W is a positive integer when divided by 5 gives remainder 1 and when divided by 7 gives remainder 5. Find W.Ans:

take the larger divisor ..here its 7 ...the number is 7k+5

now this number when divided by 5 gives remainder 1 ...so 7k + 4 is exactly divisible by 5

put k=0 , 1 ,2.....

for k=3 we have 7.3 + 4 divisible by 5 ....thus we have have found k = 3

put it in our original number 7k+5 to get 26

A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?Ans:

Call: N=NumberD=DivisorWe have:N=x*D+24 (1)2N=y*D+11 (2)D>24

(1)(2)-> 2(x*D+24)=y*D+11--> (y-2x)D=37(3) --> D=37The sum of n different positive integers is less than 100. What is the greatest possible value for n?1011121314Ans:

The greatest value of n will be obtained when the first number of the series is the smallest possible, because in that case we will need more numbers to reach the required guarism (<100).

The sum of the n first natural numbers could be obtained by using this formula : n(n+1)/2

Applying this to our case:

n(n+1)/2 < 99 ==> n^2+n < 198. The greatest value of n that fills this restriction is n=13.Which of the following CANNOT be a factor of 2^i and3^j , where i and j are positive integers?

1

(A) 6(B) 8(C) 27(D) 42(E) 54Ans:

Because of 6=(2^1)*(3^1)8=(2^3)*(3^0)27=(2^0)*(3^3)54=(2^1)*(3^3)But 42=2*3*7answer is (D) 42.Find the number of two digit numbers between ,whose remainder is 1,when divided by 10 & remainder is 5 ,when divided by 6.

Ans: 10x + 1 = 6y + 5We can simplify to 10x - 6y = 4 or 5x - 3y = 2.

And then start plugging in numbers starting with x = 1, we get y = 1.X = 2, not possiblex = 3 nox = 4, yes, y = 6x = 5 nox = 6 nox = 7 yes

If the product of 3 consecutive even integers is 960, then the sum of the three integers is:A)10B)12C)22D)30E)40

Ans: While the answer is indeed 8, 10, 12, yet I believe that more important here is the approach to the question and not the answer. Without any trial you should be convinced that the answer should be 30 and the trial should be pretty much just to confirm your thought.

Since 960 is just about equal to 1000 and 1000 = 10 x 10 x 10 so you would know that the three numbers should be in the range of 10. This points immediately to 30 since diving 30 into 3 "almost equal" numbers would also give something like 10 + 10 + 10.

Glancing quickly at other options, 22 would give something in the range of 7 x 7 x 7 which would be much less than 1000 and 40 would give something in the range of 13 x 13 x 13 which is obviously larger.

Each of the following numbers has two digits blotted out. Which of the numbers could be the number of hours in x days, where x is an integer?

A. 25_,_06

2

B. 50_,_26C. 56_,_02D. 62_,_50E. 65_,_20Ans:the required number should be a multiple of 24, which has 4 in the last digit. if the number formed by last two digit of a number is divisible by 4, then it is divisible by 4. only E satisfies the condition, as 20 is divisible by 4.

Find the square root of 64009.Ans:64009 = 62500 + 900 + 100 + 400 + 100 + 9 = x^2x = 250 + 30 + 10 + 20 + 10 + 3

If x is a positive integer, what is the units digit of (24)^(2x+1) (33)^(x+1) (17)^(x+2) (9)^(2x)?Ans: 

remember the units digit of these numbers

4^1 = 44^2 = 64^3 = 4so the units digit for 4 is either 4 (when power os odd) or 6(power is even)

for 33^1 = 33^2 = 93^3 = 73^4 = 13^5 = 3so units digit of 3 follows the above patternsimilarlly for 7 7^1 = 77^2 = 97^3 = 37^4 = 17^5 = 7

and for 9 9^1 = 99^2 = 19^3 = 9

just apply the above rules

for the above it can be

4^odd = 4 9^even = 1

4 * 9 *3 *1 = 84 * 7 *1 *1 = 84 * 1 *7 *1 = 84 * 3 *9 *1 = 8for 3 and 7 it can any of the values but the product is only 8so the ans is 8

3

Find the max value of axbxc if 3a+4b+5c=720.

Ans:Arithmetic mean >= Geometric mean

(3a+4b+5c)/3 >= (3a*4b*5c)^1/3720/3 >= (60abc)^1/3240 * 240 * 240 / 60 >= abci.e abc <= 230400A certain integer n is a multiple of both 5 and 9. Which of the following must be true?I) n is an odd integerII) n is equal to 45III) n is a multiple of 15A) III onlyB) I and II onlyC) I and III onlyD) II and III onlyE) I, II, and III Ans: A 

If r and s are 2 positive integers greater than 1,and if11(s-1) = 13( r-1 ), what is the least possible value of r+s.Ans:11(s-1) = 13(r-1). here if u see carefully 11 and 13 are both prime integers and r and s are given to be +ve integers. so the two sides can be equal only if (s-1) = 13k and (r-1) = 11k where k is an +ve integer. smallest value for k is 1 therefore smallest value of s+r = 14+12 = 26.

Given that 5x=3y=z where x, y, z are integers, each choice given below must be integer except?A) z/15B) z/5C) z/3D) z/xyE) x/3

Sol: Guys get ur basics into the act to solve questions like theseGiven 5x=3y=z.Let 5x=3y=z= kHence x= k/5 y=k/3 and z= kie x:y:z = 3:5:15So we can take x =3n y=5n and z=15n (So z must be a multiple 15 and not simply5)Substitute these values in the given options, clearly z/xy = 1/n which need not be an integer

Hence the answer is D.

how many digits are there in 40^37 ?Ans:take log=> 37log(40)=> 37 ( log 4 +log10)=> 37 (2log2 + log10)as log2=0.301log10= 1=> 37 ( 2*0.301 +1 )

4

=>37*1.602=>59.274No. of digits =59+1=60(ANS)

OR,

40^37=(4*10)^37=(4^37)*(10^37)=A*BB=10^37 will contribute 37 digitsA=4^37=2^74=(2^70)*(2^4)=((2^10)^7)*16=((1024)^7)*16approximately, (1000^7)*16=( 21 digit number) *16=22 or 23 digit numberso, A*B=(37+22)=59 or (37+23)=60

when, k=sum of the digits of nif k<10, #(n) = kotherwise, #(n) = #(k)

for example,if n=52 then k=5+2=7 so #(52)=7if n=5678 , k=5+6+7+8=26, #(5678)=#(26)=8if n=8929 , k=8+9+2+9=28 , #(8929)=#(28)=#(10)=1#(9*35727) = ?Ans:

#(9*1)=9#(9*2)=#(18)=9#(9*5)=#(45)=9........surprisingly its always 9so we don't need to multiply result is 9

There is a 5digit no. 3 pairs of sum is eleven each. Last digit is 3 times the first one. 3 rd digit is 3 less than the second. 4 th digit is 4 more than the second one. Find the digit.

Ans:

first digit --> x2nd----->y3rd----->y-34th----->y+45th----->3xnow it's a combination problem and if we look only at the y's (assumingthat we are lucky enough and the magic number 11 is formed)and pick 3rd and 4th digits we have y-3+y+4=11 => 2y=10 => y=5looking at our 5 digits numbers , the only one that satisfies, should contain the digits 6,5,2,9,2, that is,

65292

5

or rearranged as

25296,

if E is taken as the last digit or A is taken as the last digit, respectively.

In a certain game, a large container is filled with red, yellow, green and blue beads worth, respectively, 7,5,3 and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?Ans:147,000 = 147*10^3 = 2^3 * 3 * 5^3 * 7^2

You can see that Red beads removed are 2

For which values of n the expression [(100^n) - 1] is divisible by 11. N is a +ve integer.a) all even b) all odd c) all prime d) any value e) no valueAns:100^n - 1 = (100^n -1) = (100 -1)*(100^(n-1) + ...) = 99*(...)so it's a multiple for 11 for every +ve integer since (a^n -b^n) = (a-b)*(a^(n-1)+....) holds for all +ve integers.

OR,

(100^n) will always have odd number of digits of the form 100, 10000, 1000000,..[(100^n) - 1] will be even no of digits of the form 99, 9999, 999999, etc.

Divisibility rule of 11 is sum of even digits - sum of odd digits = 0. Any no with even number of digits and all digits same will always be divisible by 11. So this is divisible for all +n.

What is the minimum value (greater than 1) that can divide [2^(4n+2)] + 1(n is positive integer)a) 2 b) 3 c)4 d) 5 e)9Ans:2^(4n+2) + 1 = k => 4*16^n +1 = k => 16^n = (k-1)/4 for n=1 => 16 = (k-1)/4 => k=65 or k = 5*13

If x, y, and z are nonzero integers and x>yz, which of the following must be true?1) x/z>z2) x/z>y3) x/(yz)>1A) none of the aboveB) 1onlyC) 3onlyD) 1 and 2 onlyE) 1 ,2 and 3 onlyAns:A it isTry using the two combinationsx,y,z=-3,4,-1 and -3,-1,4

which is bigger a>b>c>d>0a-d b-cAns:

6

a>b>...-d>-c>...add and --> a - d > b – c

Note:2 choices: 1. the quantity that is subtracted, consider it as being added. 2. Change sides.

x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many x are there?Ans:

First find the least number which satisifies this equation = 8So numbers are in arthimetic progression with a =8 and d = 21 (7*3)

8 + (n-1) * 21 =< 500

(n - 1) =< 492/21

n-1 <= 23.xxxn <= 24.xxxn = 24

Note: to determine the first term we need to list the numbers meeting the condition for divisibility by 7 and that for 3. Then we need to find the common one.

