Question

Which least number must be subtracted from 99,99,999 so that it will come to the multiple of 125.
[a] 124
[b] 4
[c] 24
[d] None of these.

Answer 1

Hint: The least number to be subtracted from a number “b” such that it leaves it comes to be a multiple of another number “a” is the remainder of b obtained on division by a. So, find the remainder on dividing 9999999 on dividing by 125. The result is the answer.

Complete step-by-step answer:

We know the least number to be subtracted from a number “b” such that it leaves it comes to be a multiple of another number “a” is the remainder of b obtained on division by a.
Taking b = 99,99,999 and a = 125, we get
Since 1,00,00,000 is a multiple of 125, we get
1,00,00,000 = 125k, where k>1.
So we have 99,99,999 = 1,00,00,000-1 = 125k-1 = 125(k-1) +125-1 =125(k-1)+124
Hence the remainder obtained on dividing 99,99,9999 by 125 is 124.
Hence the least number to be subtracted from 99,99,999 so that it comes to be a multiple of 125 is 124.
Note: If b = aq+ r where $0\le rAlternatively, we can check each of the options and check which of the options is correct.
Option [a]:
We have 99,99,999-124 = 99,99,875
Now $9999875=125\times 79999$
Hence 9999875 is divisible by 125
Option [b]:
We have 9999999-4 = 9999995
Now, we have
$9999995=125\times 79999+120$
Hence 9999995 is not divisible by 5
Option [c]:
We have 9999999-24=9999975
Now, we have
$99999975=125\times 79999+100$
Hence, 9999975 is not divisible by 125
Hence of all the options subtracting option [a] leads to a number divisible by 125
Hence option [a] is correct.