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-   -   Science Need help from the CP math whizzes (https://www.chiefsplanet.com/BB/showthread.php?t=270657)

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.


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