If m and 3m have the same digit sum, then m is a multiple of:1) 62) 33) 94) None of the above

Ans:

I think this is one of the many of questions, which requires that you plug in the possible answers to find out which one is correct.

The answer should be a nine (9). By this the initial condition of the sum of the digits of m and 3m will be satisfied.

i.e the numbers will be m= 9 and 3m = 3*9 =27 summing the digits (2+7 = 9)

9 is a multiple of 9 and 27 - second condition also satisfied.

A three (3 ) answer will satisfy the second part of the question alone (ie 3 being a multiple of 3 and 9) but not the m and 3m condition.

ie m=3 and 3*3 = 9 (3 and 9 are not the same).

Note: Take small and simple value to plug in.

sqrt(a^2*b) + sqrt(a*b^2) could be

1) a*sqrt(b) + b*sqrt(a)2) -a*sqrt(b) - b*sqrt(a)3) -a*sqrt(b) + b*sqrt(a) 4) a*sqrt(b) - b*sqrt(a)I 1) alone

7

II 1) + 2) III 1) + 3) IV 1) + 4) V all answers are correctAns:value of a and b < 0 are ruled out, because statement says square roots exist.

So, both a and b should be +ve and hence I is the answer

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is:

A - 6B - 12C - 24D - 36E - 48Ans:48^2 = 2304 = 72*32 so I think ans. is 48or following aashu's logic n^2 = 2^3*3^2*kand if k =32 = 2^5 then n^2 = 2^8*3^2 = (2^4*3)^2 = (16*3)^2 = 48^2

If x < 0, then sqrt((-x|x|)) is

A. -xB. -1C. 1D. xE. √xAns:also sqrt(x^2) = |x|.

Now the question, since x < 0 => |x| = -x

so, sqrt((-x|x|)) = sqrt(-x * (-x)) = sqrt(x^2) = |x| = -x

n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?Ans:I guess it doesn't matter, just plug in n=1 because whatever integer you plug in will always give you the third in the pattern, the 7-ending . . . (e.g. n=1, means 3^(16+3), which means 19/4, remainder three so it's the third (7) in the powers of three series

OR,

three ain't bad, bro, just see that you have 3^11 . . . 3 to the power of anything always goes in the pattern 3,9,7 (27), 1 (81) . . . so the units digit will be 7 since we're in the third of a series of four, and add two (like the problem says) and you have 9 in the units, divided by 5 that's remainder 4

(12345678987654321)^(1/2)=?Ans:

here's a general trend:

(12345678987654321)^(1/2) = 111111111 (*:no. of 1's = middle digit)

8

(123456787654321)^(1/2) = 11111111

(1234567654321)^(1/2) = 1111111

(12345654321)^(1/2) = 111111

(123454321)^(1/2) = 11111.........(121)^(1/2) = 11.

if the sum of 2 numbers is 216 and their HCF is 27.then the numbers are?Ans:216 = 27*8. combination s of 8 are (4,4), (3,5), (6,2) and (7,1). answer can be (27,189) and (81,135) using combinations of (7,1) and (3,5) respectively.

Three strings of a musical instrument vibrate 6,8,12 times a second respectively. If all the three begin to vibratre simultaneously, find the shortest time interval before all three vibrate together againAns:according to me answer is after (1/2) second(HCF of 6,8,12 is 2)Find the greatest number when divided by 12,19,18 will leave the remainder 7,11,13.Ans:

ans for 6th question is 859the short cut is {Lcm*d-c.v)

12-7=5;1-11=5;18-13=5

144*6-5=859

10^22+1 mod 11 VS 2

Ans:

100^n/11 has remainder 1 eg 100/11 =99 remainder 1 1000/11 =999 remainder 1 Hence 10^22+1 mod 11=(1+1)=2

Alternatively,(a+b)%c = ( a%c + b%c ) % c(a*b)%c = (( a%c) * (b%c)) %c(a^b)%c = ((a%c)^(b%c)) %c

by that formula, 10^22+1 mod 11 should be[(10mod11)^22mod11 + 1mod 11] mod 11 ,i.e. [10^0+1]mod11 = [1+1]mod11 = 2mod11 = 2.So the answer is C.A K-number is a positive integer with the special property that 3 times its unit digit is equal to 2 times its tens digit

COLUMN A The number of K-numbers between 10 and 99

COLUMN B 3Ans:let xy be the two digited number.Given 3y =2x i.e y/x = 2/3

9

i.e the digits y and x must be in the ratio 2:3hence the values of (y,x) are (2,3) (4,6) & (6,9) .so the numbers are 32, 64 and 96 . Hence the answer is Cn is positive.column A the value of units digit in 2^n column B1Ans:n is positive but it is not written that it is an integer. what happens if n is 0.5?sqrt(2) = 1.414unit digit = 1

220. if a,b,c are consecutive non-positive integers and a< b < c ,which of the following must be trueI -a< -cII abc<0III abc is even

a) none b) II only c) III only d)II n III only e) I II and III Ans:The answer ought to be c , watch that a, b, c are non-positive, that means they could be -2, -1, 0If n is an integer greater than 6, which of the following must be divisible by 3?A) n(n+1) (n-4) B) n(n+2) (n-1) C) n(n+3) (n-5) D) n(n+4) (n-2) E) n(n+5) (n-6)Ans:If n = 7 then a, b, eIf n = 8 then aABC is a three digit number and A>0. The value of ABC is equal to the sum of the factorials of three digits. The value of B isA) 19B) 8C) 4D) 2Ans:Let us proceed by process of elimination.6! = 7207! = >1000As 3 digit number 19 , 8 are not valid choiceABC A = 1B= 4C =5ABC = 145 = 1 + 24 + 120

10

What is the least integer n such that 54/n^3 is positive.A) -3 B)-1 C)1 D)2 E)3Ans:54/n^3= (3^3 * 2)/n^3 It should be C (1).Two different numbers when divided by same divisor leave remainders 4 and 3 respectively and when their sum is divided by the same divisor, remainder is 1. What is the divisor?Ans:Let the two numbers be d1 and d2.given, d1+d2=(q)*n +1

d1= q1*n+4d2=q2*n+3Adding the above two eqnsd1+d2=(q1+q2)*n +7d1+ d2= ((q1+q2)*n +6)+1n=6.

If n and y are positive integers and 450Y=n^3, which of the following must be an integer?I. y/(3 x 2^2 x 5)II. y/(3^2 x 2 x 5)III. y/(3 x 2 x 5^2)Ans:Factorize 450 ->(3 x 2^2 x 5^2)y=n^3for this to hold true y should be a multiple of k^3(3x2^2x5) where k is an integer, since this factor makes n^3 a whole cube.

Therefore, y/(3x2^2x5) is an element of Integers

The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?

Ans:since m and v are 3-digit positive integers, they can be represented as:m = 100a1 + 10b1 + c1v = 100a2 + 10b2 + c2n = 100x + 10y + zIt is given that f(m) = 9f(v)=> 2^a1*3^b1*5^c1 = 9*2^a2*3^b2*5^c2=> 2^a1*3^b1*5^c1 = 3^2* 2^a2*3^b2*5^c2=> 2^a1*3^b1*5^c1 = 2^a2*3^(b2+2)*5^c2

This means that m-v will differ by 2*10 (difference in ten's place) = 20The product of all the prime numbers less than 20 is closest to which of the following powers of 10?A) 10^9 B) 10^8 C) 10^7 D) 10^6 E) 10^5 Ans:

11

2*5 = 1019 close to 20= 2*103*7=21 close to 20 = 2*1011 close to 1013*17 = 221 close to 2*10^2 soabout (rough approx) 8*10^6 close to 10^7If y ≠ 3 and 3x/y is a prime integer greater than 2, which of the following must be true?

I) x = y

II) y = 1III) x and y are prime integers.

A) None

B) I only

C) II only

D) I and II only

E) III only

Ans:If x = y = 1 then III falseIf x = 10, y = 6 then I & II false Answer is A

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, the p isa. between 2 and 10b. between 10 and 20c. between 20 and 30d. between 30 and 40e. greater than 40Ans:h(n) = 2.4.6....n h(100) = 2.4.6...100 = 2. (2.2). (2.3).(2.4).(2.5).(2.6).....(2.50) => h(100) is divisible by all numbers from 2 to 50, => when h(100) + 1 is divided to each of these numbers, there is always a remainder = 1, in other words, h(100) +1 is not divisible by all numbers from 2 to 50.h(100) + 1 is divisible by p => p is not any number from 2, 3, 4, 5, 6, 7...50 => p> 40 --> answer is e.

if x (not equal) 0 then sq. root(x^2)/x is equal to?a -1b 0c 1d xe |x|/xAns: eif n & y are positive integers and 450y=n^3 which of the following must be an integerI y/3*2^2*5II Y/2*3^2*5^2III Y/3*2*5^2

options are:

12

noneIIIIIIAns:2*5^2*3^2*y=n^3now the left hand side needs 2^2*5*3 to become a perfect cubeso I and fraction is 1What's the remainder when ( 13^7 + 14^7 + 15^7 + 16^7 ) is divided by 58?Ans:13^7 + 14^7 + 15^7 + 16^7 = (13^7 + 16^7) + (14^7 + 15^7 )

both of the expressions with in brackets are divisible by 29 (13+16 OR 14+15).

