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the greatest prime factor of n t

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by sanju09 » Wed Apr 08, 2009 4:57 am
If n and t are positive integers, what is the greatest prime factor of n t?

(1) The greatest common factor of n and t is 5.

(2) The least common multiple of n and t is 105.



OA B
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Source: — Data Sufficiency |

by bluementor » Thu Apr 09, 2009 4:43 am
Statement 1: GCF(n,t) = 5

if n=15, t=10, greatest prime factor of nt = 5
if n=21, t=15, greatest prime factor of nt = 7. Hence insufficient.

Statement 2: LCM(n,t) = 105 = 3x5x7

7 is the greatest prime factor for 105.

either n or t or both must contain 7 as a factor. Therefore 7 is the greatest prime factor of nt. Sufficient.

Choose B.

-BM-
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by TheDeliveryGuy » Mon Aug 23, 2010 4:15 pm
Bluementor, I think you overlooked an important fact in statement 1:

(1) The greatest common factor of n and t is 5.

Although your answer is correct and your though process is aligned with what the question intends. The statement specifies that both n and t have 5 as GCF. N=21 (1*3*7) does not have 5 as a factor and can't be used in this case. An alternative example (and easier to compute mentally) would be n=5 and t=5 versus n=35 (equivalent to 7*5) and t=5. The GCF would indeed be 5 for both, but the product of n and t in the second case would yield 175, which is divisible by 1, 5, 7, and 7 being the largest prime factor. Therefore insufficient.

Notice that we could have used the next highest prime factor, in this case 11. We would have obtained n=55 (5*11) and t=5. The GCF is 5, but the greatest prime of their product is now 11. When confronted with these type of questions that involve prime, common, and least common factors, it is almost always easier to visualize and decompose the numbers in terms of factors. Instead of thinking about 4 (visualize 2*2) or instead of 12 (visualize 1*2*3). This enables you to see the big picture and prevents you from committing "silly" mistakes.

(2) The least common multiple of n and t is 105.

Assume n=105 (3,5,7) and lets say we double this and use t=210 (2,3,5,7) as well. 7 remains the largest prime factor. We could even use n=105 (3,5,7) and t=1155 (3,5,7,11), but still 7 will be largest prime in common. Hence sufficient.
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