what are the three consecutive lowest numbers that are divisible by cubes of 5 , 2 and 3 ?
Answers
Given : numbers that are divisible by cubes of 5 , 2 and 3
To Find : three consecutive lowest numbers
Solution:
2 ,3 and 5 are prime numbers
Hence their cubes would be co prime with each other
2³ = 8
3³ = 27
5³ 125
Lowest number which will be divisible by these will be LCM of these numbers
LCM ( 8 , 27 , 125) = 8 * 27 * 125
= (30)³
= 27000
Lowest number = 27000
next two numbers can be 27000 * 2 and 27000 * 3
Hence 27000 , 54000 and 81000 are three consecutive lowest numbers that are divisible by cubes of 5 , 2 and 3
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