CAT Remainders Questions & Solutions

A sample of real CAT Remainders past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 26 Remainders questions in all — sign up free to practise them timed.

1. Q1.CAT 2024

   When $3^{333}$ is divided by 11, the remainder is

   • 5

   • 10

   • 1

   • 6

   Show solution

   There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator. 

   Noting down the powers of 3, 3, 9, 27, 81, 243

   243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11. 

   243 is 3 raised to power 5; we can rewrite the given term as  $\dfrac{3^{330}\times\ 3^3}{11}$
   The overall remainder will be  $\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

   $\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$
   $\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$
   $1^{66}\times\ \left[\dfrac{27}{11}\right]_R$
   $1\times\ 5$
   $5$

   Therefore, Option A is the correct answer. 

2. Q2.CAT 2024

   If $10^{68}$ is divided by 13, the remainder is

   • 5

   • 8

   • 9

   • 4

   Show solution

   There are multiple ways of solving such questions involving remainders; one easy way is to look for a power of numerator that leaves a remainder of 1 or 01 when divided by the denominator. 

   In this instance, 1000, when divided by 13, leaves a remainder of -1

   We can rewrite the numerator as  $\frac{10^{66}\times\ 100}{13}$

   The remainder would be  $\left[\frac{10^{66}}{13}\right]_R\times\ \left[\frac{100}{13}\right]_R$
   $\left(-1\right)^{22}\times\ 9$
   9

   Therefore, Option C is the correct answer. 

3. Q3.CAT 2018

   The number of integers x such that $0.25 \leq 2^x \leq 200$ and $2^x + 2$ is perfectly divisible by either 3 or 4, is

   Answer: 5

   Show solution

   At $x = 0, 2^x = 1$ which is in the given range [0.25, 200]

   $2^x + 2$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution. 
   

   At $x = 1, 2^x = 2$ which is in the given range [0.25, 200]

   $2^x + 2$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution. 

   At $x = 2, 2^x = 4$ which is in the given range [0.25, 200]

   $2^x + 2$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution. 

   At $x = 3, 2^x = 8$ which is in the given range [0.25, 200]

   $2^x + 2$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution. 

   At $x = 4, 2^x = 16$ which is in the given range [0.25, 200]

   $2^x + 2$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution. 

   At $x = 5, 2^x = 32$ which is in the given range [0.25, 200]

   $2^x + 2$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution. 

   At $x = 6, 2^x = 64$ which is in the given range [0.25, 200]

   $2^x + 2$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution. 

   At $x = 7, 2^x = 128$ which is in the given range [0.25, 200]

   $2^x + 2$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution. 

   At $x = 8, 2^x = 256$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.

   Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values. 

4. Q4.CAT 2005

   Let $n!=1*2*3* ...*n$ for integer $n \geq 1$.

   If $p = 1!+(2*2!)+(3*3!)+... +(10*10!)$, then $p+2$ when divided by 11! leaves a remainder of

   • 10

   • 0

   • 7

   • 1

   Show solution

   According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!.  So p = 11! - 1! = 11! - 1.  p + 2 = 11! + 1 .So when it is  divided by 11! gives a remainder of 1. Hence, option 4.

5. Q5.CAT 2003

   How many even integers n, where $100 \leq n \leq 200$ , are divisible neither by seven nor by nine?

   • 40

   • 37

   • 39

   • 38

   Show solution

   Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 - (7+6-1) = 39 

   There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

6. Q6.CAT 2002

   When $2^{256}$ is divided by 17, the remainder would be

   • 1

   • 16

   • 14

   • None of these

   Show solution

   $2^4 = 16 = -1$ (mod $17$ )
   So, $2^{256} = (-1)^{64}$ (mod $17$ )
   $= 1$ (mod $17$ )
   Hence, the answer is 1. Option a).

7. Q7.CAT 2001

   Let $b$ be a positive integer and $a = b^2 - b$. If $b \geq 4$ , then $a^2 - 2a$ is divisible by

   • 15

   • 20

   • 24

   • All of these

   Show solution

   
   We know that a= $b^2-b$ .

   So $a^2-a$ = b( $b^3-2b^2-b+2$ ) . = (b - 2)(b - 1)( b)(b + 1)

   The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)

   In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).

   Thus, for any value of b >=4, $a^2-4$ would be divisible by 3 x 2 x 4 = 24.

   Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)

8. Q8.CAT 2000

   Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?

   • 0

   • 9

   • 3

   • 6

   Show solution

   The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

   5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

9. Q9.CAT 2000

   Let N = $55^3 + 17^3 - 72^3$. N is divisible by:

   • both 7 and 13

   • both 3 and 13

   • both 17 and 7

   • both 3 and 17

   Show solution

   $55^3 + 17^3 - 72^3$ = $(55-72)k + 17^3$ . This is divisible by 17

   Remainder when $55^3$ is divided by 3 = 1

   Remainder when $17^3$ is divided by 3 = -1

   Remainder when $72^3$ is divided by 3 = 0

   So,  $55^3 + 17^3 - 72^3$ is divisible by 3

   So, the answer is d) 3 and 17

10. Q10.CAT 1998

    A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29.

    • 5

    • 4

    • 1

    • Cannot be determined

    Show solution

    Let's say N is our number
    N = (899K + 63) or N = ( $29 \times 31$ K) + 63
    So when it is divided by 29, remainder will be $\frac{63}{29}$ = 5

11. Q11.CAT 1997

    If m and n are integers divisible by 5, which of the following is not necessarily true?

    • m - n is divisible by 5

    • $m^2$ - $n^2$ is divisible by 25

    • m + n is divisible by 10

    • None of these

    Show solution

    Let's say m=5k and n=5t
    So m-n = 5(k-t) will be divisible by 5.
    $m^2 - n^2 = 25(k^2 - t^2)$ will be divisible by 5.
    $m+n = 5(k+t)$ will be divisible by 5 but not necessarily with 10.

12. Q12.CAT 1991

    Let k be a positive integer such that k+4 is divisible by 7. Then the smallest positive integer n, greater than 2, such that k+2n is divisible by 7 equals

    • 9

    • 7

    • 5

    • 3

    Show solution

    let's say k+4 = 7m
    k = 7m-4
    Now for k+2n or 7m+(2n-4) is also  multiple of 7.
    or 2n-4 should be a multiple of 7 
    So 2n-4 = 7p
    or 2n = 7p+4
    For p=2; n=9 (p cannot be 1 as n is an integer )

CAT Remainders previous year questions with solutions — AthenaPrep