# When a smallest possible number is successively divided by 4 and 6 the

| When a smallest possible number is successively divided by 4 and 6 the respective remainders are 2 and 5. If the number is successively divided by 5 and 3, the respective remainder will be:

A. 0 and 1

B. 2 and 1

C. 1 and 0

D. 3 and 1

### Right Answer is: B

#### SOLUTION

Successive division means on division by 4, the quotient we get is further divided by 6

Let the number be n

N = 4q+2 [q is the quotient we get on dividing N by 4]

Here q = 6q’ + 5 [q’ is the quotient we get on dividing q by 6]

Thus N = 4(6q’+5) + 2

For the number to be smallest, we take the value of q’ = 0

N = 4(0+5) +2 = 22

Now on successively dividing N by 5 and 3, we get the following remainders;

22/5 => 2 and quotient = 4

4/3 => 1

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When a smallest possible number is successively divided by 4 and 6 the