# Math Problem

**URL:** https://boards.straightdope.com/t/math-problem/505332
**Category:** Factual Questions
**Created:** [August 4, 2009, 8:15pm UTC](https://boards.straightdope.com/t/math-problem/505332 "2009-08-04T20:15:52Z")
**Posts on this page:** 6
**Page:** 1

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### Author: ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)
#### Post date: [August 4, 2009, 8:15pm UTC](https://boards.straightdope.com/t/math-problem/505332/1 "2009-08-04T20:15:52Z")

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What is the easiest way to solve a problem like this:

k is a positive integer and 225 and 216 are both divisors of k.

If k=2^a x 3^b x 5^c, where a,b,and c are positive integers, what is the least possible value of a+b+c?

(A) 4  
(B) 5  
© 6  
(D) 7  
(E) 8

So right off the bat you can eliminate (A) and (B). Must you find the least common multiple of 216 and 225, then factor that? Is there a quicker way to do this? Thanks in advance.

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### Author: ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)
#### Post date: [August 4, 2009, 8:32pm UTC](https://boards.straightdope.com/t/math-problem/505332/2 "2009-08-04T20:32:15Z")

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225=3^2 x 5^2  
216=2^3 x 3^3

So k must have 2^3 x 3^3 x 5^2 as a factor.

So the least possible value of a+b+c is 3+3+2 = 8.

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### Author: ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)
#### Post date: [August 4, 2009, 8:33pm UTC](https://boards.straightdope.com/t/math-problem/505332/3 "2009-08-04T20:33:42Z")

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Break 225 and 216 down to prime factors:

3x3x5x5  
2x2x2x3x3x3

The smallest set of prime factors that includes both of those is

2x2x2x3x3x3x5x5

That’s three 2s, three 3s, two 5s, for a total of eight.

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### Author: ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [August 4, 2009, 8:37pm UTC](https://boards.straightdope.com/t/math-problem/505332/4 "2009-08-04T20:37:44Z")

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> [@brickbacon](#):
>
> Must you find the least common multiple of 216 and 225, then factor that? Is there a quicker way to do this?

Yes. No.

On preview: What they said.

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### Author: ![brickbacon](https://avatars.discourse-cdn.com/v4/letter/b/898d66/32.png) [@brickbacon](https://boards.straightdope.com/u/brickbacon)
#### Post date: [August 4, 2009, 9:21pm UTC](https://boards.straightdope.com/t/math-problem/505332/5 "2009-08-04T21:21:42Z")

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Ah, thanks

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### Author: ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)
#### Post date: [August 4, 2009, 9:44pm UTC](https://boards.straightdope.com/t/math-problem/505332/6 "2009-08-04T21:44:14Z")

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> [@brickbacon](#):
>
> Must you find the least common multiple of 216 and 225, then factor that? Is there a quicker way to do this?

Note that, in doing what they did, **Giles** and **Ximenean** _did_ find the least common multiple (in factored form—they didn’t find the l.c.m. and _then_ factor it).