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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.

Questions? Ask us in the comments!


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