Finding the perfect square, perfect cube and perfect fifth

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Hello,

Can you please assist with this:

Given that N = a^3b^4c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

a) a^3b^4c^5
b) a^5b^4c^3
c) a^2b^3c^5
d) a^7b^6c^5
e) a^27b26c^25

oa: e


I tried to solve this by taking a = 2, b = 3 and c = 5 => N = 2^3 3^4 5^5

a) 2^3 3^4 5^5 x 2^3 3^4 5^5 = 2^6 3^8 5^10 - Not all three
b) 2^5 3^4 5^3 x 2^3 3^4 5^5 = 2^8 3^8 5^8 - Not all three

This way I tested all the way till D and when none of them satisfied all three conditions I selected E. I was wondering if there is a better method for solving this. Thanks for your help.

Best Regards,
Sri

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by [email protected] » Tue Apr 29, 2014 12:35 am
Hi Sri,

This question is built around a few Number Properties involving exponents.

For a number to be a perfect square, all of the "exponent terms" must be EVEN.

for example....
25 is a perfect square because 25 = 5^2
16 is a perfect square because 16 = 4^2 = 2^4

For a number to be a perfect cube, all of the "exponent terms" must be A MULTIPLE OF 3.

8 is a perfect cube because 8 = 2^3
64 is a perfect cube because 64 = 4^3 = 2^6

For a number to be a perfect fifth power, all of the "exponent terms" must be A MULTIPLE OF 5.

32 is a perfect fifth power because 32 = 2^5
1024 is a perfect fifth power because 1024 = 4^5 = 2^10

Here, we need each exponent to become a multiple of 2, 3 and 5. The prompt refers to the SMALLEST number, so we need the Least Common Multiple of 2, 3 and 5 ......which is 30. The correct answer will multiply by the given prompt to equal 30.

Since we're starting with (A^3)(B^4)(C^5), we'll need to multiply with something that will end with (A^30)(B^30)(C^30).

Final Answer: E

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by Bill@VeritasPrep » Wed Apr 30, 2014 3:14 pm
Rich's explanation is great. I think this question is a great example of using simple examples (like testing simple perfect squares to see what the rule is) to unlock a harder question.
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by Bill@VeritasPrep » Wed Apr 30, 2014 3:17 pm
EDIT: double post. Thanks, Uverse :shock:
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by unknown13 » Thu May 01, 2014 7:25 pm
IMO correct answer is E