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Both 5^2 and 3^3 are factors of n 2^5 6^2 7^3 where n is a

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Manager
Joined: 10 Dec 2007
Posts: 53

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Both 5^2 and 3^3 are factors of n 2^5 6^2 7^3 where n is a [#permalink]

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30 Dec 2007, 14:40
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Both 5^2 and 3^3 are factors of n × 2^5 × 6^2 × 7^3 where n is a positive integer. What is the smallest possible positive value of n ?
25
27
45
75
125

[spoiler]OA is D[/spoiler]

Kudos [?]: 21 [0], given: 0

VP
Joined: 09 Jul 2007
Posts: 1098

Kudos [?]: 144 [0], given: 0

Location: London
Re: Factors of [#permalink]

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30 Dec 2007, 15:12
AVakil wrote:
Both 5^2 and 3^3 are factors of n × 2^5 × 6^2 × 7^3 where n is a positive integer. What is the smallest possible positive value of n ?
25
27
45
75
125

[spoiler]OA is D[/spoiler]

5*5 and 3*3*3

6=2*3*2*3

n*2*2*2*2*2*2*3*2*3*7*7*7/3*3*3 so we need one more 3
n*2*2*2*2*2*2*3*2*3*7*7*7/5*5 here we need 25 for it to be factor.

25 itself cannot be divided by 3 thus 25*3=n=75

D

Kudos [?]: 144 [0], given: 0

Director
Joined: 01 May 2007
Posts: 793

Kudos [?]: 386 [0], given: 0

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30 Dec 2007, 15:26
5^2 = 5 * 5

3 ^ 3 = 3 * 3 *3

Now factor out: n × 2^5 × 6^2 × 7^3

n * 2*2*2*2*2 * 2*3*2*3 * 7*7*7

So we said 5^2 was a factor of n × 2^5 × 6^2 × 7^3, so it takes two 5s. There are none, so we know those two 5s came from n.

Then we said 3^3 was a factor of n × 2^5 × 6^2 × 7^3, so it takes 3 threes, but we only have 2. So one of those 3s must be from n.

So...5*5*3 = 75. That is the lowest possible value of n.

Kudos [?]: 386 [0], given: 0

30 Dec 2007, 15:26
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Both 5^2 and 3^3 are factors of n 2^5 6^2 7^3 where n is a

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