the number of triangles that can be drawn is (n-2) because from any vertex, you can draw diagonals to every other vertex except the two adjacent ones.

ABC

D

E FG

Diagonals of a Polygon

The number of diagonals in a polygon can be found by using the formula n (n-3)/2 where n is the number of sides of the polygon. For example, say you had a pentagon. Then, n=5 so when you plug it into the formula, you get 5*2/2 = 5. According to the formula, a pentagon should have 5 diagonals. As shown below, a pentagon really does have 5 diagonals.

Now the reason for the formula is because when you draw a diagonal, you are merely choosing 2 points and connecting them. So, in essence, drawing all the diagonals and sides consists of taking all the vertices and connecting them in every way.

This is the same as n choose 2 (explained in the combinations and permutations section). The formula for n choose 2 is n (n-1)/2 which looks similar to our formula above but when you do n choose 2, you are including the sides so in order to get the number of diagonals, you have to subtract n and you have n (n-1)/2 –n which when multiplied out becomes

(n2 – n –2n)/2 which is equal to (n2-3n)/2 which is the formula for the number of diagonals of a polygon.

Alternative explanation: Given an n-gon. From each vertex you can draw (n-3) diagonals (excluding itself and its two adjacent vertices). This would yield n(n-3) diagonals. However, each diagonal has been counted twice, once from each endpoint. Therefore, divide by 2 to get the result: n(n-3)/2, same as above.


Completing the Square

( a + b )2 = a2 + 2ab + b2

The method of completing the square is a technique used in a variety of problems to change the appearance of quadratic expressions.

The basic idea is to make a perfect binomial square show up by manipulating the original expression.

The method is based on the simple observation that, while x2+10x is not a perfect square, x2 + 10x + 25 is. "Completing the square" means putting in the missing 25 that is needed to form the perfect square.

Two obvious concerns are "How do you know you need 25?" and "How can you just add 25 any place you want?"

Here are the steps used:

Start with a quadratic expression like x2+10x Square half of the linear coefficient 10. That would be 52=25. Simultaneously add and subtract 25 and factor the perfect square. So x2 + 10x = x2 + 10x + 25 - 25 = (x+5) 2-25. Note that the constant in the binomial turns out to be half of the linear coefficient.

That's the main idea. Of course there are some variations to encounter in different types of problems.

Here are three more examples of completing the square. Make sure you understand the steps involved. Look especially at Example 3 in which the leading coefficient is not one.

Example 1x2 + 8xx2 + 8x + 16 – 16 since (8/2) 2 = 42 = 16(x + 4) 2 - 16 by factoring x2 + 8x + 16

Example 2x2 - 3x + 1 x2 - 3x + 9/4 – 9/4 + 1 since (-3/2) 2 = 9/4(x – 3/2) 2 - 5/4 by factoring x2 - 3x + 9/4

Example 33x2 + 5x + 63(x2 + (5/3)x) + 6 factor leading coefficient 3 from first 2 terms3(x2 + (5/3)x + 25/26 – 25/26) + 6 since (5/6) 2 = 25/263(x2 + (5/3)x + 25/26) – 25/12 + 6 separating –25/26 from the parentheses3(x + 5/6) 2 + 47/12 by factoring x2 + (5/3)x + 25/36


Factoring

Factoring has confounded mathematicians since the inception of number theory. Specifically, they have endlessly tried to determine an effective method for factoring large numbers into prime components. However, it may seem unclear what practical value such factorization would produce. Quite simply, many number theoretical theorems depend upon the ability to factor numbers into primes. Without such information, they would become merely intellectual exercises. Both of the following theorems, popular in mathematical contests, require prime factorization. Keep this in mind.THEOREM: Let n be any positive integer. Also, let , where are the different prime factors of n. Then, d(n), the number of divisors in n, is determined by

Most mathematical theorems are written in this format. Its conciseness helps for reference. See if you can understand it. In plain language, this theorem says that the number of divisors in some number can be determined by factoring it into its prime components, adding one to their powers, and multiplying them together.Example: Find the number of divisors of 24 by hand and through the theorem.Solution: By hand, 24 has the following divisors: 1, 2, 3, 4, 6, 8, 12, 24, making 8. By the theorem, since 1, .The justification for this rests in the multiplicative nature of d(n); that is, d(ab) = d(a) d(b).

