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 Argo

Some kids are ****ing smart.

My youngest son took first overall in the state for the individual portion (his team took 2nd for the team part) when he was in 5th grade /proud papa


I went up through Differential Equations in college and look at these questions now and feel stupid.

[Flachief58]

[Frosty]

Originally Posted by cdcox

If a kid had played with numbers a ton, they might be able to remember Patteeu's trick.

Is there an explanation for it or it just a pattern to remember? The "10's " are trivial and I remembered the "9's" trick from school but explaining the pairs was trickier other than "it just is".

[AustinChief]

Originally Posted by Frosty

Is there an explanation for it or it just a pattern to remember? The "10's " are trivial and I remembered the "9's" trick from school but explaining the pairs was trickier other than "it just is".

What trick? That any 2 digit number whose digits add up to 9 is also divisible by 9?

[cdcox]

The relationship in the pairs is:

(2a*10+a)/3a = 7 and (a*10+2a)/3a = 4 for a = 1,2,3,4

That's the why, but the way you would "just know it" would be playing with numbers all the time.

[Frosty]

Originally Posted by AustinChief

What trick? That any 2 digit number whose digits add up to 9 is also divisible by 9?

More like the digits of the multiples of 9 from 1 to 10 add up to 9. Not really a trick, I guess. I think the back of my mind was also thinking about how transposed numbers are divisible by 9 when I wrote "trick".

[Frosty]

Originally Posted by cdcox

The relationship in the pairs is:

(2a*10+a)/3a = 7 and (a*10+2a)/3a = 4 for a = 1,2,3,4

I think we'll just go with "because it is" on that one.

[Sofa King]

TLR

[Frosty]

Originally Posted by Sofa King

TLR

I don't blame you. There were a lot of numbers and stuff.

[pr_capone]

Originally Posted by cdcox

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.

Wifey took a look and stated that this would be the best approach.

[Frosty]

I have a new one. It's driving us nuts because it seems straightforward but we are not getting the answer they give so we must be missing something.

On a straight road, an inspecting officer traveled from the rear to the front of an army column, and back, while the column marched forward its own length. If the officer and and the column maintained steady (but different) speeds, what was the ratio of their speeds, faster to slower?

Thanks for any help. Remember - you're doing it for the kids.

[theelusiveeightrop]

Drinking does not make this easier.

[cdcox]

Are you getting 1+sqrt(2) ?

[Frosty]

Originally Posted by cdcox

Are you getting 1+sqrt(2) ?

No but that's the answer they give. Do you mind letting me know how you got it (I was getting sqrt(3)).

[cdcox]

Let a be the speed of the column, b the speed of the officer, and x be the the length of the column. Let t' be the time for the officer to reach the front of the column and t be the total time. We then get the following relationships:

at = x

bt' = at' + x

bt = 2at' + x

Do a bunch of algebra and get b/a = 1+sqrt(2)

Let me know if it isn't working out.