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GRE Math: Prime Factorization Solution

September 11, 2013 by

Prime FactorizationPrime FactorizationGreat work, all! Thanks for giving our practice GRE All-That-Apply problem a try. (In case you haven’t tried it already, visit this blog entry to give it a go.) Here is the full explanation:

In order to quickly determine which numbers are factors of both 252 and 770, take the prime factorization of each. 252 can be rewritten as: 2*2*3*3*7, and 770 breaks down to 2*5*7*11.

Now, see whether you can find sets of factors from each number that multiply together to each answer choice:

  • Choice (A), 2, is definitely a factor of both numbers. Keep it.
  • Choice (B), 4, is a factor of 252 (it’s the product of the two 2s). However, there’s no way to get a product of 4 by multiplying together any of the prime factors of 770, so it’s not a factor of this number. Eliminate.
  • Choice (C), 14, breaks down into 2*7. The prime factorizations of 252 and 770 each contain both a 2 and a 7, so 14 is a factor of each number. Keep this choice.
  • Choice (D), 15, is not a factor of either number – neither prime factorization has both a 3 and a 5. Eliminate.
  • Finally, let’s look at choice (E): 21 breaks down to 3*7. We have both 3 and 7 in the prime factorization of 252, but there’s no 3 in the prime factorization of 770, so it’s not a factor of both numbers. Eliminate.

So the correct answers are (A) and (C). Remember: Whenever you’re asked to work with factors of large numbers, use prime factorization to break numbers down into smaller components in order to take control of the information.

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