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Problem Solving - Factorization Question??

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by JohnQ2011 » Fri Jan 28, 2011 12:41 pm
Check this out when you have a minute and let me know if you know how to solve it properly...

If both 5^2 (25) and 3^3 (27) are factors of n x 25 x 62 x 73, what is the smallest possible positive value of n?

a) 25
b) 27
c) 45
d) 75
e) 125

Thanks!
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by tomada » Fri Jan 28, 2011 12:52 pm
IMO, the answer is (B). We're given that 27 is a factor of n x 25 x 62 x 73, which means that the factorization of n x 25 x 62 x 73 must include at least instances of 3 as a factor.

The number 62 is factored as 31x2, and can't go any further.
The number 73 is already prime, so no factorization possible.
The number 25, as we know, is 5x5.

We still need the three instances of 3 as a factor. The smallest number that provides these is 27.[/b]
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by JohnQ2011 » Fri Jan 28, 2011 1:03 pm
ANSWER IS {D} - IF ANYONE CAN HELP WITH DETAILED METHODOLOGY AND WORK THAT WOULD BE GREAT!
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by aleph777 » Fri Jan 28, 2011 2:03 pm
CL3AV3R,

The answer D is incorrect. What's your source for this?

If we're looking for the smallest possible value of n, the only primes in n x 25 x 62 x 73 that we could overlook are the two fives in 25, since we know they're accounted for with the 5^2, which we know is a factor already.

The only factors left primes left are 2, 3, 3, 3, 31, and 73. And since we know 3^3 is a factor, 27 is smallest factor we can possibly create.

Answer: B
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by JohnQ2011 » Fri Jan 28, 2011 2:12 pm
Hi,

Could be an error from the publisher

Answer in the back of the book is described as:

(d) The approach here is not to multiply out the numbers, but rather to completely factor the large number, then compare its factors to 52 and 33. Any 5s or 3s that can't be factored out of 25 x 62 x 73 (or 25 x 22 x 32 x 73) will have to be factors of n. Since we can account for two 3s, but no 5s in the large number, 52 and the remaining 3 must be factors of n.
Therefore n is equal to 5 x 5 x 3.[/spoiler]
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by Anurag@Gurome » Fri Jan 28, 2011 2:14 pm
CL3AV3R wrote:If both 5^2 (25) and 3^3 (27) are factors of n x 25 x 62 x 73, what is the smallest possible positive value of n?

a) 25
b) 27
c) 45
d) 75
e) 125
n*25*63*73 = 25*(n*2*31*73)

As 25 is already accounted for and the rest of the product has no 3 in it, all the 3's of 3^3 must come from n itself. Hence, minimum possible value of n is 3^3 = 27

The correct answer is B.
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by JohnQ2011 » Fri Jan 28, 2011 2:14 pm
*** Should read: then compare its factors to 5^2 and 3^3***[/b]
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by JohnQ2011 » Fri Jan 28, 2011 2:17 pm
Hi, Anurag! Thanks for your time. Can you give me more detail about this question and take me through the step by step process broken down?? I am not sure if the incorrect answer is throwing me off or I can not remember how to solve this type of question...it has been a few years...

Thanks again![/b]
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by aleph777 » Fri Jan 28, 2011 2:25 pm
CL3AV3R wrote:Hi,

Could be an error from the publisher

Answer in the back of the book is described as:

(d) The approach here is not to multiply out the numbers, but rather to completely factor the large number, then compare its factors to 52 and 33. Any 5s or 3s that can't be factored out of 25 x 62 x 73 (or 25 x 22 x 32 x 73) will have to be factors of n. Since we can account for two 3s, but no 5s in the large number, 52 and the remaining 3 must be factors of n.
Therefore n is equal to 5 x 5 x 3.[/spoiler]
Looks as if there might have been a typo in your question up top, then. 52 and 33 don't appear in your question.
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by JohnQ2011 » Fri Jan 28, 2011 2:33 pm
*** Should read: then compare its factors to 5^2 and 3^3***

wasn't showing exponents properly in that response...
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by ikaplan » Sun Feb 27, 2011 8:05 am
I would go with B.

the multiple n x 25 x 62 x 73 contains no 3 - since 3^3 is a factor of the multiple, then n must contain at least 3^3
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by gopinathhyd » Tue Nov 01, 2011 1:03 pm
I go with B

Since none of the numbers are multiples of 27, n should be 27
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by ronnie1985 » Thu Apr 26, 2012 10:58 am
(B) QED
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by karthikpandian19 » Mon Jun 04, 2012 7:01 pm
25 & 33 are factors of "nx25x62x73" is given.

And from the question you can find out that 25 is already allocated in that number. Now look out for the 27 which is three 3's. There is no 3's in the "nx25x62x73", therefore the n"n" should be 27 in order to satisfy the factor

Answer is B
JohnQ2011 wrote:Check this out when you have a minute and let me know if you know how to solve it properly...

If both 5^2 (25) and 3^3 (27) are factors of n x 25 x 62 x 73, what is the smallest possible positive value of n?

a) 25
b) 27
c) 45
d) 75
e) 125

Thanks!
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by Mboy » Sun Jul 22, 2012 1:26 am
Misread it to say 5^2 * 25 and 3^3 * 27
:( :(
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