Thanks again. Got it now.

How long would they get to work a problem like that?

Thats fairly easy twice the speed or 2x

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.

Nope

eleventy billion

[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.

That's what I thought but cdcox set me straight.

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

Is the answer 70?

Now I'm going with 91.

91

Too quick for me. That's right. Is there an easy way to explain how to do it?

Can the kids factor 2007 into 3x3x223 quickly?

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