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
Learn More:
1. Check the divisibility of the following numbers by 2, 3, 9 and 11 a ...
brainly.in/question/12343368
If a 9 digit number 32x4115y2 is divisible by 88, then the value of (4x ...
brainly.in/question/13567692
If a six digit number 93p25q is divisible by 88 then find the value of p ...
brainly.in/question/3788301
If the 8-digit number 2074x4y2 is divisible by 88, then the value of ...
brainly.in/question/12253210