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by sivaelectric » Sat May 28, 2011 8:59 am
Find the product of all the factors of 432.
  • A. 512^20
    B. 216^20
    C. 432^20
    D. 432^10
    E. 216^20
Can someone please post an explanation.
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by SoCan » Sat May 28, 2011 9:12 am
The prime factorization of 432 is (3^3)(2^4), so the total number of factors is (3+1)(4+1) = 20. You therefore have 10 factor pairings whose product is equal to 432. If you multiply all ten of these pairings, that's the equivalent of 432^10.
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by sameerballani » Sat May 28, 2011 9:49 am
sivaelectric 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
Can someone please post an explanation.
Answer=D

432=2^4 * 3^3
Total number of factors: 5*4=20
So basically product of two factors(pair)=432
This means 10 pair the product will be: 432^10
Factor1*Factor2=Product=432
1*432=432
2*216=432
3*144=432
And so on we will get 10 pairs each with product =432. Multiplying all =432^10 !!
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by sivaelectric » Sat May 28, 2011 10:08 am
OA D. Thanks for the explanation guys.
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by cans » Sat May 28, 2011 10:37 am
Good one. For a number, say x, with no. of factors = n,
product of all factors is x^(n/2)
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