Divisibility Test - MCQ Solved MCQs UPSC Prelims Paper 2 CSAT

Answer : B

Explanation:A number is exactly divisible by 8, then the last 3 digits of the numbers must be divisible by 83

Here the last 3 digits are 8x4.

Put each value in given options in the place of x and check whether it is divisible by 8 or not .

 Option b ,  824 which is exactly divisible by 8. 

So answer is option b.

Answer : A

Explanation:

if a number is divisible by 9, the sum of its digits must be a multiple of 9.

Here, 3+7+*+1+2+4=17+*

Here the value of * must be 1 because the next multiple of 9 is 18.

Answer: C  

Explanation:

 For a number to be divisible by 88, the number must be divisible by 8 and 11.

Write  88 as product of two factors :  22 ,2

                                                   11 ,8

                                                   44,2

                Of these pairs , 11 and 8 are co primes. 
   So the number must be divisible by 8 and 11.

Answer : B

Explanation:

For divisibility by 11, the difference of sums of digits at even and odd places must be either zero or divisible by 11.

For 8261955, Difference =(8+6+9+5) -(2+1+5)=28-8=20.

The units digit is at odd place. So we add 2 to the number  

 => 8261955 +2 = 8261957

Now , (8+6+9+7) -(2+1+5)=30-8=22  => 22 is a multiple of 11 and hence 8261957 is also divisible by 11.

Answer : A

Explanation:

Divisibility by 6 requires that the number be divisible by 2 as well as 3 , i.e, the following 2 conditions must be met

i) Unit digit be Zero or even

ii) Sum of digits be divisible by 3

The given number is 9724*

Sum of the digits =9 +7 +2 +4 +* =22+*

The digit which on being added to 22 will give the sum divisible by 3 are 

22+2 =24 and 22 +5=7.

2 and 7 satisfy the condition ii.

The blank space is at unit's place. So the missing digit must satisfy the condition (i) also

      Out of 2 and 5, only 2 is even.