What least number must be subtracted from 1936 so that the remainder when divided by 9 10 and 15 will leave in each case the same remainder 7?

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What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?

LCM of 9, 10 and 15 = 90⇒ The multiple of 90 are also divisible by 9, 10 or 15.∴ 21 × 90 = 1890 will be divisible by them.∴ Now, 1897 will be the number that will give remainder 7.

∴ Required number = 1936 – 1897 = 39

The answer for this question is 39.

Explanation :

Step 1 - Take LCM (9,10,15) = 90

Step 2 - Divide 1936 by 90.

1936÷90 = 21 (quotient) and 46(remainder)

Step 3 - To get 7 as remainder you must subtract 39.

As 46–39 = 7

Result : 1936–39 = 1897

1897÷9 = 210(quotient) + 7(remainder)

1897÷10 = 189(quotient) + 7(remainder)

1897÷15 = 126(quotient) + 7(remainder