Question

Find the smallest number which when divided by 24 and 36 leaves a remainder
9.

Answer 1

Answer:

Let's factorize, 

24 = 2x2x2x3

36 = 2x2x3x3

54 = 2x3x3x3

The smallest number divisible by all 24, 36, 54 is:

LCM(24, 36, 54) = 216

Thus, to get 5 as remainder, we need to add 5 to 216.

 (as 216 is the smallest number which gives remainder 0 when divided by 24 or 36 or 54)

216 + 5 = 221

For 12, 221 gives 18 as quotient and 5 as remainder

For 36, 221 gives 6 as quotient and 5 as remainder

For 54, 221 gives 4 as quotient and 5 as remainder

Hence, 221 is required number

I HOPE IT IS HELPFUL ❤