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If both 5 squared and 3 cubed...

Expert replies
by houstonrockets16 » Mon Feb 02, 2009 4:58 pm
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?

A) 25
B) 27
C) 45
D) 75
E) 125

I am having trouble with this question....


I keep thinking it is A...

the correct answer is... D

Please help! Thanks!
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Source: — Problem Solving |

by houstonrockets16 » Mon Feb 02, 2009 5:01 pm
cetmk wrote:n x 2^5 x 6^2 x 7^3 can be written as
n x 2^5 x (3 x 2)^2 x 7^3
= n x 2^7 x 3^2 x 7^3

This should contain atleast 2 factors of 5 (5^2) and 3 factors of 3(3^3). Since the number already has 2 factors of 3, the smallest possible value of n is 5^2 x 3 = 75
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prime factor

by NavdeepG » Tue Feb 03, 2009 10:39 am
since n multiplied by 2^5 x 6^2 x 7^3 already has 2 3's (6^2) as its prime factors, the only factors missing are one 3 and 2 5's.
Hence 3 * 5 * 5 = 75.

What is OA
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by houstonrockets16 » Thu Feb 05, 2009 7:03 pm
75 is correct. Thanks
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by clooner » Tue Aug 02, 2011 9:17 am
Anyone can explain this question to me? the solution in the book is to brief for me to understand. I mean I understand the math involved but I can't get my head around the first step!
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by MBA.Aspirant » Tue Aug 02, 2011 10:52 am
houstonrockets16 wrote: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?

A) 25
B) 27
C) 45
D) 75
E) 125
prime factorize

n x 2^5 x 6^2 x 7^3

n x 2^5 x (2^2 x 3^2) x 7^3

since 3^3 and 5^2 are factors of this, and 3^2 is also included in it, then what remains is 3 and 5^2

so n = 3 * 5^2 = 75
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by clooner » Tue Aug 02, 2011 11:32 am
MBA.Aspirant wrote: prime factorize

n x 2^5 x 6^2 x 7^3

n x 2^5 x (2^2 x 3^2) x 7^3

since 3^3 and 5^2 are factors of this, and 3^2 is also included in it, then what remains is 3 and 5^2

so n = 3 * 5^2 = 75
Thanks for the quick reply. My first language is not English and I think the problem is not that I don't get the math but that I don't understand what is asked. The question doesn't make sense to me at all. Maybe you or someone has another question with the same problem which I can try to solve. I'm starring blind on this one.
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