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If 5400mn = k^4

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by Anurag@Gurome » Sun Feb 27, 2011 3:48 am
Baten80 wrote:If 5400mn = k^4, where m, n, and k are positive integers, what is the least possible value of m + n?
A. 11
B. 18
C. 20
D. 25
E. 33
5400 = 2^3 * 3^3 * 5^2
5400mn = 2^3 * 3^3 * 5^2 * mn
Since 5400mn = k^4, so 2, 3, and 5 should be all 4th powers
2^3 * 3^3 * 5^2 * mn = k^4 implies mn = 2 * 3 * (5^2) = 150, which is the least value of mn.
15 = 15 * 10 (15 and 10 are the least possible factors of 150)
Now to find the least possible value of m + n = 15 + 10 = 25

The correct answer is D.
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by lenagmat » Tue Oct 25, 2011 7:21 am
Please could you explain how does it implies mn = 2 * 3 * (5^2) = 150 from 2^3 * 3^3 * 5^2 * mn = k^4

Anurag@Gurome wrote:
Baten80 wrote: Since 5400mn = k^4, so 2, 3, and 5 should be all 4th powers
2^3 * 3^3 * 5^2 * mn = k^4 implies mn = 2 * 3 * (5^2) = 150, which is the least value of mn.
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by rooster » Tue Oct 25, 2011 9:14 pm
lenagmat wrote:Please could you explain how does it implies mn = 2 * 3 * (5^2) = 150 from 2^3 * 3^3 * 5^2 * mn = k^4

Anurag@Gurome wrote:
Baten80 wrote: Since 5400mn = k^4, so 2, 3, and 5 should be all 4th powers
2^3 * 3^3 * 5^2 * mn = k^4 implies mn = 2 * 3 * (5^2) = 150, which is the least value of mn.
There are a few hints that will let us know. First you need to break this into primes

5400mn=k^4 and k is a number

so that means the fourth root of 5400mn is a number

5400mn= (3*3*3*2*2*2*5*5)mn

In order for this to be a perfect fourth root you need 3*2*5*5 which means mn needs to have a combination of those 4 numbers.

15 +10 satisfies this, therefor the answer is D
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by GmatMathPro » Wed Oct 26, 2011 6:07 am
lenagmat wrote:Please could you explain how does it implies mn = 2 * 3 * (5^2) = 150 from 2^3 * 3^3 * 5^2 * mn = k^4
By itself it doesn't imply that exactly. The idea is that 5400mn=k^4 implies that 5400mn can be represented as some integer raised to the power of 4. For that to be true, all of the exponents in the number's prime factorization must be multiples of 4. For example, 2^12*3^8*5^4=(2^3*3^2*5)^4=360^4. This is only possible because the exponents in the prime factorization(12,8,4) are all multiples of 4.

In this problem, 5400mn=2^3*3^3*5^2mn, so we need to choose mn so that it makes the exponents on the right side all multiples of 4. If mn=2*3*5^2, then the expression becomes 2^4*3^4*5^4=(2*3*5)^4=30^4. However, if we are only interested in making it a 4th power, we also could choose mn=2*3^5*5^6, so the expression becomes 2^4*3^8*5^8=(2*3^2*5^2)^4=450^4. We have infinitely many choices for mn to make 5400mn equal to an integer raised to the fourth power. However, in this problem we also want to minimize m+n which implies that we should choose the lowest possible value of mn.
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