Question

find the greatest number of 3 digits which when divided by 20,24 and 45 leaves a remainder 9 in each case.

Answer 1

Given numbers,
20 , 24 and 45
Prime factorization of
20=2^2×5

24=2^3×3

45=3^2×5

LCM=product of each prime factor of highest power
LCM=2^3×3^2×5=360

Greatest four digit number=9999
greatest four digit number divisible by all given numbers=9999-remainder when 9999 is divided by LCM of given numbers
greatest four digit number divisible by given numbers=9999-279=9720
Given that,
required number when divided by 20 , 24, 45 leaves remainder 18.
Therefore,required number=9720+18=9738

Hence greatest four digit number when divided by 20,24 and 45 is 9738.

Answer 2

Given numbers,
20 , 24 and 45
Prime factorization of
20=2²×5

24=2³×3

45=3²×5

LCM=product of each prime factor of highest power
LCM=2³×3²×5=360

largest 3 digit number = 999
largest 3 digit number which divisible by 20,24 and 45 = 720
for remainder 720+9 = 729