What least number must be subtracted from 1936 so that the remainder when divided by 19 15 will leave in each case the same remainder 7?

1. 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