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If both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what

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If both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what [#permalink] New post 23 Dec 2003, 19:44
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A
B
C
D
E

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If both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what is the smallest possible positive value of n ?

25
27
45
75
125
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shubhangi


Last edited by shubhangi on 23 Dec 2003, 20:00, edited 1 time in total.
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 [#permalink] New post 23 Dec 2003, 19:46
I think the question should read 5^2 and 3^3 in which case the answer should be D)
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 [#permalink] New post 23 Dec 2003, 19:50
i am sorry yeah..its square.. but iknow the answers..paul ..i need to know the working.. so pl..if u dont mind.. could u work out for me
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 [#permalink] New post 23 Dec 2003, 19:59
i agree! it occured to me after it took me more than 2 minutes to realize there was something wrong the question.

Last edited by Titleist on 23 Dec 2003, 20:01, edited 1 time in total.
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 [#permalink] New post 23 Dec 2003, 20:00
sure, but I just thought the working would be in the Kaplan book though.
Here it is:

To be the multiple of a factor, a given number must include all of the prime numbers of the factor to their respective power.

2^5 x 6^2 x 7^3 is missing factors 5^2 and 3^3
But 6^2 = (2x3)^2 = 2^2 x 3^2

Thus, the only factors that the number 2^5 x 6^2 x 7^3 is missing is 5^2 and 3. The result would therefore be 5^2 x 3 = 25 x 3 = 75.

Hope I was clear in my explanation...
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 [#permalink] New post 23 Dec 2003, 20:13
types of problems are very simple once you break everything down to prime factorization. :wink:
  [#permalink] 23 Dec 2003, 20:13
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If both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what

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