Need help from the CP math whizzes

[Frosty]

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:

The sum of the digits of my 2-digit counting number can be divided into my number with no remainder. How many different numbers could I be thinking of?

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?

[Frosty]

Originally Posted by cdcox

Can the kids factor 2007 into 3x3x223 quickly?

This is 7th grade level, so they should be able to, once they figure out that 223 is prime.

[Frosty]

Originally Posted by cdcox

If so 2007^3^2 = 2007^6

now factoring into primes

2007^6 = 223^6 * 3^2^6 = 223^6 * 3^12

once you've expressed a number as a product of primes raised to exponents, there is a trick where you add 1 to each exponent and multiply to get the number of factors

(6+1) * (12+1) = 7*13 = 91

That you kind sir.

Most of these problems come down to knowing the "tricks" like this. Unfortunately, math class was a long time ago.

[cdcox]

Originally Posted by Frosty

This is 7th grade level, so they should be able to, once they figure out that 223 is prime.

Yeah, the factoring of 2007 is the hard part. If they guess it is divisible by 3, you get 669, which is clearly divisible by 3. Then they just have to go with their gut that 223 is prime.

It is pretty easy to see that 223 isn't divisible by 2, 3, 5, 7, or 11, so guessing that it is prime is pretty safe at that point.

[Setsuna]

I'm a Math major but....F this question.

[patteeu]

Originally Posted by cdcox

Yeah, the factoring of 2007 is the hard part. If they guess it is divisible by 3, you get 669, which is clearly divisible by 3. Then they just have to go with their gut that 223 is prime.

It is pretty easy to see that 223 isn't divisible by 2, 3, 5, 7, or 11, so guessing that it is prime is pretty safe at that point.

They should be able to recognize that it's divisible by 3 if they know the simple "trick" that any number whose digits add up to a number divisible by 3 is, itself, divisible by 3. I'm sure you and frosty are well aware of this, but just in case, I thought I'd mention it.

2+0+0+7=9

9 is divisible by 3, therefore 2007 is divisible by 3.

[aturnis]

Originally Posted by patteeu

They should be able to recognize that it's divisible by 3 if they know the simple "trick" that any number whose digits add up to a number divisible by 3 is, itself, divisible by 3. I'm sure you and frosty are well aware of this, but just in case, I thought I'd mention it.

2+0+0+7=9

9 is divisible by 3, therefore 2007 is divisible by 3.

Never seen this trick before. Why aren't we taught all the tricks in school? Might help.