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Joined: 15 Feb 2009
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2 questions [#permalink] New post 12 Jun 2009, 16:59



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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?


[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
\(5^2\) 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:



\(\frac{n\times 2^5\times6^2\times7^3}{5^2\times3^3}\)
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 \({5^2\times3}\).


Hence the answer is D.

Also here is a good overview of the basics of factors. ... 3L1GL.html
Re: 2 questions   [#permalink] 12 Jun 2009, 18:01
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