smartyman wrote:
Dear mike,
I still have problem understanding the following question, could you please further elaborate? Thanks
Question:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?
I still have problem understanding the following question, could you please further elaborate? Thanks
Question:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?
Dear smartyman,
First of all, understand that this is a very hard question, perhaps an 800-level question. It involves several sophisticated concepts, including
(a) prime factorization
http://magoosh.com/gmat/2012/gmat-math-factors/
(b) counting & the FCP
http://magoosh.com/gmat/2012/gmat-quant-how-to-count/
Insight #1 --- the question asks for number of possible values of b, but it's much much easier to count the number of possible values for (a^3), and every value of (a^3) will be paired with a unique value of b, so we will count the (a^3)'s as a way to get the answer.
Insight #2 --- a^3 is a perfect cube, and every prime factor in a perfect cube must appear either three times or some number of times that is a multiple of three. Thus, (a^3) simply could be 1, and all the factors could be in b. If (a^3) has any factors of 2, it only could have 2^3, because that's all the factors of two available. If (a^3) has any factors of 3, it only could have 3^3, because the available powers of 3 don't go up as high as any other multiple of 3. If (a^3) has any factors of 5, it could have either (5^3) or (5^6) --- we have enough factors of 5 to construct either one of those, so we have both of them as options.
Insight #3 --- how many combinations in (a^3)?
For the factors of 2, we have two choices ---- 2^0 = 1 or 2^3
For the factors of 3, we have two choices ---- 3^0 = 1 or 3^3
For the factors of 5, we have three choices ---- 5^0 = 1 or 5^3 or 5^6
Any option in any one category could be matched with any option from any other category, so this is a case in which we can employ the FCP:
total number of combinations = 2*2*3 = 12
There are 12 possibilities for (a^3), which means there are 12 possibilities for b.
Because there are only 12, I will list everything for clarity, to demonstrate that the FCP works. Notice, in the (a^3) term, all prime factors always have powers divisible by 3.
(1) a = 1, b = (2^3)(3^4)(5^7)
(2) a = (2^3), b = (3^4)(5^7)
(3) a = (3^3), b = (2^3)(3^1)(5^7)
(4) a = (5^3), b = (2^3)(3^4)(5^4)
(5) a = (5^6), b = (2^3)(3^4)(5^1)
(6) a = (2^3)(3^3), b = (3^1)(5^7)
(7) a = (2^3)(5^3), b = (3^4)(5^4)
(8) a = (3^3)(5^3), b = (2^3)(3^1)(5^4)
(9) a = (2^3)(3^3)(5^3), b = (3^1)(5^4)
(10) a = (2^3)(5^6), b = (3^4)(5^1)
(11) a = (3^3)(5^6), b = (2^3)(3^1)(5^1)
(12) a = (2^3)(3^3)(5^6), b = (3^1)(5^1)
Does all this make sense?
Mike