Another theorem is closely related to this.THEOREM: Let n be any positive integer. Also, let , where are the different prime factors of n. Then, (n), the sum of all the positive divisors of n, is determined by

Do any of the terms in this product look familiar? They are all reminiscent of the sum of a geometric series. This is no coincidence, as shown in the following example.

Example: Find the sum of the divisors of 24 by hand and through the theorem.Solution: 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60.

Through the theorem,

. Interestingly, though, observe that 1 + 2 + 3 + 4 +

6 + 8 + 12 + 24 can also be written as

.

Of course, this is not a proof by any means, but it should strengthen your belief that the theorem holds true.


Greatest Common Factor

The greatest common divisor, abbreviated GCD, of two positive integers is the largest number that divides both of them (GCD is at times depicted as GCF, the greatest common factor). For example, the GCD of 8 and 12 is 4. Notationally, the GCD of two numbers is expressed with parentheses, such as (8, 12) = 4. Do not confuse this with ordered pairs. For small numbers, finding the GCD does not pose very large of a problem. Expressing the two numbers in their prime factors and choosing the common factors suffices to determine the GCD. For example, to find the GCD of 16 and 1000, one expresses 16 as 24 and 1000 = 103 = (5*2)3 = 53*23. The number 23 is common to both numbers, so 23 = 8 is the GCD.

Example: Find the GCD of 1800 and 168.

Solution: 1800 = 18*100 = 6*3*52*22 = 3*2*3*52*22 = 23*32*52.168 = 2*84 = 22*42 = 23*21 = 23*3*7. 23*3 = 24 is common to both numbers, so 24 is the GCD.

With large numbers, finding the GCD becomes considerably difficult using the prime factor method. However, a method called the Euclidian algorithm circumvents determining the factors. The application is not very hard, but the proof requires knowledge of mathematical induction and some experience. If desired, any number theory book should explain all the steps of the proof. The Euclidian algorithm involves dividing numbers, finding their remainders, taking the result, and reapplying the method. Though hard to explain, an example clearly illustrates this process.

Example: Find the GCD of 12345 and 54321.

Solution: First, divide the larger (the dividend) by the smaller (the divisor) and look at the remainder. Here, 54321 = 4*12345 + 4941. Then, express the divisor in terms of the remainder. Here, 12345 = 2*4941 + 2463. Repeat the process.

4941 = 2*2463 + 152463 = 164*15 + 315 = 5*3 + 0.

When the remainder becomes 0, the GCD is the last quotient, 3 in this case. When two numbers have a GCD of 1, they are called relatively prime. The Euclidian algorithm is the quintessential algorithm, a series of steps that, when performed a finite number of times, yields an answer. Usually, algorithms are used heavily in computer programs, where calculations can be accomplished very quickly.


Related to the GCD is the LCM, or the least common multiple. You may know it as the least common denominator when adding fractions. Typically, the LCM can be calculated by factoring both numbers into primes. The notation for LCM is [a, b]. Take all the factors that are either in a or b, and multiply them together. For example, to determine [6, 8], express 6 as 2*3 and 8 as 23. Then, [6, 8] = 23*3 = 8*3 = 24. this is difficult with large numbers. However, a relationship exists between GCDs and LCMs: (a, b)*[a, b] = ab. Convince yourself that this is true. (a, b) contains the factors common to both a and b,

whereas [a, b] contains the rest. Written in another form, .

Example: Calculate the LCM of 3542 and 6783.Solution: (3542, 6783) = 7, so [3542, 6783] = 3542*6783 / 7 = 3432198.


