AMC 10/12 Marathon - Finished

[Sillysidley]

Same as the Mathcounts Marathon, except AMC 10 and 12 level.

First Problem
How many cubes are a factor of 3!*5!*7!

[Temperal]

Hmmm.... interesting question.

Well, it's prime factors are

2(3)(2)(3)(4)(5)(2)(3)(4)(5)(6)(7)=2^3(3^3)(4^2)(5 ^2)(6)(7)

That produces that at least two are factors. But, 2^2 is equal to 4, so 4^3 is also a factor.

So three cubes are a factor of it

Second problem:

A sequence of positive integers includes the number 68 and has arithmetic mean 56. When 68 is removed the arithmetic mean of the remaining numbers is 55. What is the largest number than can occur in the sequence?

[Sillysidley]

No, there are more than three cubes.

Like you said, 3!*5!*7!=2(3)(2)(3)(4)(5)(2)(3)(4)(5)(6)(7)=2^3(3^ 3)(4^2)(5^2)(6)(7)=2^8(3^4)(5^2)(7).

To be a cube, it the power must be a multiple of 3. So exponent of 2=0,3,6; exponent of 3 must be 0, or 3.
That's 3*2=6 cubes.
(They are 1, 8, 64, 27, 216, 1728)

Oh, this was an AMC 10 #15 (or 14) from 2005.

[Sillysidley]

I haven't looked at this for a while.
BTW, I just bombed the 2003 AMC 12A with a 108 (if you use the 1.5 per blank question) or 116 (if you use the old 2.5 per blank that was in effect in 2003).
16 right, 8 left blank.

I'll look at your problem.