Also sum of the expressions provide even number (sum of two odd nrs and two even number is even)

so both of them together divisible by 58If A=2^35, B=5^15, C=6^14 arrange them in increasing order.Ans:First step is to find out factors of the powers

7*5, 5*3 and 7*2

Then find out HCF of these roots.

Here, since all of them do not have one common factor, arrange them into two groups

2^35, 6^14 (A,C) - HCF of 35 and 14 is 732^7, 36^7 - mean C is grater than A

2^35, 5^15 (A,B) - HCF of 35 and 15 is 5128^5, 125^5 - means A is grater than B

So B, A and C are in ascending order.

If the powers are in fractions, we need to find the LCM and raise them to equal powers for comparison.M and R are two different two-digit positive integers. the sum of the tens digit of M and R is 14. What is the greatest possible sum of M and R?Ans:max M+R = 14*10+2*9=158max M*R = 79*78If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?I.3 II.4 III.6Ans:m, m+1, m+2 one has to be a mult. of 3if m is a mult of 3 then k is mult of 3if m+1 is a mult of 3 then m+1+3=m+4 a mult of 3 and k mult of 3similarly if m+2 is a mult of 3 then m+5 is a mult of 3 and k mult of 3but product is even since either m+4, or m+5 is even (consecutive integers) so 3,6 will always divide k evenly.

13

Moreover, set m = 4 in the above equation then 4 should divide it evenly. So 4 could divide k evenly.

Hence all of the choices are correct.

we have a 4 digit number 2pqr (r is units digit) and we multiply it by 4and number reverses to rqp2. Then p+q is Ans:r = 8 - easy to findoutnow (2pq8)*4 = 8qp2since pq*4 value is not spilling over to the 1000th digit, we can saypq *4 + 3 = qp40p+ 4q+3 = 10q+p39p-6q+3 = 0p = 1 and q =7Which one can be the number of small cubes that could be formed from a big cube? A)8, B)27, C)81, D)5, E)16If you put together 5 cubes, you won't get a cube.

Both A and B. Volume of a cube has to be a perfect cube. Both 8 and 27 are perfect cubes. So if V and v are the volumes of big and small respectively,

V = n * v, For V to be a perfect cube, n and v both have to be perfect cubes. Hence 8, 27.

If X<0,then sqrt(-X|X|) is

A. -xB. -1C. 1D. xE.√xAns:sqrt(x^2) = |x|

Since x < 0, ans is -x.

D is the answer

Number 3 can be written as a sum of consecutive integers in 2 wayslike 3, 1+2 (take number itself as 1 way). In how many ways is 21 written as

a sum of consecutive integers?I) 3II) 4III) 5IV) 6Ans:Sum of consequtive integers = 21. (arthematic progression)

21 = n/2 * (2a+ (n-1) * 1) as d = 1.

n^2+ (2a-1)n - 42 = 0

n(n+2a-1) = 42

14

Factors of 42 are (7*3*2)

So 42 can be expressed as two factors in (2*2*2)/2 ways = 4 ways

To solve the question completely

Factors of 42 are (7*3*2)

1*42, 2*21, 6*7 and 3*14

so n can be 1, 2, 3 or 6

To give exact series,

For n = 1 => 2a-1+1 = 42 -> a = 21

21

For n = 2 => 2a-1+2 = 21 -> a = 10

10, 11

For n = 3 => 2a-1+3 = 14 -> a = 6

6,7,8

For n = 6 => a = 11,2,3,4,5,6

For which values of n the expression [(100^n) - 1] is divisble by 11. N is a +ve integer.a) all even b) all odd c) all prime d) any value e) no valueAns:100^n - 1 = (100^n -1) = (100 -1)*(100^(n-1) + ...) = 99*(...)so it's a multiple for 11 for every +ve integer since (a^n -b^n) = (a-b)*(a^(n-1)+....) holds for all +ve integers.

OR,

(100^n) will always have odd number of digits of the form 100, 10000, 1000000,..[(100^n) - 1] will be even no of digits of the form 99, 9999, 999999, etc.

Divisibility rule of 11 is sum of even digits - sum of odd digits = 0. Any no with even number of digits and all digits same will always be divisible by 11. So this is divisible for all +n.

What is the minimum value (grater than 1) that can divide [2^(4n+2)] + 1(n is positive integer)a) 2 b) 3 c)4 d) 5 e)9Ans:2^(4n+2) + 1 = k => 4*16^n +1 = k => 16^n = (k-1)/4 for n=1 => 16 = (k-1)/4 => k=65 or k = 5*13

sqrt(a^2*b) + sqrt(a*b^2) could be

1) a*sqrt(b) + b*sqrt(a)

15

2) -a*sqrt(b) - b*sqrt(a)3) -a*sqrt(b) + b*sqrt(a)4) a*sqrt(b) - b*sqrt(a)I 1) alone II 1) + 2) III 1) + 3) IV 1) + 4) V all answers are correctAns:value of a and b < 0 are ruled out, because statement says square roots exist.

So, both a and b should be +ve and hence I is the answer

n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?Ans:I guess it doesn't matter, just plug in n=1 because whatever integer you plug in will always give you the third in the pattern, the 7-ending . . . (e.g. n=1, means 3^(16+3), which means 19/4, remainder three so it's the third (7) in the powers of three series

OR,

three ain't bad, bro, just see that you have 3^11 . . . 3 to the power of anything always goes in the pattern 3,9,7 (27), 1 (81) . . . so the units digit will be 7 since we're in the third of a series of four, and add two (like the problem says) and you have 9 in the units, divided by 5 that's remainder 4

In the addition table shown below, what is the value of m+n?

Ans:m+n= (z+4) + (y+e) = (z+e) + (y+4) = 10 + (-5) = 5

If , what is the unit’s digit of ?

Ans:

16

The unit’s digit of the left side of the equation is equal to the unit’s digit of the right side of the equation (which is what the question asks about). Thus, if we can determine the unit’s digit of the expression on the left side of the equation, we can answer the question.

Since , we know that 13! Contains a factor of 10, so its

unit’s digit must be 0. Similarly, the unit’s digit of will also have a unit’s digit of 0. If we subtract 1 from this, we will be left with a number ending in 9.

Therefore, the unit’s digit of is 9. The correct answer is E.

If P represents the product of the first 15 positive integers, then P is not a multiple of:a) 99 b) 84 c) 72 d) 65 e) 57

Ans: If P represents the product of the first 15 integers, P would consist of the prime numbers that are below 15.

2,3,5,7,11,13

Any value that has a prime higher than 13 would not be a value of P.

57 = 3, 19

19 is a prime greater than 13, so the answer is E.

If y ≠ 3 and 2x/y is a prime integer greater than 2, which of the following must be true? I. x = yII. y = 1III. x and y are prime integers.(A) None(B) I only(C) II only(D) III only(E) I and IIAns:

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If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?I.3 II.4 III.6

Soln: The idea is to find what are the common factors that we get in the answer.

m = 1, k = 30 which is divisible by 1,2,3,5,6,10, 15 and 30m = 2, k = 84 which is divisible by 1,2,3,4,6, ....

As can be seen, the common factors are 1,2,3,6

So answer is 3 and 6

What is the remainder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?Ans: Remainder of 9*odd /6 is 3remainder of 9*even/6 is 0

9^1 + 9^2 + 9^3 + ...... + 9^9=9*(1+9+9^2+.....9^8)

1+9+9^2+.....9^8 is odd.

Thus we obtain 3 as a remainder when we divide 9*(1+9+9^2+.....9^8) by 6.

Another way: We get the value as 9*odd/6 = 3*odd/2 Since 3*odd = odd; odd/2 = XXXX.5so something divided by 6, gives XXXX.5, hence remainder is 6*0.5 = 3

The numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x < y, which of the following must be true?

I. The units digit of x + y is greater than the units digit of either x or y.II. The tens digit of x + y equals 2.III. The hundreds digit of y is at least 5.

A. II onlyB. III onlyC. I and IID. I and IIIE. II and IIIAns:

x= abc y= def

x = a7c y= b5f

x > y and x+y = wxyz.

I. The units digit of x + y is greater than the units digit of either x or y.

It can carryover one digit. False

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II. The tens digit of x + y equals 2.

It can be 2 or 3. False

III. The hundreds digit of y is at least 5.

a+b+1 >= 10a >b so a at least 5. True.

Ans: b

How many integers less than 1000 have no factors (other than 1) in common with 1000?Ans:1000 - multiples of 2 and/or 5

multiples of 2 = 500 (all even #)multiples of 5 = (995 -5)/10 + 1 [ Using AP formula]= 100

Answer = 1000 - (500 + 100)= 400

You cannot calculate for all multiples of 5 because you have already removed all even integers (including 10, 20, and 30). The difference in the AP series should be 10 instead of 5 because you're looking for the integers that have 5 as a unit’s digit. Therefore we divide by 10 and not 5.