Geometric Series

A geometric sequence is a sequence in which each term differs from the previous by a constant ratio. That is, for a sequence a0r = a1, a1r = a2, … a n+1 = anr.Explicitly, an = a0rn. (Convince yourself that this works)

Example: John has 1000 dollars. He places it in a bank that returns 5% interest yearly. How much money will John have in 5 years? (to the nearest cent)

a0 = 1000, the initial principal. The goal is to find a5, which equals a0r5 = 1000r5. The interest is 5%, which means the money gains 0.05 of its value each year. The amount at the end of each year is computed if the amount each year is multiplied by 1.05. This is because 5% is equal to 5/100 = 0.05 so you’re adding the original amount (1) plus the interest 0.05 all multiplied by the original amount. Thus, r = 1.05, and a5 = 1000(1.05)5 = 1276.28…

A geometric series is the summation of a geometric sequence, such as a0 + a1 + a2 … = a0 + a0r + a0r2 + … Through clever symbol manipulation, a general formula can be determined for such a series.

Let Sn = a0 + a0r + a0r2 + … + aorn. rSn = a0r + a0r2 + … + a0r n+1

Subtract rSn from Sn.

Sn = a0 + a0r + a0r2 + … + a0rn

-rSn = -a0r – a0r 2 -…….- a 0r n – a 0r n+1 Sn-rSn= a0 -a0r n+1

Sn-rSn = a0 – a0r n+1

Sn(1-r) = a0 (1-r n+1)

Admittedly, this formula appears awkward. Nonetheless, practice should help you memorize it. This formula works even for r < 0.

Example: A runner starts by running 3 miles on Sunday. He increases his distance by 10% every day. How many miles will he have run by the end of Saturday?

a0 = 3 and r = 1.1, clearly, n = 7, the number of days he has run.


Infinite Geometric Series

An infinite geometric series is one in which there are an infinite number of terms. Basically, it is a geometric series (explained above) that doesn’t end.

Obviously, if |r| > 1, calculating our infinite geometric series will be a hopeless endeavor. However, if |r| < 1, should the series still be meaningless if the terms keep decreasing? Examining the formula for summation provides the answer.

As n grows, r n+1 shrinks, if |r| < 1. (Convince yourself). Indeed rn+1 goes towards 0, even if it never reaches there! Committing a technical symbolic error (don’t tell your teacher), we can write

Weird, but true.

Example: The Greek philosopher Zenos once stated the following dilemma.

Say a person is running a marathon. Once she reached the halfway point, she would have to reach the halfway point of the remaining distance, and then the halfway point of the remainder, and so forth. How is it possible she will reach the end?

Solution: Although there are an infinite number of points, the total distance is finite. Let the entire length of the route equal 1 unit. She runs ½ unit initially, so let a0 = ½. The length to each point is halved each time, resulting in the infinite geometric series ½ + ½ (1/2) + ½ (1/2)2 + … Using the formula previously derived, such a sum equals

just as said earlier.


Permutation and Combination

Permutation and combination are methods that are used to find how many ways something can be done. The formal definitions are:

A permutation: an arrangement of a group of objects in a definite order.A combination: a group of objects in which the order or arrangement is not

considered.

Permutations of Objects that are all Different

The number of permutations of 7 things taken 4 at a time is represented by 7P4. Following the same form, the number of permutations of n things taken r at a time is represented by nPr. Suppose you had 4 digits and you wished to make as many 4 digit numbers as possible; how many could you make? Well, since there are 4 possibilities for the first digit, 3 for the second, etc. the answer is 4*3*2*1 = 24 = 4P4. Assuming that the digits were 1, 2, 3, and 4, this is how it would be shown diagrammatically:

As you can see, there are 24 different numbers. Following the reasoning above, we can conclude that

nPr = n(n-1)(n-2) … (n-r+1) there are always r factors. Along the same line:

nPn = n(n-1) … 2*1 = n!This applies only when each of the objects being chosen are different and each object can only occupy one space.

So, for example, if you had 4 terms and you wanted to pick 2 with order as a factor, it would be:

4P2 = 4*3 = 12


23

12

2

2

3

3

4

4

4

1

1

12 3

3

34

4

41

1

2

2

3

4

4

4

1

1

2

3

34

1

1

2

23

1

2

3

4

2

1

3

4

3

1

2

4

4

1

2

3

Other Permutations

If each object could be used multiple times, then, in the above example with the 4-digit numbers, the solution would be 4*4*4*4 = 256 because then each of the slots can have each of the 4 digits filling it.

