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# Number System Questions Asked in CAT Exam:

We have given some important Number system questions asked in previous CAT exam question papers. All the questions are provided with detailed explanations and solutions.

CAT Number System questions set-2 PDF

Number System Set-3 PDF

This blog is the Set-4 in the number system questions

Question 1: A two digit number has different digits. For how many such two digit positive numbers is the difference between the number itself and the number formed on reversing its digits is the perfect square of an integer?

a) 20
b) 29
c) 38
d) None of these

Question 2: The number A39K2 is completely divisible by both 8 and 11. Here both A and K are single digit natural numbers. Which of the following is a possible value of A+K?

a) 8
b) 10
c) 12
d) 14

Question 3: What is the number of trailing zeroes in 1000! ?

a) 254
b) 250
c) 248
d) 249

Question 4: How many values of ‘p’ exist such that p, p+3, p+5 are prime numbers?

a) 0
b) 1
c) 2
d) more than 2

Question 5: X is the smallest number which leaves a remainder of 2 when divided by 7 and a remainder of 1 when divided by 19. What is the remainder when X is divided by 23?

a) 12
b) 9
c) 13
d) 17

Question 6: Two numbers are such that they are 30% and 44% less than a third number. The second number is what percent less than the first number?

a) 20%
b) 14%
c) 16%
d) 25%

Question 7: Find the sum of the last two digits of 1! + 4! + 9! + 16! + 25! . . . . . . . + 961!.

a) 3
b) 5
c) 6
d) 8

Question 8: What is the remainder when $$1.1!+2.2!+3.3!+4.4!+.…+11.11!$$ is divided by 12?

a) 0
b) 1
c) 10
d) 11

Question 9: If A=7!+ 8!+ 9!+ 10!+ 11!, which of the following is a factor of A?

a) 1001
b) 897
c) 1771
d) 2261

Question 10: The HCF of two numbers is 8 and the difference between the two numbers is also 8. The LCM of the two numbers is a 3 digit number. What is the product of the digits of the largest possible value of the smaller number?

a) 0
b) 72
c) 64
d) 80

Solutions for Number System Questions Asked in CAT Exam:

Solutions:

Let the two digit number be ‘xy’
The difference between the number and the number obtained when its digits are reversed is,
(10y-x)-(10x-y) = 9(y-x)
This must be a perfect square.
So it can only take the values of 9, 36 and 81.
It cannot take the value 0 because it is given that the digits of the two digit number are distinct.
If 9(y-x) = 9, then (y-x)=1. There will be 17 numbers such that either y-x = 1 or x-y = 1.
If 9(y-x) = 36, then (y-x)=4. There will be 11 numbers such that either y-x = 4 or x-y = 4.
If 9(y-x) = 81, then (y-x)=9. There will be only 1 number such that x-y = 9, that is 90.
So the total number of numbers possible are 17+11+1 = 29.

The number is divisible by 11, so the difference between the sum of the digits at the odd places and the digits at the even places is either 0 or a multiple of 11.
Let the difference be a 0, so
11 + A = 3 + K
=> K – A = 8, the only possible value is 9,1
Now we have to check if it satisfies the divisibility by 8 test.
K= 9 makes the last 3 digits 992. This is divisible by 8.
Let’s check for other cases when the difference is 11
11 + A – 3 – K = 11
=> A – K = 3
The possible values in this case are (9,6), ( 8,5), (7,4), (6,3), (5,2), (4,1)
Among these cases only (8,5) and (4,1) will be divisible by 8. So the possible values of sum are
13, 5 and 10
Now difference between the sum of odd and even places cannot be 22
11 + A – 3 – K = 22
=> A – K = 14
Since both A and K are single digit natural numbers, this is not possible.
Thus the only possible values of sum are 5, 10 and 13.
In the given options only 10 is there. So it is the correct choice.

Number of trailing zeroes in 1000 factorial would be
[1000/5] + [1000/25] + [1000/125] + [1000/625]
=> 200 + 40 + 8 + 1 = 249

If ‘p’ is odd, then p+3 and p+5 will be even. There is only one even prime which is possible i.e 2. Thus ‘p’ cannot be odd.
Now if ‘p’ is even, the only possible value of p can be 2 (Since 2 is the only even prime number)
So if p = 2, we have p+3 = 5 and p+5 = 7
Thus, it satisfies the given conditions. Hence there is only one possible value of ‘p’

X on division by 7 leaves a remainder of 2. So X is of the form 7k+2
Same number on division by 19 leaves a remainder of 1, So X is also of the form 19m + 1
Thus
7k+2 = 19m + 1
If we put m = 3, we get a number which is of the form 7k+2. Thus the smallest number which fits the given criteria is 19*3+1 =58
When 58 is divided by 23, the remainder would be 12.

Let the third number be X
Then the first number would be .7X and the second number would be .56X
Therefore, the second number is (.14x/.70x)*100 = 20% less than the first number.

Last two digits are nothing but remainder when the number is divided by 100.
All the terms after 10! will be divisible by 100. So we need to find the remainder of first 3 terms with 100.
So required remainder is
1 + 24 + 80 = 105
=> 105 mod 100 = 05
So last two digits are 05.
Hence the sum is 5.

Let x = 1.1!+2.2!+3.3!+4.4!+.…+11.11!
Adding (1!+2!+3!+…+11!) on both sides we get,
x + 1!+2!+3!+…+11! = 1.1!+2.2!+3.3!+4.4!+.…+11.11! + 1!+2!+3!+…+11!
x + 1!+2!+3!+…+11! = 2.1!+3.2!+4.3!+….+12.11!
x + 1!+2!+3!+…+11! = 2!+3!+4!+…+12!
x = 12! – 1
Thus the remainder when x is divided by 12 is 11.

=> $$7 \times 720 \times 8721$$
=> $$3^{3} \times 720\times 7 \times 19\times17$$
=> $$3^{3}\times 720 \times 2261$$