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# 2 questions

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Intern
Joined: 15 Feb 2009
Posts: 12
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Kudos [?]: 0 [0], given: 0

2 questions [#permalink]  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:

(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:

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
Current Student
Joined: 03 Aug 2006
Posts: 116
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Kudos [?]: 156 [1] , given: 3

Re: 2 questions [#permalink]  12 Jun 2009, 18:01
1
KUDOS
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
$$\frac{5^2\times3}{5^2\times3}=1$$

$${5^2\times3}=75$$

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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