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Help with nasty factors problem

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Re: Help with nasty factors problem

by ssmiles08 » Sat Jun 27, 2009 5:32 am
oldschool wrote:Both 5^2 and 3^3 are factors of n * 2^5 * 6^2 *7^3. n is a positive integer. What is the smallest possible n?

A. 25
B. 27
C. 45
D. 75
E. 125

Any ideas? :roll: It's probably simple, but I can't figure it out.
you can see that 6^2 can be broken down to its primes. 2^2 * 3^2

now you know there are two 3's in the numerator and three 3's in the denominator. so two of the three 3's cancel out in the denominator and you are left with 5^2 * 3

since you can't cancel any further, the smallest n has to be is 5^2 * 3 for the product to be a factor of the numerator.

the question is basically asking what will it take for the denominator to get canceled completely.

n = 75
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Re: Help with nasty factors problem

by sudi760mba » Wed Jul 01, 2009 11:40 pm
Here's how I solved it:

n(2^5)(6^2)(7^3)
_______________
(5^2)(3^3)

since you can breakdown (6^2) to (2 * 3)^2 to (2^2)(3^2)

so n(2^5)(2^2)(3^2)(7^3)
_______________________
(5^2) (3^3)

so (3^2) - (3 ^3) = (3)

so n(2^5)(2^2)(7^3)
_______________________
(5^2) (3)

since nothing else would be cancelled n would have to be a minimum (5^2) * 3 = 75 (D)



ssmiles08 wrote:
oldschool wrote:Both 5^2 and 3^3 are factors of n * 2^5 * 6^2 *7^3. n is a positive integer. What is the smallest possible n?

A. 25
B. 27
C. 45
D. 75
E. 125

Any ideas? :roll: It's probably simple, but I can't figure it out.
you can see that 6^2 can be broken down to its primes. 2^2 * 3^2

now you know there are two 3's in the numerator and three 3's in the denominator. so two of the three 3's cancel out in the denominator and you are left with 5^2 * 3

since you can't cancel any further, the smallest n has to be is 5^2 * 3 for the product to be a factor of the numerator.

the question is basically asking what will it take for the denominator to get canceled completely.

n = 75
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simplify the numbers

by ahmad.kadry » Thu Jul 02, 2009 2:22 am
I agree with the above solutions .. that is definitely a primes problem...an easy way to look at this is to put it that way:

X = 5^2 = (5 * 5)
Y = 3^3 = (3 * 3 * 3)

Z = n * 2^5 * 6^2 *7^3 is ... n * (2 * 2 * 2 * 2 * 2) * (2 * 3 * 2 * 3) * (7 * 7)

for the first two number X and Y to be factors of the given forumla, Z, the formula should contain at least (5 * 5) and (3 * 3 * 3) .. the known number of the formula contain only (3 * 3) .. so it needs at least one more 3 and two fives (5 * 5) to be multiplied in order to be evenly divisable by the given two numbers.

So, n must be at least 5*5*3 -> 5^2 * 3 = 75 (D)
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