OR,

Multiples of 2 = 500Multiples of 5 = 200Multiples of 10 = 100

Answer = 1000 – (500+200-100)

Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

36, 28, 12, 9 or none

Soln: Let the divisor be a.

x = a*n + 11 ---- (1)y = a*m + 21 ----- (2)also given, (x+y) = a*p + 4 ------ (3)adding the first 2 equations. (x+y) = a*(n+m) + 32 ----- (4)

equate 3 and 4.a*p + 4 = a*(n+m) + 32or a*p + 4 = [a*(n+m) + 28] + 4cancel 4 on both sides.u will end up with.a*p = a*(n+m) + 28.

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which implies that 28 should be divisible by a. or in short a = 28 works.

Another method: I think the easiest (not necessarily the shortest), way to solve this is to use given answer choices. Since the remainders are given as 11 and 21, therefore the divisor has to be greater than 21 which leaves with two choices 28 and 36. Try 28 first; let the two numbers be 28+11= 39 and 28+21= 49. Summing them up and dividing by 28 gives (49+39=88), 88/28 remainder is 4, satisfies the given conditions. Check for 36 with same approach, does not work, answer is 28

Find the number of positive integers less than 2003 that are divisible by5,7 or 11Ans:No. of factors of 5 in 2003= 400No. of factors of 7 in 2003= 286No. of factors of 11 in 2003= 182No. of factors of 35 in 2003= 57No. of factors of 55 in 2003= 36No. of factors of 77 in 2003= 26No. of factors of 385 (5*7*11) in 2003= 5

Total no of factors = fac of 5+fac of 7+fac of 11 - fac of 35 -fac of 55 -fac of 77 + fac of 385

=400+286+182-57-36-26+5 = 754.

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?Ans:990 = 2*3^2*5*11

The greatest prime factor of 990 is 11, so the product should at least has 11 as a factor. Now, since other factors of 990 are also included in the product of all integer from 1 to 11, the least value of n is 11.

If 9 is a factor of 2x, then which of the following may not be an integer:a) 6x/54 + 2x/3b) (4x-18)/9c) (2x-27)/9d) (81-4x^2)/81e) (2x-3)/3Ans:Substitue 2x = 9y in the equations, and you find that for first one, you will get 7y/2 - which MAY NOT be a integer. All others are integers.

If the prime factorization of the integer q can be expressed as , where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?

(A) 3j + 4, where j is a positive integer(B) 5k + 5, where k is a positive integer(C) 6l + 2, where l is a positive integer(D) 9m + 7, where m is a positive integer(E) 10n + 1, where n is a positive integerAns:

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no. of divisors = (2x+1)(x+1)3x = 3 x (x+1) (2x+1)

that means it shud be

1. even 2. multiple of 3

now, (A) 3j + 4, where j is a positive integer

its not a multiple of 3

(B) 5k + 5, where k is a positive integer

can be even as well as multiple of 3

(C) 6l + 2, where l is a positive integer

every multiple of 6 is a multiple of 3, and 6l +2 cant be multiple of 3 even though its even

(D) 9m + 7, where m is a positive integer

every multiple of 9 is a multiple of 3, and 9m +7 cant be multiple of 3

(E) 10n + 1, where n is a positive integer

cant be even

so left with B only

In a certain deck of cards, each card has a positive integer written on it. in a multiplication game, a child draws a card and multiplies the integer in the card by the next larger integer. if each possible product is between 15 and 200, the least and greatest integers on the cards could be?Ans:n(n+1) > 153(4) = 12 <15 so 4(5) = 20 OKn(n+1) <20013*13 =169 and 14*14 = 216 so it has to be 13..13*14=182<200so 13.

x, a, and b are positive integers. When x is divided by a, the remainder is b. When x is divided by b, the remainder is a – 2. Which of the following must be true?

A. a is even.B. x + b is divisible by a.C. x – 1 is divisible by a.D. b = a – 1E. a + 2 = b + 1Ans:

Need to look at the answer choices for some sort of direction.Choice 1: Can't see a reason straight away why this must necessarily be trueChoice 2: No

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Choice 3: Why?Choice 4: Hold on a minute ... If b is the remainder when x is divided by a, then b < aIf a-2 is the remainder when x is divided by b, then a-2 < bSo, a-2 < b < aSince we are talking about integers here, b = a-1

if n is a positive less then 200 and 14n\60 is an integer, then n has how many different positive prime factors?23568Ans:14n/60 reduces to 7n/30Since 7 is a prime then the only way for 7n/30 to result in an integer is if n is a multiple of 30.

The prime factors of 30 are 2,3, and 5,

for n < 200 we can have n = 30, 60, 90 ..... 180In each case the distinct prime numbers is still 2,3 and 5

<x> denotes the remainder when 3x is divided by 2. Are the following numbers equal to 1?

1. <3x>2. <2x+1>3. <2x>+1Ans:1 - Can be 0 (if n is odd) or 1 (if n is even)2 - Yes3 – Yes

A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?Ans:A=3m/s, B=6m/s and C=18m/s A covers x => B covers 2x and C covers 6xi.e. in a circular path if x statring point for first time then B is there for second time and C is there for 6th timeA takes 750/3=250s to be at starting point again

A) How many numbers between 1 and 100 are not divisible by 2 and 3?B) How many numbers between 1 and 100 are not divisible by 2 and 3?Ans:Div by 2 = 50Div by 3 = 33Div by 6 = 16A) 100 – 16 = 84B) 100 – (50+33-16) = 33

The sum and product of the 5 numbers from -5 to 5, inclusive, both are integers greater than 0. What is the possible number of the negative numbers?

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a.2 b. 3 c. 4 d.5 e.6Ans:For the product to be greated than 0, we cannot choose the number 0, and the no. of negative values can either be 2 or 4, so that the product remains > 0

For the sum to be greater than zero, Absolute value of sum of positive numbers should be greater than Absolute value of sum of negative numbers, Here assuming that the same value cannot be chosen twice, the number of negative values cannot be 4,

so answer should be 2 i.e. B

M = 4^1/2 + 4^1/3 + 4^1/4 , value of M is :

- less than 3- equal to 3- between 3 and 4- equal to 4- more than 4

Ans:4 ^ 1/4 = 1.414

so min would be 1.414 + 1.414 + 1.414 > 4

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer?A. a/2B. b/2 C. (a+b) /2D. (a+2) /2 E. (2+b) /2Ans:a-b = even -- (1)=> a & b are both even or both are odda/b = even-- (2)=> a = even and a = 2n*b=> a and b are both even from (1)

Skim thru' the ans choices..(a+2)/2 = (2nb+2)/2 = nb + 1=> Since b is even, nb+1 will be oddAns D.whereas evaluating Choice E:(b+2)/2 => b/2 + 1 can be odd or even depending on whether b/2 is even or odd...Insufficient

There are 7 digits in the expression 452*17*36.When how many digits were added by 1, the increasing value of the expression is less than 1000?

A.Only one digit B.TWO CTHREE D.FOUR E.FIVEAns:only case possible is 17 * 36 * (453 - 452) < 1000

Two positive integers A and B have a least common multiple of 120. What is the greatest possible value of their common factor?

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(a) 20 (b) 18 (c) 15 (d) 12 (e) 10 Ans:120 = 2 * 2 * 2 * 3 * 5

possible GCF are

a) 2*2*2*3*5b) 2*2*3*5c) 2*2*5 (Which is the answer since there is no option greater than 20 which might be a possible GCF).

x/3 is between 3 and 100, and it is the square of a prime number. How many such x are possible

12340Ans:3<x/3<100

Let y = sqrt(x/3) = prime

==> sqrt3 < y < 10 and y is a prime

==> y = 2, 3, 5 or 7

Hence x can have 4 values.

Remainder of A/5 is 3, A/7 is 4, B/5 is 4, B/7 is 1. which number of below is factor of A-B?

a)12b)28c)48d)58e)64

Ans:

From first equation A = 5*x + 3=> A = 3,8,13,18,23,28,33,38,43,48,53,58,63,68,73,78,83,8 8....

From second A = 7*y + 4=> A = 4,11,18,25,32,39,46,53,60,67,74,81,88.......

=> A = 18,53,88....difference of 35

hence A's vale follows the sequence A = 18 + (n1 -1)*35 { nth term of an AP is a+ (n-1)d}

Similarly B = 29,64,99....

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=> B = 29 +(n2 -1)*35

B-A = 29-18 + 35 ( n2 - n1)=> B-A = 11+ 35(n2-n1)=> for n2-n1 = 1,2,3 we getB-A = 46,81,116,151,186,221,256...116 is divisible by 58256 is divisible by 64

if n2-n1 = -1,-2,-3B-A = -24,-59,-94,-129,-164,-199,-234...-24 is divisible by 12

If M is the least common multiple of 90, 196, and 300, which of the following is NOT afactor of M?A. 600B. 700C. 900D. 2,100E. 4,900Ans:M = 3*3*2*2*5*5*5*7*7

600 = 3*2 *5*5*2*2

Hence A is the ans

what r the last three digits of 7^7994 ?

Ans:

it can be written as 49*2401^1998 = 49 *(2400+1)^1998.NOw, as only the last 3 digits are relevent to us .. 2400^2 and greater powers do not add value as 2400^2, itself has the last 4 digits are zero.So, for the required 3 digits, we need only the last 2 terms of the expansioni.e 49 [1998c1*2400+1] = 49*1998*2400+49 = ..800+49=..849.

