mikemcgarry wrote:
I was inspired by another, easier question to create this question, just for fun.
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?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 12You may find this blog on prime factors helpful in thinking about this problem:
https://magoosh.com/gmat/2012/gmat-math-factors/You may also find this blog on counting helpful.
https://magoosh.com/gmat/2012/gmat-quant-how-to-count/I will post a full solution if there's interest.
Nice question and great job! It really encompass different fields of math.
I’m too late but here is my view for the sake of diversity.
I prefer to solve it for \(a^3\) rather than \(b\) as \(a^3\) is the most restrictive part.
Let’s create different cases for different perfect cubes.
1. \(a^3\) consists of a single prime – \(p^c\), where \(c\)- multiple of 3.
We have \(2^3\), \(3^3\), \(5^3\) and \(5^6\)
\(3C1 + 1 = 4\)
2. \(a^3\) is composed from two distinct primes – \(p^cq^d\), where \(c,d\) – multiples of 3.
\(2^33^3, 2^35^3, 3^35^3\) and \(2^35^6, 3^35^6\)
\(3C2 + 1*2C1 = 5\)
3. \(a^3\) – 3 distinct primes – \(p^cq^dr^e\), where \(c,d,e\) – multiples of 3.
\(2^33^35^3\) and \(2^33^35^6\) together \(= 2\)
4. and final case when all primes will be absorbed by \(b\) – \(1\) (also a perfect cube) \(p^0q^0r^0\) = 1
Total # = \(4 + 5 + 2 + 1 = 12\)