mgmat c

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mgmat c

by resilient » Sun Jun 08, 2008 8:56 pm
How many factors does 36^2 have?



2
8
24
25
26


prime factorization tree comes to four 2's and four 3's. Trap is answe choice 8 but want a way to factor see concrete way.. I factored out fully and came to 24. MIssed something important.


qa is 25
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Missed 1!

by albertrahul » Sun Jun 08, 2008 10:23 pm
Along with 4 Threes and 4 twos, 1 is also a factor.
Hence: 4(3), 4(2) and 1.

Thus resulting to 24 + 1, 25.

Hope this helps!

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by netigen » Sun Jun 08, 2008 10:24 pm
This is a permutation combination question

prime factors of 36 = 2 x 2 x 3 x 3

prime factors of 36^2 = 2 x 2 x 3 x 3 x 2 x 2 x 3 x 3

Now, the way to find all factors is to find all possible combinations

there are 4 2's and 4 3's,

so a 2 can occur from 0-4 times (5 possibilities) and similarly 3 can occur from 0-4 times

total combinations = 5x5 = 25

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by VerbalAttack » Mon Jun 09, 2008 12:14 am
In fact there is a formula for this;

To find the number of factors (also called integral divisors) that a number has, increment the powers of the prime factors by 1 and then multiply them.

As for 36^2 = 2^4 * 3^4

The powers of the prime factors are 4 & 4. Apply the formula and you get (4+1) * (4+1) = 25.

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