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Frosty 03-02-2013 07:44 PM
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?

Bugeater 03-02-2013 07:52 PM
My brain hurts

Frosty 03-02-2013 07:55 PM
Quote:
Originally Posted by Bugeater (Post 9458355)
My brain hurts
This is for the 5th grade competition, btw.

Dayze 03-02-2013 08:06 PM
The answer is 3.14

pr_capone 03-02-2013 08:06 PM
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.

Frosty 03-02-2013 08:11 PM
Quote:
Originally Posted by Dayze (Post 9458413)
The answer is 3.14
I like Pi

Quote:
Originally Posted by pr_capone (Post 9458419)
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.

AustinChief 03-02-2013 08:15 PM
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.

cdcox 03-02-2013 08:15 PM
Quote:
Originally Posted by Frosty (Post 9458327)
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.

cdcox 03-02-2013 08:19 PM
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.

KurtCobain 03-02-2013 08:21 PM
Quote:
Originally Posted by Frosty (Post 9458365)
This is for the 5th grade competition, btw.
Some kids are ****ing smart.

AustinChief 03-02-2013 08:24 PM
Quote:
Originally Posted by cdcox (Post 9458466)
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

AustinChief 03-02-2013 08:26 PM
Quote:
Originally Posted by AustinChief (Post 9458500)
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.

patteeu 03-02-2013 08:35 PM
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).

Frosty 03-02-2013 08:46 PM
Quote:
Originally Posted by cdcox (Post 9458466)
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.

cdcox 03-02-2013 08:50 PM
Quote:
Originally Posted by Frosty (Post 9458580)
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.

Frosty 03-02-2013 08:51 PM
Quote:
Originally Posted by Argo (Post 9458486)
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. :huh:

FlaChief58 03-02-2013 08:55 PM
<iframe width="560" height="315" src="http://www.youtube.com/embed/rp2vCM9ghVI" frameborder="0" allowfullscreen></iframe>

Frosty 03-02-2013 08:58 PM
Quote:
Originally Posted by cdcox (Post 9458589)
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 03-02-2013 09:09 PM
Quote:
Originally Posted by Frosty (Post 9458613)
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 03-02-2013 09:16 PM
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 03-02-2013 09:17 PM
Quote:
Originally Posted by AustinChief (Post 9458649)
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 03-02-2013 09:34 PM
Quote:
Originally Posted by cdcox (Post 9458670)
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 03-02-2013 09:58 PM
TL:DR

Frosty 03-02-2013 10:02 PM
Quote:
Originally Posted by Sofa King (Post 9458834)
TL:DR
I don't blame you. There were a lot of numbers and stuff.

pr_capone 03-02-2013 10:21 PM
Quote:
Originally Posted by cdcox (Post 9458466)
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 03-03-2013 05:10 PM
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. :D

BlackHelicopters 03-03-2013 05:34 PM
Drinking does not make this easier.

cdcox 03-03-2013 07:38 PM
Are you getting 1+sqrt(2) ?

Frosty 03-03-2013 07:47 PM
Quote:
Originally Posted by cdcox (Post 9461426)
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 03-03-2013 07:54 PM
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.

Frosty 03-03-2013 08:42 PM
Quote:
Originally Posted by cdcox (Post 9461472)
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.

cdcox 03-03-2013 08:48 PM
How long would they get to work a problem like that?

Dave Lane 03-03-2013 09:00 PM
Quote:
Originally Posted by Frosty (Post 9461054)
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. :D
Thats fairly easy twice the speed or 2x

Frosty 03-03-2013 09:05 PM
Quote:
Originally Posted by cdcox (Post 9461610)
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.

Frosty 03-03-2013 09:05 PM
Quote:
Originally Posted by Dave Lane (Post 9461640)
Thats fairly easy twice the speed or 2x
Nope

TimeForWasp 03-03-2013 09:14 PM
eleventy billion

cdcox 03-03-2013 09:14 PM
[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.

Frosty 03-03-2013 09:16 PM
Quote:
Originally Posted by ChiefsNow (Post 9461701)
eleventy billion
That's what I thought but cdcox set me straight.

Frosty 08-31-2013 07:24 PM
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

cdcox 08-31-2013 08:25 PM
Is the answer 70?

cdcox 08-31-2013 08:32 PM
Now I'm going with 91.

Frosty 08-31-2013 08:32 PM
Quote:
Originally Posted by cdcox (Post 9930050)
Is the answer 70?
91

Frosty 08-31-2013 08:33 PM
Quote:
Originally Posted by cdcox (Post 9930067)
Now I'm going with 91.
Too quick for me. That's right. Is there an easy way to explain how to do it?

cdcox 08-31-2013 08:35 PM
Can the kids factor 2007 into 3x3x223 quickly?

cdcox 08-31-2013 08:39 PM
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

Frosty 08-31-2013 08:39 PM
Quote:
Originally Posted by cdcox (Post 9930081)
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 08-31-2013 08:42 PM
Quote:
Originally Posted by cdcox (Post 9930092)
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 08-31-2013 08:47 PM
Quote:
Originally Posted by Frosty (Post 9930094)
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 08-31-2013 08:54 PM
I'm a Math major but....F this question.

patteeu 08-31-2013 10:16 PM
Quote:
Originally Posted by cdcox (Post 9930116)
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 09-01-2013 12:54 AM
Quote:
Originally Posted by patteeu (Post 9930400)
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.


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