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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 03-02-2013, 07:51 PM   #16
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Quote:
Originally Posted by Argo View Post
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.
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Old 03-02-2013, 07:55 PM   #17
Flachief58 Flachief58 is offline
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Old 03-02-2013, 07:58 PM   #18
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Quote:
Originally Posted by cdcox View Post
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".
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Old 03-02-2013, 08:09 PM   #19
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Quote:
Originally Posted by Frosty View Post
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?
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Old 03-02-2013, 08:16 PM   #20
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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.
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Old 03-02-2013, 08:17 PM   #21
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Quote:
Originally Posted by AustinChief View Post
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".
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Old 03-02-2013, 08:34 PM   #22
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Quote:
Originally Posted by cdcox View Post
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.
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Old 03-02-2013, 08:58 PM   #23
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Old 03-02-2013, 09:02 PM   #24
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Quote:
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I don't blame you. There were a lot of numbers and stuff.
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Old 03-02-2013, 09:21 PM   #25
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Quote:
Originally Posted by cdcox View Post
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.
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Old 03-03-2013, 04:10 PM   #26
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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.


Quote:
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.
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Old 03-03-2013, 04:34 PM   #27
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Old 03-03-2013, 06:38 PM   #28
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Are you getting 1+sqrt(2) ?
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Old 03-03-2013, 06:47 PM   #29
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Quote:
Originally Posted by cdcox View Post
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)).
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Old 03-03-2013, 06:54 PM   #30
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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.
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