Question

what least number must be added to 1056 to get a number exactly divisible by 23.​

Answer 1

Answer:

⇒ When 1056÷23, we get 21 as a remainder.

⇒ So, 23−21=2 must be added to 1056 in order to sum completely divisible by 23.

∴ 1056+2=1058

∴ The least number should be added to 1056, so that the sum is completely divisible by 23 is 2.

SIMPLE!!!

Answer 2

Hey !

For the problem of the form:

Finding the least number that should be added to a number N so that it is exactly divisible by d,

Follow the below mentioned steps:

Here, note that to make N exactly divisible by d we have to add the least possible number 'x'.

=> (N+x) is exactly divisible by d

=> There exists remainder of N÷d which is non-zero, say it as r

=> To get the No. x ; x = d-r.

Two step process :

- Divide N by d to find the remainder 'r'

- Subtract the remainder 'r' from 'd'

Now,

Step-1:

Rem(1056 ÷ 23) = 21

(Refer attachment for division)

Step-2:

Required least No. = 23 - 21 = 2

•°• 1056 + 2 = 1058, which is exactly divisible by 23.

(46 times 23 is 1058)

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