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Math problem involving powers and smallest possible value [#permalink]
 24 Sep 2009, 10:53
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I'm stuck on this, please help!
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
thanks!!!
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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 11:11
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I'm getting D: 75
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?
Therefore, (n)(2^5)(3^2*2^2)(7^3)
We need two 5's and 3 3's total
There are already 2 3's so we still need two 5's and one more 3 (5*5*3) = 75
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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 11:11
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The answer is D: 75.
This is a question relating to prime factorability (is that a word?).
Basically, for any number, X, to be a factor of another number, Y, the prime factors of X must be present in the prime factors of Y. For example, if you reduce the number 50 = (5^2)*(2^1), it is a factor of any number which has these prime factors when reduced.
So, for 5^2 and 3^3 to be factors of this number, the number, when reduced to prime factors, must contain 5^2 and 3^3.
n*2^5*6^2*7^3
= n*2^5*(3^2*2^2)*7^3
= n*2^7*3^2*7^3
As seen by the above equation, we need at least one more factor of 3, as well as two more factors of 5. As a result:
n = 5^2 * 3
= 75
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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 11:14
yeah, the answer is 75, but I'm not sure how you worked that out...
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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 11:22
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Think of it this way:
You have the equation 10x. What is the smallest number x can be for the equation to be a multiple of 3 (for 3 to be a factor of the equation)?
5*2*x. So for the equation to be divisible by 3 there must be a 3 in the equation. Therefore x = 3
5*2*3 is divisible by 3
Now you have the equation 15x and want to know what is the smallest number x for it to be divisible by 3^2 (or 9)
15x = 5*3*x
There is already one three in the equation. You will need at least one more 3 to factor out the 9. The smallest value of x is 3 for the number to be divisible by 9
Therefore
5*3*3 = 45
Prime factorization
Last edited by lagomez on 24 Sep 2009, 11:23, edited 1 time in total.
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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 11:22
sorry, I got the explanation, I didn't see all the responses! thanks
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Re: Math problem involving powers and smallest possible value
[#permalink]
24 Sep 2009, 11:22
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