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

A) 6B) 12C) 24D) 36E) 48Ans:every positive integer can be represented as a product of prime numbers.Let`s take 6. 6=2*3A square of an integer have prime factors of the original number in pairs.6=2*36^2=6*6=2*3*2*3=36So, if n^2=x*72 (n^2 is divisible by 72), where x is an integer, n*n=x*2*2*2*3*3To make pairs, we have to have at least four 2s and two 3s for n^2, so n alone will have 2*2*3 as prime factors. x may be 2 or a multiple of 2 for n^2 to be divisible by 72.This makes the answer 12

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If 10^k-1 <o.ooo25 < 10^k , what is the value of k? Ans:10^(k-1) < 0.00025

the largest possible value of 10 to some power that satisfies the inequality is 0.0001 < 0.000250.0001 = 10^(-4)10^(k-1) = 10^(-4)k - 1 = -4k = -3

OR,

10^k-1 < 0,00025< 10^k thus 10^k-1 * 10^5< 0,00025*10^5<10^k*10^5so 10^(k+4) < 25<10^(k+5) so k = -3 ( because 10<25<100)

What is the number of Odd divisors of 20! ?

A. 42984B. 34560C. 2160D. 720E. 64Ans:20! odd primes = 3,5,7,11,13,17,19

20! = (3^x)*(some integer) (what is the greatest value of x ?)

=> x = [20/3]+[20/9] = 8

similarly power of 5 = [20/5] = 4power of 7 = [20/7] = 2power of 11,13,17,19 = 1

hence, no. of odd divisors = (8+1)*(4+1)*(2+1)*(1+1)^4 = 2160.

Let f(x) = ax^2 + bx + c and g(y) = ly^3 + my + n maximum number of elements in the solution set of g(f(x) is:

A) 3B) 4C) 5D) 6E) 7Ans:it says that g(y)=ly^3+my+n. question asked is g(f(x)).

subs f(x) in place of y. it becomes g(f(x))= l [f(x)]^3 +m[f(x)]+n

now, equation of f(x). g(f(x))=L[(ax^2+bx+c)^3+m[(ax2+bx+c)]+n

from above equation we know that highest powers will be x^6. so, we will have 6 solutions

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Which of the following will have a zero remainder when divided by the product (17-11)(17^2 + 11^2)(17^3 + 11^3)(17^4 + 11^4)A. 17^5 - 11^5B. 17^4 - 11^4C. 17^6 - 11^6D. 17^18 - 11^18E. 17^24 - 11^24Ans:17^24 - 11^24= (17^12-11^12)(17^12+11^12)= (17^6-11^6)(17^6+11^6)(17^12+11^12)= (17^3-11^3)(17^3+11^3)(17^6+11^6)(17^12+11^12)= (17-11)(17^2+17*11+11^2)(17^3+11^3)(17^2+11^2)(17^4-17^2*11^2+11^4)(17^4+11^4)(17^8-17^4*11^4+11^8)

so this is divisible. Use a^3-b^3 = (a-b)(a^2+ab+b^2) anda^3+b^3 = (a+b)(a^2-ab+b^2) formulasNote: the highest power of 17 in the divisor is 24.

Ans:The smallest positive prime number is 2 so by multiplying 2 with 2 would give us maximum number of products... right? Right...

more like this, 2*2*2*2*2*2 = 64Now to go one step higher we will take 2*2*2*2*2*3 = 96if we go one more step higher i.e. if we take 2*2*2*2*2*5 = 160 which is not a two digit number so the right answer is Two

What is the tenths digit of 7^202?Ans:(7)*(7) * (7) *(7) ..... 202 termsMultiply 2 terms together

49 * 49 * ...... * 49 101 termsMultiply 2 terms together2401 * 2401 ..... * 2401 (total terms 25) * 49

now 25 terms = 2401 .. so multiplying them will give a tenth digit as zero only.And when this is multiplied with 49.. it gives a "4" at tenth digit place.

ABC is a three digit number and A>0. The value of ABC is equal to the sum of the factorials of three digits. The value of B isA) 19B) 8C) 4

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D) 2Ans:19 can be eliminated right away8! > 3 digit number eliminate it.

ABC

A can be 1 to 9.C can be 1 to 9.

but 9! 8! 7! are all > 999so A and C have to be less than or equal to 6.again 6! is 720. and A <6. doesn't make sense.so, A and C have to be less than or equal to 5.again. 5! = 120 2*5! = 240. so the maximum sum can be only 240 + 4! = 264=> A has to be less than or equal to 2.

againABCA <= 2B say 4.C <=5again maximum sum can be only 5!+4! + 2! < 200=> A has to be 1

againA = 1B say 4C <=5maximum sum can be 5!+24!+1 = 120+25 = 145.check out A = 1 B = 4 and C =5. works fine

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?

abdc+dbca

(A) 3689(B) 6887(C) 8581(D) 9459(E) 16091Ans:

10d + 11c < 100 - a=> 10d + 11c + a < 100

The last 2 digits of each number are dc and ca

dc + ca = 10d + c + 10c + a = 10d + 11c + a < 100

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Hence there is no carry forward to the hundred's place. ANd since sum of b+b has to be even, we can say that 8581 cant be the answer.

If n=35*3^4*2^7 and a and b are integers such that 1<a<b and n/(ab^3) is an integer, what is the sum of the possible values of a?Ans:n=35*3^4*2^7=> n = 5*7*(3^3)*3*(2^3)*(2^3)*2 = 2*3*5*7*(3*2*2)^3=> a < 12 as b can have a max value of 12 Since n/(a*(3*2*2)^3) is an integer=> a can have values of 2,3,5,6,7,10

b can also be 6. Implying that 5*7*3*(2^4) is divisible by a. and since a<b => a<6, we can get one more value of a which is not mentioned above i.e (4)

sum = 37

S is the set of positive integers n such that when 777 is divided by n, the remainder is either 7 or 77. How many elements does S contain?

(A) 11 (B) 15 (C) 19 (D) 23 (E) none of theseAns:Remainder 7:777 = an + 7 where n > 7=> an = 770=> an = 2*5*7*11=> Desirable combinations for n > 7 = 4c4+4c3+4c2+1 (for single number 11)=> 1 + 4 + 6 + 1 = 12

Remainder 77:777 = an + 77 where n > 77=> an = 700=> an = (2^2)*(5^2)*7Desirable combinations = 100, 140, 175, 350, 700

Total = 12+5 = 17

For a positive integer k written in decimal notation, #(k) is defined as the product of the digits of k. For instance, #(333)=27 and #(142)=8. If Y is the set of four digit whole numbers w such that #(w)=60, what percent of the elements of Y are greater than 5000?Ans:Total possible combinations are combinations of (2,2,3,5), (1,3,4,5), (1,2,5,6) => 4!/2! + 4! + 4!= 60

Desired Combinations =>3!/2! = 3 (from (2,2,3,5) group) + 3! = 6 (from (1,3,4,5) group) + 3! = 6(from 1,2,5,6 group -> first digit 5) + 3! = 6(from 1,2,5,6 group -> first digit 6)

Desired Prob = 21/60 = 7/20 = 35%

There are 63 oranges and 42 pears in a basket. After each of a certain number of children get n oranges and m pears, 11 oranges and 3 pears are left in the basket. What is the number of the children?Ans:

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number of children is a common divisor of (63-11, 42-3) = a common divisor of (52, 39) = 13

N = P1^a1 x P2^a2 x P3^a3 x............xPn^an

Sum of the factors: [(P1^a1+1)-1]/P1-1 x [(P2^a2+1)-1]/P2-1 x ...........x[(Pn^an+1)-1]/Pn-1

Let's look at an example 127008 = 32 x 81 x 49 = 2^5 x 3^4 x 7^2

Sum of the factors is (2^6)-1/1 x (3^5)-1/2 x (7^3)-1/6 = 63 x 121 x 57 = 434511

I think it's clear

8184 = 2^3*3*11*31Sum= (2^4)-1/1 x (3^2)-1/2 x (11^2)-1/10 x (31^2)-1/30= 15 x 4 x........So it should end with 0

If N=t^3, where t is a positive integer and 8, 9, 10 are factors of N. Which of the following must be the factor of N?A. 16B. 81C. 275D. 225E. 325 or 375Ans:LCM of 8,9 and 10 is 360360 = 2^3*5*3^2.=>t^3 > 2^3*5*3^2But N = t^3 is a perfect cube. So, N(Min) = 2^3*5*3^2 * (5^2*3) ( multiply by 5^2 and 3 to make it a perfect cube)=> N(Min) = 360 * 75and 225 = 5 * 5 * 3 *3

Hence "D" is the ans

If x=71637432/102002733, y=71637399/102002698 and z=71637456/102002798, which of the following is true?

