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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, 06:52 PM   #2
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Old 03-02-2013, 06:55 PM   #3
Frosty Frosty is offline
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Quote:
Originally Posted by Bugeater View Post
My brain hurts
This is for the 5th grade competition, btw.
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Old 03-02-2013, 07:06 PM   #4
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The answer is 3.14
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Old 03-02-2013, 07:06 PM   #5
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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.
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Old 03-02-2013, 07:11 PM   #6
Frosty Frosty is offline
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Quote:
Originally Posted by Dayze View Post
The answer is 3.14
I like Pi

Quote:
Originally Posted by pr_capone View Post
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.
Thanks. Saturday night probably wasn't the best time to post this but there isn't any hurry. The next practice is Wednesday.
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Old 03-02-2013, 07:15 PM   #7
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Off the top of my head there isn't a short cut here. Let me think after I finish dinner...

To get started.. the formula would look something like

PHP Code:
-1<x<10
-1<y<10 <- (except when x 0 then 0<y<9))

(
10x+y)/(x+y
but I don't think this will help much.
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Old 03-02-2013, 07:15 PM   #8
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Quote:
Originally Posted by Frosty View Post
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?
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.
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Old 03-02-2013, 07:19 PM   #9
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I ran our local MathCounts competion last week for 6th through 8th graders. Some of these kids are darn fast. Still the tests are designed so that very few kids will get them all right or even finish on time. That's the only way to separate kids that are all probably pretty darn good at ciphering.
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Old 03-02-2013, 07:21 PM   #10
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Quote:
Originally Posted by Frosty View Post
This is for the 5th grade competition, btw.
Some kids are ****ing smart.
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Old 03-02-2013, 07:24 PM   #11
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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.
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
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Old 03-02-2013, 07:26 PM   #12
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Quote:
Originally Posted by AustinChief View Post
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
yeah, scratch that, I don't think it will get us where we need to go.
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Old 03-02-2013, 07:35 PM   #13
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Observation: In addition to the (12,21), (24,42), and (48,84) pairs, there is also a (36,63) pair (that overlaps with two of your 9's).
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Old 03-02-2013, 07:46 PM   #14
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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.
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.
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Old 03-02-2013, 07:50 PM   #15
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Quote:
Originally Posted by Frosty View Post
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.
If a kid had played with numbers a ton, they might be able to remember Patteeu's trick.
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