The least number that must be subtracted from 13601 so that the result is exactly divisible by 87 is 29.

Step-by-step explanation

We want a number that is exactly divisible by 87 , so we first divide 13601 by 87 and look at the remainder.

  • Do the division:
    13601÷8713601\div 8713601÷87 gives:

    • Quotient = 156
    • Remainder = 29

This means:

13601=87×156+2913601=87\times 156+2913601=87×156+29

To make 13601 exactly divisible by 87, we must remove the remainder.
So, subtract 29 from 13601:

13601−29=1357213601-29=1357213601−29=13572

Now check:

13572÷87=156with remainder 013572\div 87=156\quad \text{with remainder }013572÷87=156with remainder 0

So:

  • The new number 13572 is exactly divisible by 87.
  • Therefore, the least number that must be subtracted is 29.

Quick Scoop (forum-style insight)

Many exam and practice sites frame this exact question: “What least number must be subtracted from 13601 so that the remainder is divisible by 87?”
The consistent worked-out solution is that you divide 13601 by 87, note the remainder 29, and that remainder itself is the smallest number you need to subtract.

Key points:

  • Original number: 13601
  • Divisor: 87
  • Remainder on division: 29
  • Subtract this remainder to get a clean multiple of 87.

So the final answer is: 29 TL;DR:
Divide 13601 by 87 → remainder 29 → subtract 29 to make it exactly divisible → answer = 29.

Information gathered from public forums or data available on the internet and portrayed here.