What is the largest number that must be substracted from 1160 so that the sum of 1160 and the resulting number is completely divisible by 19.

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Solution

Let the least number be n which need to be subtracted.

1160+1160-n this should be divisible by 19.

The remainder obtained on dividing 2320 by 19 is 2. [1160+1160=2320]

On subtracting 2320–2 you would get a number divisible by 19.

Therefore n is 2.

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What is the largest number that must be substracted from 1160 so that the sum of 1160 and the resulting number is completely divisible by 19.

Let the least number be n which need to be subtracted. 1160+1160-n this should be divisible by 19. The remainder obtained on dividing 2320 by 19 is 2. [1160+1160=2320] On subtracting 2320–2 you would get a number divisible by 19. Therefore n is 2.

Let the least number be n which need to be subtracted.

1160+1160-n this should be divisible by 19.

The remainder obtained on dividing 2320 by 19 is 2. [1160+1160=2320]

On subtracting 2320–2 you would get a number divisible by 19.

Therefore n is 2.