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Old 03-02-2013, 06:44 PM  
Frosty Frosty is offline
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Need (more) help from the CP math whizzes

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:

Quote:
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?

Last edited by Frosty; 03-03-2013 at 04:53 PM..
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Old 03-03-2013, 07:42 PM   #31
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Quote:
Originally Posted by cdcox View Post
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.
Thanks again. Got it now.
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Old 03-03-2013, 07:48 PM   #32
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How long would they get to work a problem like that?
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Old 03-03-2013, 08:00 PM   #33
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Quote:
Originally Posted by Frosty View Post
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.





Thanks for any help. Remember - you're doing it for the kids.
Thats fairly easy twice the speed or 2x
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Old 03-03-2013, 08:05 PM   #34
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Quote:
Originally Posted by cdcox View Post
How long would they get to work a problem like that?
That's one of 40 questions on the test that they get 35 minutes to complete. The final 10 are challenge questions where they get extra points if they complete them. That question was the final challenge question.

Here is a link to the test if you are interested:

http://www.wamath.net/contests/Mathi...IC03_78Reg.pdf

If I had graphed out that question, it would have been more obvious how to tackle it.
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Old 03-03-2013, 08:05 PM   #35
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Quote:
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Thats fairly easy twice the speed or 2x
Nope
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Old 03-03-2013, 08:14 PM   #36
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Old 03-03-2013, 08:14 PM   #37
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[quote=Frosty;9461659]That's one of 40 questions on the test that they get 35 minutes to complete. The final 10 are challenge questions where they get extra points if they complete them. That question was the final challenge question.

Here is a link to the test if you are interested:

http://www.wamath.net/contests/Mathi...IC03_78Reg.pdf

If I had graphed out that question, it would have been more obvious how to tackle it.[/QUOTE]

Sketches really help more often than not. Your wife should incorporate sketches into her teaching.

Sketch or no sketch, that isn't a one or two minute problem, at least for me.
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Old 03-03-2013, 08:16 PM   #38
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eleventy billion
That's what I thought but cdcox set me straight.
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Old 08-31-2013, 07:24 PM   #39
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Got another one we need help on -

"How many positive integer factors does the cube of the square of 2007 have?"

We can brute force out the answer but there has to an easier way to do it and to explain it to the kids.

TIA
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Old 08-31-2013, 08:25 PM   #40
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Is the answer 70?
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Old 08-31-2013, 08:32 PM   #41
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Now I'm going with 91.
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Old 08-31-2013, 08:32 PM   #42
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Quote:
Originally Posted by cdcox View Post
Is the answer 70?
91
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Old 08-31-2013, 08:33 PM   #43
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Quote:
Originally Posted by cdcox View Post
Now I'm going with 91.
Too quick for me. That's right. Is there an easy way to explain how to do it?
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Old 08-31-2013, 08:35 PM   #44
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Can the kids factor 2007 into 3x3x223 quickly?
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Old 08-31-2013, 08:39 PM   #45
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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
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