There is also a different twist to the problem if some of the objects are the same. For example, if you wanted to find the total number of permutations of the word Book, it wouldn’t just be 4P4 because if you switched two of the o’s around, the permutation would still be the same. What the answer actually becomes is

which is because there are 2 o’s. The number of permutations of n things taken n at a time with r things alike, s

things alike, and so on is:

Another aspect of permutations is circular permutations. A circular permutation is a way in which a number of different things can be arranged about a circle or any other closed curve. The difference between linear permutations and circular permutations is that when it is arranged around a closed curve, the starting place doesn’t matter so if you have n objects and the number of linear permutations taking n at a time would be n! but the number of circular permutations would be n! / n because then the first place no longer applies. For example, suppose you had 4 people sitting around a circular table.

All 4 of the above are the same permutation because with respect to the table, everyone is in the same relative position. That is the reason you divide by n when dealing with circular permutations.


A

B

C

DA

B

C

D A

B

C

DA

B

C

D

Combinations

As stated previously, a combination is a group of objects in which order does not matter. To make it clearer, the different permutations ABC, CBA, and BCA are all the same combination.

nCr = the number of combinations of n things taken r at a time.Because for each combination of r objects there are r! permutations, r!nCr = nPr. Then, since nPr = n(n-1)(n-2) … (n-r+1), by substituting, we can arrive at

nCr = n(n-1)(n-2) … (n-r+1) / r!

Another way to look at combinations is that nCr is the number of ways to take r objects from a group of n. If you take r, then you leave n-r. So what remains is the number of ways to take n-r objects from a group of n or nCn-r.

For example, if you wanted to choose 2 objects from 5, 5C2 = (5*4) / 2! = 10. You can also see that 5C3 = (5*4*3) / 3! = 10. They are the same, but is that coincidence? Actually, if you write out the formulas for nCr and nCn-r you get:

nCr = n(n-1)(n-2) … (n-r+1) / r!nC n-r = n(n-1)(n-2) … (r+1) / (n-r)!

Now to prove that those are equal, multiply the nCr equation by (n-r)! / (n-r)! to result in n! / (r!(n-r)!). Multiply the nCn-r equation by r! / r! to get the same result, which proves that nCr = nC n-r and it also gives us the formula for nCr.

nCr = n! / (r!(n-r)!)

The total number of combinations of n things means to find the total of the number of combinations taking things 0 at a time, 1 at a time, 2 at a time … n at a time. In notation, this means

Total number of combinations = nC0 + nC1 … nCn.It just so happens that the total number of combinations of n objects is 2n. This can be proven but it may be a bit complicated so for now it is just a handy fact to keep in mind. It is related as well to Pascal’s Triangle, an extremely deep topic with numerous interesting properties.


Probability

Probability is the ratio of the number of favorable events to the total number of possible events. Probability is often referred to as chance, the chance of an event occurring. As such, the probability of an event can never be less than 0 because there can never be a negative number of events. Also, the probability can never be greater than 1 because the number of favorable events cannot exceed the total number of events.

Example 1: Suppose a bag contains 7 white balls and 4 blue balls. What is the probability that, drawing one ball at random, you’ll get a blue ball?

Solution: Well, following the definition, you consider the number of favorable events. In this case, you see that there are 4 blue balls, so the number of favorable is 4. Then, you see that there is a total of 11 balls. So, the probability is 4 out of 11 or 4/11.

Example 2: Assume you have a bucket containing 5 slips of red paper and 6 slips of green paper. What is the probability that, drawing 2 slips of paper at random, you will get one red and one green?

Solution 2: There are 2 methods of approaching a problem of this type. First, you can view it as a single probability. Then, you would say that there are a total of 11 choose 2 possible events (choose, meaning combination, is explained in the section of the guide titled permutation and combination). So, you do (11*10)/(2*1) = 55. Then, you say that there are 6 ways of drawing a green and 5 ways of drawing a red so there are 6*5 = 30 favorable cases. So, the answer you arrive at is 30/55 or 6/11.

