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