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2 questions [#permalink] New post 12 Jun 2009, 16:59
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D
E

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I have the answers, but it would be great if some one can explain to me.

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

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

[Reveal] Spoiler:
the answer is 75


(I'm not good in questions asking about facots..does some one have a site of set of questions about factors so i can practice?
*********

How many different ways can 2 students be seated in a row of 4 desks, so that there is alwayes at least one emplty desk between the students?

A)2
B)3
C)4
D)6
E)12

[Reveal] Spoiler:
The correct answer is D


Thank you

Last edited by bb on 12 Jun 2009, 23:16, edited 1 time in total.
Converted to Math Symbols and hid the answer choices
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Re: 2 questions [#permalink] New post 12 Jun 2009, 18:01
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For the first one
Given
5^2 and 3^3 are factors of n\times 2^5\times6^2\times7^3
Need to find the smallest value of n.

As 5^2 and 3^3 are factors of n\times 2^5\times6^2\times7^3,
n\times 2^5\times6^2\times7^3 should be either the Least common Multiple of the two or a multiple of the LCM itself. i.e. dividing n\times 2^5\times6^2\times7^3 by the LCM should result into an integer.

Lets find the LCM of the two:

5^2=5\times5
3^3=3\times3\times3

\text{LCM}=5\times5\times3\times3\times3=5^2\times3^3

Now
\frac{n\times 2^5\times6^2\times7^3}{5^2\times3^3}
should be an integer
=\frac{n\times 2^5\times2^2\times3^2\times7^3}{5^2\times3^3}
Reducing it further
= \frac{n\times 2^5\times2^2\times7^3}{5^2\times3}
for the fraction to be an integer n should be divisible by {5^2\times3}. Ans the smallest value n can have is {5^2\times3}.
As
\frac{5^2\times3}{5^2\times3}=1

{5^2\times3}=75

Hence the answer is D.

Also here is a good overview of the basics of factors.

http://www.math.com/school/subject1/les ... 3L1GL.html
Re: 2 questions   [#permalink] 12 Jun 2009, 18:01
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