The second method is to consider the problem as 2 separate probabilities. What is the probability of drawing a green on the first draw? Well, following example 1, you find that it is 6/11. After that, there are 10 slips of paper left, 5 of which are red so the probability of drawing a red on the second draw would be 5/10 or ½. If

You multiply because you want A to occur and then B to occur in that order. 6/11*1/2 = 3/11. Then, because you can also draw the red slip first and the green second, and that also has a probability of 3/11, you add the 2 probabilities 3/11 + 3/11 and the final result is 6/11, same as the first method.


With probability, there are numerous different twists and turns that questions can take. For example, in example 2, instead of asking what the probability of drawing a red and a green in 2 draws is, what if you now had 3 draws? Or, what if in example 3, instead of coins, you were rolling die. Then, the question might be, what is the probability of obtaining a sum greater than 18? The key is to think the problem through logically so you can determine exactly what data is necessary and what to do about it.

Odds are another aspect of probability that ought to be kept in mind. Odds are the number of favorable cases versus the number of unfavorable cases or, the probability of the occurrence of the favorable event versus the probability of the occurrence of the unfavorable event. For example, if you had 7 blue balls and 5 red ones, the odds of drawing a blue ball are 7 to 5. In reality, it is (7/12)/(5/12) and the 12’s just cancel out. You could also say that the odds of not drawing a red ball is 7 to 5.


Divisibility Rules

Divisibility rules come in handy when you want to factor something quickly. There isn’t really much explanation that is required to go along with them so I will merely state them and give examples. However, for the extremely curious, these rules can be proven by Number Theory so feel free to look into that. Some will be extremely obvious and everyone will know them but others will be obscure. So everyone should be able to learn something.

Divisibility Rule for 2 = any even number is divisible by 2.

Divisibility Rule for 3 = if the sum of all the digits in the number is divisible by 3, then the number is divisible by 3.

Divisibility Rule for 4 = if the 2-digit number that is made up of the last 2 digits of the number are divisible by 4, then the number is divisible by 4.

Divisibility Rule for 5 = if the units digit of the number is 5 or 0, then the number is divisible by 5.

Divisibility Rule for 6 = just test both 2 and 3 and if the number is divisible by both, then it is divisible by 6.

Divisibility Rule for 7 = this one is pretty complicated. Starting from the units digit and moving left, you group the digits into groups of 3. Then, you take all the odd groups and subtract all the even groups or vice versa and then if the resulting number is divisible by 7, then the original number is divisible by 7. If the number does not have a multiple of 3 number of digits, that’s ok, just pretend there are zeroes added on to the left of the number.

Divisibility Rule for 8 = if the 3-digit number that is made up of the last 3 digits of the number are divisible by 8, then the number is divisible by 8.

Divisibility Rule for 9 = if the sum of the digits is divisible by 9, then the number is divisible by 9.

Divisibility Rule for 10 = if the units digit of the number is 0, then the number is divisible by 10.

Divisibility Rule for 11 = if the sum of the even placed digits minus the sum of the odd placed digits is a multiple of 11 or vice versa, then the number is divisible by 11.

Divisibility Rule for 12 = if it is divisible by 3 and 4, then it is divisible by 12.

Divisibility Rule for 13 = same as the divisibility rule for 7 except you divide the result of the odd groups minus the even groups (or vice versa) by 13 instead of 7.


Examples:

Example of 3: Let’s say your number was 137568. So, following the rule, you add the digits 1+3+7+5+6+8 = 30. Since 30 is divisible by 3, you know that 137568 is as well. 9 follows the same process.

Example of 4: To test the number 107184, you take the last two digits so you have the number 84. Since 84 is divisible by 4, 107184 is as well. The divisibility rule for 8 follows the same process except with the last 3 digits.

Example of 7: If your number is 1723575, what you do is you split it into groups of 3 so what you have is 1 | 723 | 575. Now, it is the odd groups minus the even groups or vice versa so you can do 723-575-1 = 147. Since 147 is a multiple of 7 (147 / 7 = 21), you know 1723575 is a multiple of 7. 13 follows the same process.

