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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 10:11

1

This post received KUDOS

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:

Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 10:22

1

This post received KUDOS

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, 10:23, edited 1 time in total.

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