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

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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