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What is the smallest positive integer n for which n^2 is divisible by 18 and n^3 is divisible by 640? 

 Sep 21, 2020
 #1
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a=1; c=a^2;b=a^3; if(c%18==0 and b%640==0, goto4, goto5);printc, a; a++;if(a<1000, goto1, 0)

 

The smallest n = 120

 Sep 22, 2020
 #2
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What is the smallest positive integer n for which n^2 is divisible by 18 and n^3 is divisible by 640? 

 

n^2 is divisible by 18

18=2*2*3

n must have factors 2 and 3

and

n^3 is divisible by 640

640=5*2^7

n must have factors  2*2*2*5

 

Put these together and n must have factors    2*2*2*3*5 = 120    That is the smallest possible value of n 

 Sep 26, 2020

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