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Re: If the product of the integers a, b, c and d is 546 and if 1 [#permalink]

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21 Mar 2013, 09:14

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Alright, so we know that the product of a,b,c and d is 546 and 1<a<b<c<d so the first thing we need to do is to break down 546 into a product of prime factors :

546 divided by 2 yields 273 273 divided by 3 yields 91 91 divided by 7 yields 13

Which means that \(546 = 2*3*7*13\).

So if we consider a = 2, b = 3, c = 7 and d = 13 (in respect to the second condition : 1<a<b<c<d), we have : b+c = 3 + 7 = 10 which is answer choice D.

"If the product of the integers a, b, c and d is 546 and if 1 < a < b < c < d, what is the value of b+c?"

Factors of 546:

546 | 2 273 | 3 91 | 13 7 | 7 1 | 1

If 1 < a < b < c < d, the only possible way is:

d=13 c=7 b=3 a=2

Therefore: b+c = 3 + 7 = 10

Answer D

Make prime factorization of 546: 546=2*3*7*13.

Now, since given that 1 < a < b < c < d, then a=2, b=3, c=7 and d=13 --> b+c=3+7=10.

Answer: D.

Answer is D

Hi Brunel - Can we actually expect such easy questions to come in GMAT? Just trying to make sure i dont get too excited by solving these quicker!
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