(A) x<y<z (B) x<z<y (C) z<x<y (D) z<y<x (E) y<x<zAns:x=71637432/102002733y=71637399/102002698 z=71637456/102002798

Let Numarator of x is N(x) similarly Denominator of x is D(x)....& so onz=N(x)+24 / D(x)+65So definitely x>z as we have increased N by 24 but D by 65.z=N(y)+57 / D(y)+100So again y>zNow we have to find relation between x & yx & y are close to 7/10 eachx=N(y)+33 /D(y)+35Now 33/35 is greater than 7/10..........( logic here is if we have 7/10 as a number as if we are adding 7 in N and 10 in D....then that number become same....but if we are

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adding 8 in N and 10 in D then resulting number is greater than previous)So here x>yHence ans is x>y>zHence DOR,

taking approx figures:x=430/730y=400/700z=460/800

now simply multiply deno of z with numerator of x and viseversa ...x>zsame with x and y....x>yso x>y>zz<y<xD

When a number is divided by 13, quotient is K, remainder is 2; when divided by 17, remainder is 2. What is the remainder when K is divided by 17?Ans:n = 13k+2n = 17a+2=> k = (17/13)*a=> a = 13b since k is an integer=> k = 17b which divided by 17 will have a remainder 0

When 20 is divided by k the remainder is k-2, What can be the value of k? (This one's easy just plug in)A.5 B.3. C.11 D. 14. E.6Ans:Same concept as above:20 = ak + k-2=> 22 = k(a+1)=> 2*11 = k(a+1)=> 11 is the only ans choice which satisfies this eqn

If s and t are factors of N, N^(st) is divisible by which of the followings? I. s^t II. (st)^2 III. s+tAns:N = n*s*t=> N^(st) => n^st*s^st*t^st=> I. s^t & II. (st)^2 are factors

YX7+6Y=Y7XX = ?Ans:YX7+6X=100Y+10X+7+60+Y=100Y+70+X=Y7X=>9X+Y=3if X>=1 then 9X+Y>=9>3 (eliminate)so that X=0 (Y=3).The correct answer is X=0

X^2 + Y^2 <= 100How many pairs of answers(x, y) can you have?Ans:

31

for ......0 there will be 10+10+1=21(1-10 both positive and negative and 0)wayssimilarly,1--------192--------193--------194--------195--------176--------177--------158--------139--------910-------1 ways.

total ways for 1 to 10= 148 excluding 0but they can be negative also.

so total ways= 2*148+21= 317

A 3 digit number is concatenated with itself to form a 6 digit number( if the initial number is 123 then the 6 digit number is 123123 ). Now...the 6 digit number is always divisible by which of the following numbers:1) 1112) 1573) 1434) 2565) 345Ans:let me give u the logic now.let us suppose the number is n.then 6 digit number is: 1000n + n= 1001* nnow...1001 is divisible by 143.

So...the answer is 143.

The Flying Acrobatic Team Is Made Up Of 120 Airplanes, The Teams Wants To Form A Rectangular Formation With X Planes In Row And Y Planes In Column. If No Of Airplnes In A Row Is No Less Than 4 And No More Than 30. How Many Different Combinations Of Rectangular Shape Are Possible?Ans:10...2( 4, 30; 8, 15; 6, 20; 10, 12; 5, 24)

OR, 120= 2^3*3^1*5*1no of divisors: (3+1)(1+1)(1+1) = 16subtract: 120, 1, 2, 60, 3, 40; no of divisors = 16-6 = 10

OR,

120 = 2*2*2*3*5

(5C2 ) /3!5C2 for rows, 3! for repeated 2's

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=5*4*3 / 3*2= 10

If k in an integer and 5^k < 20,000, what is the greatest possiblevalue of k ?5^k < 20000 ==> 5^k < 5^4 * 32 ==> 5^(k-4) < 32 ==> k-4=2 ==> k=6

If r,s are positive integers each greater than 1 and if 11(s-1)=13(r-1),what is the greatest possible values of r+s less than or equal to 100?Ans:s-1\ r-1 = 13\11 By trial and err consider a no: multiple of 13 in the num' and multiple of 11 in denom', such that adding both nos : gives u an ans less than 10013*4\ 11*4= 52\44since we need to sub 1 from the num' and denom' i too a no: 1 more than the 52 and 44

answer : 98

x+y+z=7x*y + y*z + z*x = 10What is the maximum value of x?Ans:(x+y+z)^ 2 = x^2 +y^2 +z^2 +2(10) = 49x^2+y^2+z^2 = 29If y,z are negative, y^2,z^2 is positive.

Maximum value of x = root(29)

How many positive integers less than 20 are equal to the sum of a positive multiple of 3 and a positive multiple of 4?Ans:multiples of 3: 3 6 9 12 15 18multiples of 4: 4 8 12 16numbers less than 20 that are equal to the sum of a positive multiple of 3 and a positive multiple of 4: make pairs 4: 3, 6, 9, 12, 158: 3, 6, 912: 3, 616: 3and 19 repeats twice so total 10 numbers

Which of the following CANNOT be expressed as the product of exactly 2consecutive integers?(A) (2) (3) (7)(B) (2) (3) (7) (11)(C) (2) (25) (13)(D) (4) (3) (5) (7)(E) (9) (11) (13)Ans:well the product of 2 consecutive integers is always an even number...

so at first glance, only E gives an odd result

5A+ BC

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-------D43---------

The numbers used are 1 to 9 (both inclusive). No repetition is allowed.

Col A - A + B + C + D

Col B - 20Ans:A+50+C+10B=43+100D==> 100D-A-C-10B=7D must equal 1. A+C+10B=93 ==> A+B+C+D = 93-10B+B+1 = 94-9B. If B can equal 9 then D, if not then A. If B=9, A+C=3, either of them will equal 1 - which is not possible because D is 1.

The solution is A.

OR,

A and C can take values 1 and 2 = 3 4 and 9 = 135 and 8 = 136 and 7 = 13accordingly B takes the values 9 or 8but for (1 and 2) and (5 and 8) either D or B takes the value 1 or 8 (repetition) and for other two pairs, (4 and 9) and (6 and 7) A + B + C + D = 22 so answer is A

Consider a positive odd integer n. Let X = twice the number of factors of n, and let Y = the number of factors of 2n. Which of the following must be true?

a. X = 4 b. X > Y c. Y > X d. X = Y e. There is not enough information to solve the problem.Ans:n can be prime factorised to give: n = (p^a)*(q^b)*(r^c)*... where p,q,r,... are distinct prime numbers other than 2 (since n is an odd integer). The number of distinct factors of n is given by (a+1)*(b+1)*(c+1)*...X = 2*(a+1)*(b+1)*(c+1)*...

Similarly, number of factors of 2n, Y = (1+1)*(a+1)*(b+1)*(c+1)*... = X

If y ≠ 3 and 3x/y is a prime integer greater than 2, which of the following must be true? Ⅰ. x = yⅡ. y = 1Ⅲ. x and y are prime integers.(A) None(B) Ⅰ only(C) Ⅱonly(D) Ⅲonly(E) Ⅰand ⅢAns:

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3x/y is a prime integer > 2 and y!=3

I. x=yif 3x/y=3=>x/y=1=>x=yif 3x/y=5=>x/y=5/3=>x!=y

II. y=1=>3x/y=3x if x=1=>3x/y=3 primeifx=2=>3x/y=6 not prime

III. x and y are prime => 3x/y cannot be an integer as no common factors (as y!=3)

Hence I, II, III are false

answer = A

((3^5)(5^7)(7^9)(11^13)) / 4Ans:3^5*5^7*7^9*11^13= 3*7*11*((4-1)^4*(4+1)^7*(8-1)^8*(12-1)^12)= 3*7*11*( multiple of 4 + 1)Remainder = mod(231,4) = 3

OR,

A quicker way: For a^n.b^n.c^n/y and the remainders when a^n, b^n, c^n leaves a remainder of Ra, Rb & Rc respectively, then Ra*Rb*Rc/y leaves the same remainder as a^n.b^n.c^n/y. So, this simplifies life (And this problem in qn)

((3^5)(5^7)(7^9)(11^13)) % 4= (3%4*5%4*7%4*11%4)= (27%4)= 3

For any number , x x denotes the least non-negative number y such that y x + is an integer. What is the value of 8.4 – 8.4?(A) -0.4(B) 0(C) 0.6(D) 7.8(E) 8.0Ans: 7.8

if 2x^2 - 3x - m is divisible by x - 3, then m =

a) -18b) -9c) -6d) 9e) 18Ans:if x-3 is a factor, then x = 3 solves the equation.if you substitute 3 for x, you get m = 9

A cube is 100 by 100 by 100. However, the estimator made a mistake and the error could be +1 or -1 on all the dimensions. What is the maximum difference between actual and estimated?Ans:

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x^3-y^3 = (x-y)(x^2+xy+y^2) = (x-y)((x+y)^2-xy)101^3 - 100^3 = (101-100)((101+100)^2 - 101*100) = 30301

How many different sums can be obtained by adding any 10 different odd numbers selected at random from the series of first 100 odd numbers. Such that no two sums have identical results.Ans:Sum of 10 least odd +ve integers = 10^2 = 100Sum of 10 highest odd +ve integers = 100^2 - 90^2 = 1900

All the potential sums will be even numbers spanning from 100 to 1900. Using AP,100 + (n-1)*2 = 1900n = 901

If a – b = a - b, Then in terms of b, a equals:

(A) b(B). b(C). b - 2b + 1(D). b + 2b + 1(E). b^2 -2b + bAns:(a-b)= [Sqrt(a) - sqrt(b)]*[sqrt(a) + sqrt(b)]

sqrt(a) + sqrt(b) = 1

=> sqrt(a) = 1-sqrt(b)squaring both sidesa = 1+b- 2sqrt(b)

a is the sum of x consecutive positive integers. b is the sum of y consecutive positive integers. For which of the following values of x and y is it impossible that a = b?(A) x = 2; y = 6 (B) x = 3; y = 6 (C) x = 7; y = 9 (D) x = 10; y = 4 (E) x = 10; y = 7Ans:The sum of a series of n consecutive integers will be:Even if n = 2*even numberOdd if n = 2*odd numberEven or Odd if n is odd

With Option D, the sums can never equal.

