# Divisions and RemaindersMathematics SSC Railways

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Question1) The five-digits number 32516 is not divisible by 3 and the remainder will be 2 ?

Solution :-

We know that, if the sum of the digits form a number and that number is divisible by 3 then the remainder will be 0. So,

The sum of the digits=(3+2+5+1+6)

=17

17 is not divisible by 3.

Yes, the five-digits number 32516 is not divisible by 3 and the remainder will be 2.

Option- ['True', 'False']

Answer: True

Explaination: True

The five-digits number 32516 is not divisible by 3 and the remainder will be 2 ?

Solution :-

We know that, if the sum of the digits form a number and that number is divisible by 3 then the remainder will be 0. So,

The sum of the digits=(3+2+5+1+6)

=17

17 is not divisible by 3.

Yes, the five-digits number 32516 is not divisible by 3 and the remainder will be 2.

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Question2) The five-digits number 32514 is divisible by 3 and the remainder will be 0 ?

Solution :-

We know that, if the sum of the digits form a number and that number is divisible by 3 then the remainder will be 0. So,

The sum of the digits=(3+2+5+1+4)

=15

15 is divisible by 3.

Yes, the five-digits number 32514 is divisible by 3 and the remainder will be 0.

Option- ['True', 'False']

Answer: True

Explaination: True

The five-digits number 32514 is divisible by 3 and the remainder will be 0 ?

Solution :-

The sum of the digits=(3+2+5+1+4)

=15

15 is divisible by 3.

Yes, the five-digits number 32514 is divisible by 3 and the remainder will be 0.

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Question3) The six-digits number 325138 is divisible by 11 and the remainder will be 0 ?

Solution :-

We know that, if the difference between sum of the digits at odd places and sum of the digits at even places form a number and that number is divisible by 11 then the remainder will be 0. So,

(Sum of its digits at odd places)-(Sum of its digits at even places)=(8+1+2)-(3+5+3)=11-11=0

Yes, the six-digits number 325138 is divisible by 11 and the remainder will be 0.

Option- ['True', 'False']

Answer: True

Explaination: True

The six-digits number 325138 is divisible by 11 and the remainder will be 0 ?

Solution :-

We know that, if the difference between sum of the digits at odd places and sum of the digits at even places form a number and that number is divisible by 11 then the remainder will be 0. So,

(Sum of its digits at odd places)-(Sum of its digits at even places)=(8+1+2)-(3+5+3)=11-11=0

Yes, the six-digits number 325138 is divisible by 11 and the remainder will be 0.

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Question4) The five-digits number 32516 is not divisible by 9 and the remainder will be 8 ?

Solution :-

We know that, if the sum of the digits form a number and that number is divisible by 9 then the remainder will be 0. So,

The sum of the digits=(3+2+5+1+6)

=17

17 is not divisible by 9.

Yes, the five-digits number 32516 is not divisible by 9 and the remainder will be 8.

Option- ['True', 'False']

Answer: True

Explaination: True

The five-digits number 32516 is not divisible by 9 and the remainder will be 8 ?

Solution :-

We know that, if the sum of the digits form a number and that number is divisible by 9 then the remainder will be 0. So,

The sum of the digits=(3+2+5+1+6)

=17

17 is not divisible by 9.

Yes, the five-digits number 32516 is not divisible by 9 and the remainder will be 8.

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Question5) The five-digits number 32517 is divisible by 9 and the remainder will be 0 ?

Solution :-

We know that, if the sum of the digits form a number and that number is divisible by 9 then the remainder will be 0. So,

The sum of the digits=(3+2+5+1+7)

=18

18 is divisible by 9.

Yes, the five-digits number 32517 is divisible by 9 and the remainder will be 0.

Option- ['True', 'False']

Answer: True

Explaination: True

The five-digits number 32517 is divisible by 9 and the remainder will be 0 ?

Solution :-

The sum of the digits=(3+2+5+1+7)

=18

18 is divisible by 9.

Yes, the five-digits number 32517 is divisible by 9 and the remainder will be 0.

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Question6) The three-digits number 325 is divisible by 5 and the remainder will be 2 ?

Solution :-

We know that, the last digits of a number are 0 and 5 then the number is divisible by 5 then the remainder will be 0. So,

325 is divisible by 5 because of a number is 5 and the remainder is 0.

The three-digits number 325 is divisible by 5 but the remainder will not be 2.

Option- ['True', 'False']

Answer: False

Explaination: False

The three-digits number 325 is divisible by 5 and the remainder will be 2 ?

Solution :-

We know that, the last digits of a number are 0 and 5 then the number is divisible by 5 then the remainder will be 0. So,

325 is divisible by 5 because of a number is 5 and the remainder is 0.

The three-digits number 325 is divisible by 5 but the remainder will not be 2.

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Question7) The three-digits number 324 is divisible by 2 and the remainder will be 0 ?

Solution :-

We know that, the last digits of a number are even numbers is divisible by 2 then the remainder will be 0. So,

324 is divisible by 2 because the last digit of a number is 4.

The three-digits number 324 is divisible by 2 and the remainder will be 0.

Option- ['True', 'False']

Answer: True

Explaination: True

The three-digits number 324 is divisible by 2 and the remainder will be 0 ?

Solution :-

We know that, the last digits of a number are even numbers is divisible by 2 then the remainder will be 0. So,

324 is divisible by 2 because the last digit of a number is 4.

The three-digits number 324 is divisible by 2 and the remainder will be 0.

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