perplexus.info :: Numbers : Perfectly divisible

Perfectly divisible (Posted on 2024-11-19)

Let n be the smallest integer that satisfies the following conditions:

n/2 is a perfect square.
n/3 is a perfect cube.
n/5 is a perfect fifth.

How many divisors does n have that are not multiples of 10?

No Solution Yet Submitted by Danish Ahmed Khan

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Solution

Comment 1 of 1

The number should be of the form (2^a)(3^b)(5^c). The conditions require

a-1, b, c all be multiples of 2.

a, b-1, c all be multiples of 3.

a, b, c-1 all be multiples of 5.

From which a=15, b=10, c=6 are the smallest positive a,b,c.

Ordinarily we'd want each of 16 possible a, 11 possible b, 7 possible c. But since multiples of 10 are not allowed, we need a=0 if c>0 and c=0 if a>0. There 16+7-1=22 ways of doing this.

The final answer is thus 11*22 = 242.

Posted by Jer on 2024-11-19 10:35:08

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