prime factors

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prime factors

by naaga » Tue Mar 03, 2009 12:06 am
What is the number of prime factors contained in the product 30^7 * 22^5 * 34^11 ?

23
44
46
48
53

do we need to calculate all the factors involved in the multiplication ?

OA is E
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by DanaJ » Tue Mar 03, 2009 12:58 am
Break eveything into prime factors:
30 = 2 * 3 * 5. This means 30 has 3 prime factors, so there will be 3 * 7 (7 being the power of 30) = 21 prime factors in 30
22 = 2 * 11 - use the same reasoning to get 2 * 5 = 10 prime factors here
34 = 2 * 17 - again, the number of prime factors will be 2 * 11 = 22.

This makes for a total of 21 + 10 + 22 = 53 prime factors.

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by avenus » Tue Mar 17, 2009 9:31 am
There's something in this question that doesn't quite add up, or at least, I don't get it.

30 has 3 prime factors, 2, 3 and 5... no problem with that, but
why does 30^7 have 21 prime factors??

2^n, 3^n, 7^n, n=2,...,7 (or any product of those) are not prime factors

To me, 30^7 has the same three prime factors that 30 does, i.e., 2, 3 and 7...

any comments?

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by Ian Stewart » Tue Mar 17, 2009 10:27 am
The wording of the question is not good - where is the question from? When we talk about the prime factors of a number, normally we mean the number of distinct prime factors. That is, 30 has three prime factors: 2, 3 and 5, and 900 also has three prime factors - 2, 3 and 5. The powers in the prime factorization aren't relevant if you're only counting prime factors.

The question really means to ask for what is known in Number Theory as the 'length' of 30^7 * 22^5 * 34^11. The question should be asking "How many primes would need to be multiplied together to get 30^7 * 22^5 * 34^11"? To answer that, we simply need to prime factorize the number, and add together the powers.

I'd add that you don't often see questions that ask you to do this on the GMAT, though there is one in the Official Guide (that's the only one I've seen, in fact).
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