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How many different positive integers are factors of 225?

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by Brent@GMATPrepNow » Sun Jan 07, 2018 2:55 pm
M7MBA wrote:How many different positive integers are factors of 225?

(A) 4
(B) 6
(C) 7
(D) 9
(E) 11
----ASIDE----------------------------------
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
-----ONTO THE QUESTION!!--------------------------------

225 = (3)(3)(5)(5)
= (3^2)(5^2)
So, the number of positive divisors of 225 = (2+1)(2+1)
= (3)(3)
= 9

Answer: D

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Brent
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by Brent@GMATPrepNow » Sun Jan 07, 2018 3:00 pm
M7MBA wrote:How many different positive integers are factors of 225?

(A) 4
(B) 6
(C) 7
(D) 9
(E) 11
Another approach:

When we scan the answer choices (always scan the answer choices before getting to work!!), we see that all 5 answer choices are pretty small.
So, we could just LIST all of the factors.
We'll do so in PAIRS of values that have a product of 225
We get:
1 and 225
3 and 75
5 and 45
9 and 25
15 and 15

So, the factors are: {1, 3, 5, 9, 15, 25, 45, 75 and 225}

Answer: D

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Brent
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by Scott@TargetTestPrep » Sun Jun 09, 2019 6:06 pm
M7MBA wrote:How many different positive integers are factors of 225?

(A) 4
(B) 6
(C) 7
(D) 9
(E) 11
To determine the number of positive integer factors of 225, we first factor 225 as:

225 = 15^2 = 5^2 x 3^2,

We see that the exponent of 5 is 2, and the exponent of 3 is 2. We add 1 to each of these exponents and then calculate the product. Thus, there are (2 + 1)(2 + 1) = 3 x 3 = 9 positive integer factors of 225.

Answer: D

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