CAT Number Systems Questions & Solutions

A sample of real CAT Number Systems past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 137 Number Systems questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is

   Answer: 6

   Show solution

   According to question, in the given 3-digit number N, the digits are non-zero and distinct.

   So, the possible digits in 3-digit number =  $2,3,5,6,7,8$

   It is also given that only one of the digits is a prime number.

   So, the minimum possible value of N = 268

   (2 is the smallest prime digit, and the non-prime digits has to be 6 and 8)

   Now,  $268=4\times\ 67=2^2\times\ 67$

   So, the number of factors =  $\left(2+1\right)\left(1+1\right)=3\times\ 2=6$

2. Q2.CAT 2024

   The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is

   Answer: 70

   Show solution

   We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84

   Let us write $x+y+z=84$
   x, y, z being the three distinct integers in ascending order. 

   If the smallest number is increased by 7 and the largest number is reduced by 10
   $(x+7)+(y)+(z-10)=81$

   New arithmetic mean will be  $\frac{81}{3}=27$
   And this is said to be 2 more than the middle number, meaning
   $27-2=y=25$
   

   $x+z=59$

   We are given that difference between the largest and the smallest numbers becomes 64, 
   $(z-10)-(x+7)=64$
   $z-x=81$

   Adding the two equations we get,  $2z=140$
   $z=70$
   

3. Q3.CAT 2022

   For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is

   • 4

   • 7

   • 6

   • 5

   Show solution

   To find the largest possible value of n, we need to find the value of n such that n! is less than 15000.

   7! = 5040

   8! = 40320 > 15000

   This implies 15000! is not divisible by 40320!

   Therefore, maximum value n can take is 7.

   The answer is option B.

4. Q4.CAT 2020

   How many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ ?

   • 2018

   • 2019

   • 2017

   • 2020

   Show solution

   $ab\ =\ 4^{2017}=2^{4034}$

   The total number of factors = 4035.

   out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

   And since the given number is a perfect square we have one set of two equal factors.

   .'. many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ = 2018. 

5. Q5.CAT 2017

   If $a, b, c,$ and $d$ are integers such that $a+b+c+d=30$ then the minimum possible value of $(a - b)^{2} + (a - c)^{2} + (a - d)^{2}$  is
   

   Answer: 2

   Show solution

   For the value of given expression to be minimum, the values of $a, b, c$ and $d$ should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get
     $(8 - 8)^{2} + (8 - 7)^{2} + (8 - 7)^{2}$
   => 1 + 1 = 2

6. Q6.CAT 2005

   If R = $(30^{65}-29^{65})/(30^{64}+29^{64})$ ,then

   • 0&lt;R\leq0.1

   • 0.1&lt;R\leq0.5

   • 0.5&lt;R\leq1.0

   • R&gt;1.0

   Show solution

   
   $\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\frac{(30^{64}+30^{63}*29+....+29^{64})}{(30^{64}+29^{64})}$ , which is greater than 1 . Hence option D.

7. Q7.CAT 2003

   A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals

   • 31

   • 63

   • 75

   • 91

   Show solution

   Since  in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1. Hence option D.

8. Q8.CAT 2001

   Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?

   • $y(x-z)^2$ is even

   • $y^2(x-z)$ is odd

   • $y(x-z)$ is odd

   • $z(x-y)^2$ is even

   Show solution

   Take x=3 , z=2 , y=5.

   $y(x-z)^2 = 5(3-2)^2 = 5$

   Option A gives 5 which is odd.

9. Q9.CAT 2000

   Let S be the set of integers x such that:

   1) 100 <= x <= 200

   2) x is odd

   3) x is divisible by 3 but not by 7.

   How many elements does S contain?

   • 16

   • 12

   • 11

   • 13

   Show solution

   Number of multiples of 3 between 100 and 200 = 66 - 33 = 33
   Number of odd multiples = 16
   Number of odd multiples of 21 = 3 (105, 147, 189)
   So, the required number = 13

10. Q10.CAT 1999

    If n = 1 + x, where x is the product of four consecutive positive integers, then which of the following is/are true?

    A. n is odd

    B. n is prime

    C. n is a perfect square

    • A and C only

    • A and B only

    • A only

    • None of these

    Show solution

    Let the four consecutive positive integers be $a,a+1,a+2$ and $a+3$ .
    Therefore, $n=1+a(a+1)(a+2)(a+3)$
    Or, $n = 1+(a^2+3a)*(a^2+3a+2)$
    Or, $n = (a^2+3a)^2 + 2*(a^2+3a)+1 = (a^2+3a+1)^2$
    Hence, n is a perfect square and therefore not a prime.

    The product of four consecutive positive integers is always even. Hence, n is always odd.
    Therefore, from the given statements, only A and C are true.

11. Q11.CAT 1997

    If n is an integer, how many values of n will give an integral value of $\frac{(16n^2+ 7n+6)}{n}$ ?

    • 2

    • 3

    • 4

    • None of these

    Show solution

    Expression can be reduced to 16n + 7 + $\frac{6}{n}$
    Now to make above value  an integer n can be 1,2,3,6,-1,-2,-3,-6
    Hence answer will be D).

12. Q12.CAT 1991

    What is the greatest power of 5 which can divide 80! exactly?

    • 16

    • 20

    • 19

    • None of these

    Show solution

    The highest power of 5 in 80! = [80/5] + [ $80/5^2$ ] = 16 + 3 = 19

    So, the highest power of 5 which divides 80! exactly = 19

CAT Number Systems previous year questions with solutions — AthenaPrep