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Really tricky PS question.

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by Anurag@Gurome » Sun Jun 03, 2012 10:25 pm
myselfhari wrote:Find the product of all the factors of 432.
The prime factorization of 432 is (3^3)*(2^4)

Hence, total number of factors of 432 = (3 + 1)*(4 + 1) = 20.
Therefore, we have 10 factor pairs and product of each pair is equal to 432. If we multiply all ten of these pairs or all 20 factors, we will get 432^10.

The correct answer is D.
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by Anurag@Gurome » Sun Jun 03, 2012 10:28 pm
We can write the concept as the following formula,
  • For any integer N, with number of factors m, the product of all factors of N = N^(m/2)
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by Abhishek009 » Mon Jun 04, 2012 9:19 am
myselfhari wrote:Find the product of all the factors of 432.

A. (512)^20
B. (216)^20
C. (432)^20
D. (432)^10
E. (216)^20

I need a easy method to solve this tricky one.

Please help
Express 432 as a product of its prime factors.

(3^3)*(2^4)

Increment the powers of each of the prime factors.

(3^4 )* (2^5)

Product of the incremented numbers( powers )

4*5 = 20

The product of all factors of 432 is

432^(number of factors/2)

432^(20/2)

432^10

Hence answer is D...



The tricks for solving such problems is -

1. First find out the prime factors of the number

2. Increase the powers of the factor by 1 and multiply them.

3. Raise the no ^( Increased the powers )/2
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