The Sum of n consecutive integers with a, the smallest integer = (n/2)*(2a+n-1) when n is evenNow, (2a+n-1) = even + even - 1 = odd (always)Consequently, if n = 2*(even number) the Sum will be even, and if n is 2*(odd number), the Sum will be odd.

In n is odd,Sum = [(n-1)/2]*(2a+n-2) + (a+n-1)The product will be even or odd depending on the (n-1) termThe last term will be even or odd depending on a.The Sum can be odd or even.

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OR,

Consecutive positive numbers represented = n+(n+1)+(n+2)+(n+3) ......+n+(n-1)

plug in choices and check

X= 2 , 2n+1 = always oddy= 6 ,6n+15 =always odd

possible set

X= 3 ; 3n+3 = even if n=odd, odd if n=oddY = 6, 6n+15 =always oddat time possible .keep this choice

x = 7; y = 9 x=7; 7n+21 = depends at time odd and at times even x=9 ;9n+45 = dependsat times possible

x = 10; y = 4 x=10; 10n+55 =always oddy =4; 4n+6 = always even never possible .stop here

4^(a-5+b) = 2^(a+b) * 2^b * 2^(a-4) -63Find the sum of a & b.Ans:4^(a-5+b) = 2^(a+b) * 2^b * 2^(a-4) -63=>2 ^ ( 2a – 10 + 2b ) = 2 ^ ( a + b + b + a – 4) – 63 = 2 ^( 2a + 2b – 4 ) – 63=>2^ (2a + 2b – 4 ) * 2^ (-6) = 2 ^( 2a + 2b – 4 ) – 63=>63 = 2^(2a + 2b – 4) * (1 – 2^(-6) )=>63 = 2^(2a + 2b – 4) * 63 / 64=>64 = 2^ (2a +2b – 4)

2a + 2b – 4 = 6 ; a+b = 5

R = (30^65-29^65)/(30^64+29^64)

1) 0<R<0.1 2) 0.1<R<0.5 3)0.5<R<1.0 4)1.0<RAns:In general when n is odda^n - b ^n = ( a-b) * ( a^n-1 + b^n-1 + a*b^n-2+.....+ a^n-2 * b)

for example a^3 - b^3 = (a-b) (a^2 +b^2 + ab)a^5 -b ^5 = (a-b) (a^4 + b^4 + a*b^3 + a^2*b^2 + a^3*b )

Therefore 30^65 - 29^65 = (30-29) * (30^64 + 29^64 + some positive term) = (30^64 + 29^64 + some positive term)

Now (30^65 - 29^65 ) / (30^64 + 29^64 ) = 1 + (some positive term)/(30^64 + 29^64 )

Therefore R > 1

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Nine numbers: 0,1,1,4,5,6,8,8,9,can be compose to three number (all three figures), every number can be used one time, what is the maximum value of the sum of the three numbers?Ans:Arrange the digits in decreasing order, i.e.

9,8,8,6,5,4,1,1,0

As we have to find the greatest sum, so we need to find the three greatest 3digit number from these digits, so take first three digits as hundreds digit for the three numbers.

so,numbers will be , 900 , 800 & 800Next again take the next three digits in decreasing order and add themm to 900, 800 & 800, i.e., 960, 850,840Now same as above & add the next 3 digits, so numbers will be

961 + 851+ 840 =2652.what would be remainder for (2^91)/7?Ans:case1: a^n/(a+1) leaves a remainder of

a if n is odd1 if n is even

case2: (a+1)^n /a always leaves a remainder of 1.

In the above problem,2^91/7=> 2* (2^90/7)=> 2* (8^30/7)=> 2* ( (7+1)^30/7) ( (7+1)^30/7) always leaves a remainder of 1=> 2*1=2

OR,

2^91 = 2 * (2^3)^30 = 2 * (7 +1 ) ^ 30 = 2 * ( 7^30 + a* 7^29 + b * 7^28........+ n* 7 + 1^30) = 2 ( 7 *m + 1) = 14 * m + 2

a,b,c...are coefficients of the algebraic expression (x+y)^n. But the point is every term in the expression is divisible by 7 except the last term which is one.

Therefore the remainder is 2.

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Ans:I. x^2 < 2x < 1/x x^2 < 2x for all positive x < 2.2x < 1/x for all positive x < (sqrt2)/2.x^2 < x for all positive x < 1.Thus statement I is true for all positive x < (sqrt2)/2For example, if x = 0.5:x^2 = 0.252x = 11/x = 2

II. x^2 < 1/x < 2xx^2 < 1/x for all positive x < 1.1/x < 2x for all x > (sqrt2)/2.x^2 < 2x for all positive x < 2.Thus statement II is true for all (sqrt2)/2 < x < 1.For example, if x = 0.8:x^2 = 0.641/x = 1.252x = 1.6

III. 2x < x^2 < 1/x2x < x^2 for all x > 2.x^2 < 1/x for all positive x < 1.Contradiction. Statement III can never be true.

I and II only. The answer is D.

The average of the five consecutive integers is an odd number. Which of the following must be true?

I. The largest of the integers is even.II. The sum of the integers is odd.III. The difference between the largest and smallest of the integers is an

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even number.

A)I onlyB)II onlyC)III onlyD)I and II E)II and IIIAns:n + (n+1) + (n+2) + (n+3) + (n+4) = 5n+10avg = n+2 -> oddso n -> oddI) n+4 -> oddII) n odd => 5n+10 -> oddIII) n+4 – n = 4 -> even

Hence E

If x, y and z are positive integers such that 0<x<y<z and x is even, y is odd and z is prime, which of the following is a possible value of x+y+z?

A)4B)5C)11D)15E)18Ans:X = 2, y = 3c) 6d) 10e) 13

hence e

-1/2 <= x <= -1/3 and -1/4 <=y <= -1/5, what is the minimum value of xy^2?a) -1/75b) -1/50c) -1/48d) -1/32e) -1/16Ans:for xy^2 to be minimum, y^2 should be greater of the given options of y, and x should be lesser of the given options of x==> -1/2 * (-1/4)^2 = -1/32

When a four digit number is multiplied by N, the four digit number repeats itself to give an 8 digit number .If four digit number has all distinct digits then N is a multiple of ?a)11 b) 37 c)73 d) 27Ans:X * N = X * 10^4 + X

X * (N -1) = X * 10^4

N - 1 = 10000; N = 10001 = 73 * 137

so the answer is 73.

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Find the greatest and least values of f(x) = x^4 +x^2+5 / (x^2+1)(x^2+1)for real x.Ans:x^4 + x^2 + 5 = x^4 + 2x^2 + 1 - ( x^2 - 4) = (x^2+1)^2 - (x^2-4)

Y = x^4+x^2+5 / (x^2+1)^2 = 1 - (x^2-4) / (x^2+1)^2 = 1 - A

A= (x^2-4) / (x^2+1)^2 , since x^2 is positive , this is minimum when x = 0 and A = -4

A= (x^2-4) / (x^2+1)^2 = (x^2 - 4)/ (x^2 - 4 + 5 )^2, this is maximum when x^2 – 4 = 5 or x^2 = 9 and A = 0.05

Therefore the max and minimum values are 1+4 (=5) and 1-0.05 (=0.95).

A positive three-digit number X is such, when divided in two unequal three-digit numbers; the larger part is the arithmetic mean of X and the smaller part. How many values can X take?Ans:Let us say the number is x and it is divided into two parts a and b.

x = a + b and b = (x + a)/2 or x = 3a , b = 2a

So we need to divide x into 1:2 ratio so that a and b are also three digit numbers. This puts lower a limit on x to be 300 as we cannot divide any number less than 300 into 1:2 ratio without one of the numbers being less than 100.

Now the question is how many values that x cane take between 300 and 999 (inclusive of both) and divisible by 3.

Since between 301-997 there are 232 numbers divisible by 3 and both 300, 999 being divisible by 3, the total numbers are 234.

The number 444444.... (999 times) is definitely divisible by:a) 22 b) 44 c) 222 d) 444Ans:Let us take 444444444 (9 times)

This same as 444000000 + 444000 + 444 = 444 * 10^6 + 444* 10^3 + 444

Each of the three terms are divisible by 444.

Now extend this for 999 times.

OR,take 44,if we divide 44444............(999 times) by 44in 44 there are two digit means it will divide 44444.......(998 times) as 998/2=499 so will be left with 1 4.so not divisible.Take 444,444 has 3 digits, so 999/3=333 means it will be divisible by 444.Take 22, again we have 2 digits and 44/22=2, 4444/22= 202, 444444/22=2020 so if there are even no. of digits in 44444...... then only it will be divisible by 22. But there are odd number of digits i.e. 999 so not divisible.

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Take 222,3 digits, means 444/222=2, 444444/222 = 2002, 444444444/222 =2002002,3,6,9,12,------999=3+(n-1)3=3+3n-3n=333means it will also be divisible by 222

Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?

abdc+ dbca

(A) 3689(B) 6887(C) 8581(D) 9459(E) 16091Ans:Sum: 1000(a+d) + 100(b+b) + 10(d+c) + (a+c)From given, 10(d+c)+(a+c) < 100; that is, no carryover to the hundredth digit. Therefore, hundredth digit has to be even (see bold).

C is the Answer.