Example of 11: Assume the number in question is 93159. It’s the odd placed digits minus the even placed digits or vice versa so you can do 9 - 3 + 1 – 5 + 9 = 11. Since 11 is a multiple of 11, we know that 93159 is a multiple of 11.


Possible Paths

The following is a useful method to employ when encountering a problem of this specific type. I think the best way to teach this method is by example so here we go.

Problem: How many different paths are there from point A to point B, provided that you can only move downwards?

Solution: Begin at point A and give it a value of 1. Then, for every following intersection, label it the sum of the values above it that connect to it. For example:

C’s value would be 2 + 3 + 2 = 7 because those are the lines that are above it and connect to it. Likewise, D’s value would be 2 + 1 = 3. You begin with point A and just use this method throughout the entire diagram and your result, the value at point B, is the number of possible paths from point A to point B.

So, there are a total of 174 possible paths from A to B =)


A

B

C D

223

1

A

B

1

1

11

1

1

41

2

11

4

15

5 57

10

30 27

174

72

715

27

72

Checkerboards Squares/Rectangles

How many squares can be formed from:

The answer is 1.

Well, how many squares can be formed from:

How many of you answered 5?Well, the correct answer is 6.

Well, now what if there are 16 dots?How many squares are there in this?

Well, let’s see, shall we?

So, 9 + 4 + 1 + 4 + 2 = 20. There are 20 different squares in 16 dots.


There are 4 small ones

There is one big one

There is one diagonal one

+4 11 + 6=

There are 9 small squares

There is 1 square 3 by 3

There are 4 squares that are 2 by 2 (only 1 shown)

There are 4 diagonal squares that are 2 by 2

There are 2 other different diagonal squares (only 1 shown for simplicity)

Ok, so the sequence is 1, 6, 20… Does anyone see a pattern?Let’s take a look at Pascal’s triangle.

If you take a look at the circles, you’ll note that the first circle is around 1, the second is around 1 and 5, and the last circle is around 5 and 15. 1 + 5 = 6 and 5 + 15 = 20, fascinating, isn’t it?

So, in order to find the number of squares, all you have to do is find that diagonal in Pascal’s triangle. If you had 25 dots, then there would be 15 + 30 =45 squares and so forth.

This sequence can also be represented by the formula:

where n is the number of squares and x is the number of dots that comprise one side of the largest square.

Another point to note is that there is another pattern if you disregard the diagonal squares (you only count the squares whose sides would be drawn if there was a cartesian grid). Then, for 4 dots, there would still be only 1, but for 9 dots, instead of 6, there would only be 5 and for 16 dots, instead of 20, there would only be 14. Now, look at the diagonal in Pascal’s triangle right above the one we were just using and you’ll see that 1 = 1, 1 + 4 = 5, and 4 + 10 = 14 so if you disregard the diagonal squares, you could find the number of squares by applying the same procedure to the diagonal above.


11

11

11

11

1

111

1

32

36

11

4 45 510 10

7 76 615 1520

30 3021 21

Rectangles:

There exists an interesting method for finding the number of rectangles given vertical and horizontal lines (rectangles whose sides are formed by the lines, no diagonal ones). For example, say you had 5 vertical lines and 4 horizontal lines.

The method involves using combination. To find the number of rectangles, all you have to do is:

(Number of vertical lines choose 2) multiplied by (Number of horizontal lines choose 2)

In this case, it would be:

(5 choose 2) x (4 choose 2) = 10 x 6 = 60 rectangles.

If you’d like to test this for correctness, I’d advise choosing a simple diagram, applying this method, and then counting to check.

The reason this method works is because if you think about it, a rectangle is composed of 2 vertical lines and 2 horizontal lines. So, by doing (number of vertical lines choose 2), you’re getting all the different combinations of the 2 vertical lines that could possibly form the vertical sides of the rectangle. You do the same with the horizontal sides and then when you multiply them, that gives you the total number of possibilities of different rectangles.


Factors of Factorials Technique

The Factors of Factorials Technique is another handy trick that comes in useful at times on math competitions. It can be shown by example better than anything else so that is the method I’ll use to explain it.

