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
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Really tricky PS question.
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myselfhari
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Anurag@Gurome
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The prime factorization of 432 is (3^3)*(2^4)myselfhari wrote:Find the product of all the factors of 432.
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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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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Abhishek009
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Express 432 as a product of its prime factors.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
(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
Abhishek
