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

Hey

Here is your answer,

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.

Hope it helps you!