Example: If 7n is a factor of 400!, what is the largest n?

Well, what you would do is “scan”.

(1) You divide 400 by 7 (rounded down) to get the number of numbers in 400! that are divisible by 7 and that is 57.

(2) Then, you take the quotient, 57, and divide by 7 (rounded down) to get the number of numbers in 400! That are divisible by 49 since those are the ones with at least 72 as factors. 57 / 7 = 8.

(3) Then, take the quotient, 8, and divide that by 7 rounded down to get 1 and that is the

number of numbers in 400! that contain 73. (4) So, to find n, all you have to do is add up the numbers, 57+8+1 = 66. There are 66

sevens in 400!.

Now, if you were trying to find the number of 10s in 400!, or any other number that is composite, you would follow a slightly different process.

(1) Since 10 is 5*2, first you must decide which factor there will be less of. In this case, it is fairly obvious to know that there will be more factors of 2 than 5 in 400!. So, all you need to do is scan for 5’s.

(2) Following the same procedure as the previous problem, we get (400 / 5 = 80), (80 / 5

= 16), and (16 / 5 = 3 ) => 80 + 16 + 3 = 99. Since you know that for every 5 there is at least a 2, you know that there are 99 tens in 400!


Interesting Formulas

For all formulas given below:an = nth term n = number of termsr = ratiod = common difference

An arithmetic sequence is a sequence of numbers in which there is a constant difference between each term and its following term. Meaning that a n+1 = an + d

A geometric sequence is one in which each number differs from the following number by a common ratio. Meaning that a n+1 = an * r

(1) Nth term of an arithmetic sequence: an = a1 + d(n-1)

(2) Sum of first n terms of an arithmetic sequence: sum = n (2a1 + d(n-1))/2

(3) Using the formula given in (2), we can figure out that if you want to add 1 + 2 + 3… n, the formula is: sum = n(n+1)/2. This is because in a sequence like 1 + 2 + 3… the difference is 1 and the first term is 1 so your 2a1 + d(n-1) term simplifies to 2 + n – 1 = n + 1.

(4) Nth term of a geometric sequence: an = a1 * r n-1

(5) Sum of first n terms of a geometric sequence: sum = a1(1-rn)/(1-r)

(6) As n nears infinity, if r > 1, then the sum goes to infinity as well. However, if r < 1, then the formula for the sum of the terms of an infinite geometric series is: a1/(1-r) = sum

(7) There is a special formula for the sum of a sequence consisting of 12 + 22 + 32 … and that is: n(n+1)(2n+1)/6 = sum of first n terms.

(8) For 13 + 23 + 33 … the formula is (the square of the formula for 1 + 2 + 3…):

(9) The formula for the distance from a point (c , d) to a line ex + fy + g = 0 where e, f, and g are constants that define the line is:


JHMC Guide Table of Contents

Title: Page Number(s): Author(s):

Work/Rate/Speed 1 - 3 Cindy Xi

Interest 4 - 5 Cindy Xi

Triangle Tips 6 – 9 Lucy Lu and Cindy Xi

Shoelace 10 – 11 Cindy Xi

Circles 12 Lucy Lu

Interior Angles 13 Lucy Lu

Diagonals of a Polygon 14 Denny Tu

Completing the Square 15 – 16 Lucy Lu

Factoring 18 Eric Syu

Greatest Common Factor 19 Eric Syu

Geometric Series 20 – 21 Eric Syu

Combination & Permutation 22 – 24 Denny Tu

Probability 25 – 26 Denny Tu

Divisibility 27 – 28 Denny Tu

Possible Paths 29 Denny Tu

Checkerboards 30 – 32 Denny Tu

Factors of Factorials 33 Denny Tu

Interesting Formulas 34 Denny Tu

Authors: Lucy Lu, Tao Luo, Eric Syu, Denny Tu, and Cindy Xi

Inspired by Cindy XiEditors: Mr. Ron Vavrinek and Denny Tu

Produced by Denny Tu



Please send any comments, suggestions, or questions to [email protected] or Denny Tu at IMSA. Thank You again, best of luck with the competition! =)

mailto:[email protected]

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