At a book store " MS BOOK STORE" the name is flashed using Neon lights. The words are individually flashed at intrevals of 2 4 and 5 seconds respectively and each word is put off after a second. What is the least time at which the full name of the book store can be read again?Ans:We need to take LCM of 3,5,6 as they represent net cycle time for each of the lights to complete one full ON/OFF operation.

For instance the MS will be lighted 2 seconds and be in ON position for 1 sec. This process repeats every three seconds.

Therefore MS will be lighted at 2,5,8,11,14,17,20,23,26,29 seconds.BOOK will be lighted at 4,9,14,19,24,29 seconds.and STORE will be lighted at 5,11,17,23,29 seconds.

This is nothing but finding LCM of 3,5,6 instead of 2,4,5.

How many integral solution(x,y) exist satisfying the equation |x|+|y|<=4?Ans:x = 0y = 0,1,2,3,4 and negative of all those no.==> 9 (x,y) setsx=1y = 0,1,2,3 and negative of those no.==> 7 setsx=2y=0,1,2 and negative==> 5 setsx=3

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y=0,1 and negative==> 3 setsx=4y=0==> 1setTotal = 25 setsAnother 16 sets for x=-1,-2,-3 and -4.Total 41

Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?Ans:5a + 10b = 460 .. least value of a such that a>blet a =b==> a = 30.66so the least value of a will be > 30 = 32(as a should be even)

abc is a three digit natural number so that abc=a!+b!+c!.What is thevalue of (b+c)^a?Ans:abc=100a+10b+c=a!+b!+c!now,6!=6*5*4*3*2*1=7207!=5040If 7 is one of the digits, then the sum of the factorials become a 4 digit number.hence the numbers 7,8,9 can be neglected.Considetr 6!=720, but number 7 cannot be there in hundred's place.henece we can neglect 6 also.Now,!5=120,!4=24,!3=6,!2=2,!1=1 &0!=1To get a 3digit number, 5 has to be present in the number.But 5 cannot be in hundred's place as then the number becomes greater than 500 which cannot be obtained as the sum of factorials.Max. possible number is 5!+4!+3!=120+24+6Also , a cannot be 0 as it is a 3 digit number. Hence a=1Then different Possible cases are,154,153,152,125,135,145.only 145 satisfies the following condition as,145=100+40+5=!1+4!+5!=1+24+120=145so,(b+c)^a=(4+5)^1=9

What is the remainder of (6^83 + 8^83) / 49?Ans:(x+y)^n = x^n + n x^n-1 y + n(n-1) /2! x^n-1 y^2 + ...... n! / (n-1)!*x^1*y^n-1 + n!/n! * x^0 * y^n

(7+1) ^ 83 = lots of terms divisible by 49 except last two + 83 * 7 + 1(7-1) ^ 83 = lots of terms divisible by 49 except last two + 83 * 7 - 1

Remainder is (83*7*2)/49 = (83*2)/7 = 166/7 = 23 5/7 ==> remainder above 7 is 5 or above 49 is 35.

When n is divided by 9, remainder is 7.. The remainder when n is divided by 3Ans:

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answer for the first one is : 1

e.g: 27 + 7 = 34

rem (34/9)= 7rem(34/3)=1

what is the remainder when 6^3 is divided by 8 ?Ans:6^3=(2*3)^3

now it implies: 2^3 * 3^3

clearly u have 8 *27 .

so 8 is a factor. thus no remainder.

what if it asked for the remainder when you divide 6^5 by 5?Ans:remainder is 1 since any power of 6 ends on 6

On the number line above, the segment from 0 to 1 has been divided into fifths, as indicated by the large tick marks, and also into sevenths, as indicated by the small tick marks. What is the LEAST possible distance between any two of the tick marks?Ans:Creating a scaled up number line, lets expand the number line such that 1 maps to 35 (L.C.M of 7, 5 ) ..0 on the new scale maps 0 on the old scale.35 on the new scale maps to 1 on the old scale

So the division marks for 5 on the new scale are at 5 , 10 , 15 , 20 , 25 , 30, 35

And the division marks for 5 on the new scale are at

7 , 14 , 21 , 28 , 35 ..

Smallest dist between the markings on the new scale between 15 and 14 = 1 ..Now, 35 on the new scale maps to 1 on the old scaleHence the diff of 1 unit on the new scale corresponds to 1/35 on the original scale

What is the smallest possible common multiple of two integers, which are both greater than 250. 1) 2512) 2523) 5024) 7505) 884Ans:252 = 2 * 126502 = 2 * 251

answer – 502

The product of two numbers is 16200. If their LCM is 216, find their HCF.a)75b)70

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c)80d) Data inconsistentAns:Given a correct LCM and HCF of two numbers, it is true the product of the numbers is same as the product of LCM and HCF.

However, the LCM and the product of the two numbers given in this problem are inconsistent. As you rightly said, the problem leads us to believe HCF of 75. That is where the incosistency lies.

By the basic definition of HCF and LCM, an LCM should always be divisible by HCF. Here 216 cannot be divided by 75.How many zeroes will be there at the end of the product(2!)^(2!) * (4!)^(4!)*(6!)^(6!)*(8!)^(8!)*(10!)^(10!)a)10!+6!b)2(10!)c)10!+8!+6!d)6!+8!+2(10!)Ans:If a number 'x!' in the series contains a number that is divisible by 5 , it contributes x! zeros

So in this series 6! has one 5, 8! has one 5, and 10! has 5 & 10.

Therfore the number of zeros = 6! + 8! + 10! + 10!

If f(x)=sum of all the digits of x, where x is a natural number, then what is the value of f(101)+f(102)+f(103)+------+f(200)a)1000b)784c)999d)1001e)865Ans:In unit's place we will have all the numbers from 0 to 9 ten times each.In tens place again we will have all the the numbers from 0 to 9.In hundred's place we will have 99 one's and 1 two'sSo,sum= 10(0+1+2+3+4+5+6+7+8+9)+10(0+1+2+3+4+5+6+7+8+9) +99*1+1*2= 2*450+101=1001

If n is a positive integer, what is the remainder when [7^(4n + 3)](6^n) is divided by 10 ?Ans:equation can be written as

7^4n * 7^3 * 6^n

the units digit of 7^4n, for all values of n, is 1the units digit of 7^3 is 3the units digit of 6^n, for all values of n, is 6

so the final units digit will be 8 since 6*3 = 18.

so the remainder would be 8.

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What's the sum of all 5-digit numbers formed using the digits 0,2,3,4,5?(assuming repetitions are allowed)?Ans:625(2+3+4+5)(10000)+500(2+3+4+5)(1111)

Mason and Kathy, who work in the evening wish to arrange for an evening off together. Each evening that Mason is off is followed by 3 evenings that he is at work, and each evening that Kathy is off is followed by 5 evenings that she is at work. If Mason is off this evening and Kathy will be off tomorrow evening, how many evenings must pass before they have an evening off together?Ans:Well, pattern for Mason is 1+4n (1,5,9 and so on)pattern for Katty is 2+6m (2, 8, 14 and so on)1+4n=2+6m => n=(1+6m)/4 This equation does not have any solutions for integer n&m.

N is a positive integer and N < 10000. How many prime factors does N have?Ans:while multiplying the primes like 2*3*5*7*11 we get 2310. The next prime is 13, if we multiply 2310 by 13 it will exceed 10000, which is a violation. Max 5

If n is a positive integer, Which of the following cannot be a factor of 3n + 4 ?. a) 4b) 5c) 67) 78) 8Ans:3n + 4= 3(n + 1) + 1= 1 mod 3

In other words, 3n + 4 will never be a multiple of 3, since there will always be a remainder of exactly 1.

If 3n + 4 is never a multiple of 3, it will also never be a multiple of 6 (since 6 = 2 x 3).

If it's not a multiple of 6, then 6 can't be a factor of it, thus C.

A restaurant pays a seafood distributor d dollars for 6 pounds of Maine lobster. Each pound can make v vats of lobster bisque, and each vat makes b bowls of lobster bisque. If the cost of the lobster per bowl is an integer, and if v and b are different prime integers, then which of the following is the smallest possible value of d?(A) 15(B) 24(C) 36(D) 54(E) 90Ans:d dollars will pay for 6 * v * b (bowls)

Using the lowest prime numbers for v and b (must be different)d = 6 * 2 * 3 = 36

Among 1 to 2000,wat r the nos that are perfect square and a perfect cube?

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Ans:Need integers 'n' of the form 1 <= (n^3)^2 <= 2000=> 1 <= n <= 3i.e the possible numbers are 1, 64 and 729

If n is a positive number less than 200 and 14n / 60 is an integer, then n has how many different positive prime numbers ?(A) 2 (B) 3 (C ) 5 (D) 6 (E) 8Ans:So 'n' can have values 30,60,90,180.30 = 2*3*560=2*2*3*590=3*2*3*5120=2*2*2*3*5180=2*3*2*3*5

hence 3

In the correctly worked mult. table below, each letter represents a distinct digit from 0 through 9.what digit must D represent?

ADAA-----BABA

02459Ans:505* 5------2525

D = 0

What is the greatest possible common divisor of two different positive integers which are less than 144?

a) 143b) 142c) 72d) 71e) 12Ans:The greatest divisible number less than 144 is 142. Half of that number is 71. So if our two numbers are 142 and 71, then their greatest common divisor is 71.

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