Originally Posted by Bugeater (Post 9458355)

My brain hurts

Originally Posted by Dayze (Post 9458413)

The answer is 3.14

Originally Posted by pr_capone (Post 9458419)

My wife is a 4th/5th grade teacher with math certifications coming out the ears. If you don't have an answer by the time she gets home, I'll see if she wants to take a crack at it.

Originally Posted by Frosty (Post 9458327)

My wife volunteers at the middle school to coach kids for a local math contest called Math Is Cool. For the contest, kids have to take a variety of tests, some individual and some team tests, as well as a "College Bowl" type team to team competition.

Anyway, to prepare for teaching the kids, she works through the problems on old tests so she can explain to the kids at practice. She ran into this problem and couldn't figure out a quick solution:



Now you can brute force the answer by listing out all 90 numbers but these are timed tests and that would take too long. She did it that way to see if there was a pattern but we couldn't see one that could she could explain.

First of all, you have the nine "10's": 10, 20, 30, 40, etc

Then you have the nine "9's": 18, 27, 36, 45, etc

After that, there are only six other numbers:
two "3's": 12, 21, two "6's": 24, 42, and and two "12's: 48, 84

(when I say a number is a "9's", I mean the digits add up to 9: 18 > 1+8=9)

That adds up to 24. However, 90 is on both the 10's and 9's list so you have to subtract one to get 23.

We also noticed that all multiple of 12 and of 21 are on the list but are not sure how that helps.

So - is there a quick way to do this problem to explain to the kids if they run into something like this?

Originally Posted by Frosty (Post 9458365)

This is for the 5th grade competition, btw.

Originally Posted by cdcox (Post 9458466)

You're doing it wrong.

Go through the sums, starting with zero.

0, none
1, 1+0, none
2, 1+1 and 2+0, bam 20
3, 1+2, 2+1, 3+0 bam 12, 21, and 30
4, 1+3, 3+1, 2+2, 4+0, bam 40

that is going to be quicker than looking at all 90. For instance 5 you can see in about an instant that 50 is the only one. Its reasonable to do the calculations in a minute or so if you have the concept down.

Originally Posted by AustinChief (Post 9458500)

In the end I think this is the only route to go.

Again, have to finish dinner then I'm thinking their may be a formulaic way to do this but not anything that would be any faster.

This is the next step btw... (10x+y) = p^n * q^m and (n+1)(m+1) = # of factors

Originally Posted by cdcox (Post 9458466)

You're doing it wrong.

Go through the sums, starting with zero.

0, none
1, 1+0, none
2, 1+1 and 2+0, bam 20
3, 1+2, 2+1, 3+0 bam 12, 21, and 30
4, 1+3, 3+1, 2+2, 4+0, bam 40

that is going to be quicker than looking at all 90. For instance 5 you can see in about an instant that 50 is the only one. Its reasonable to do the calculations in a minute or so if you have the concept down.

Originally Posted by Frosty (Post 9458580)

Thanks. This looks like the most likely solution. Most of the questions have a little trick to them at this level. We were just coming at it from the wrong side.