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# m02#24

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Retired Moderator
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05 Nov 2008, 08:21
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How many distinct integers are factors of 90?

(A) 6
(B) 8
(C) 9
(D) 10
(E) 12

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I don't quite agree with the OA.
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05 Nov 2008, 09:37
Is OA not 12?

90 = 9*10 = 2*3*3*5

distinct factors:
1, 2, 3, 5,
3*3, 3*2, 3*5, 2*5,
3*3*2, 3*3*5, 3*2*5,
3*3*2*5
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05 Nov 2008, 09:51
The OA is 12.

the factors are 90, 45, 30, 18, 15, 10, 9, 6, 5, 3, 2, 1

But wouldn't "distinct integers" mean only 1,2,3,6,5,9? Hence 6...
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05 Nov 2008, 12:16
The question is asking for "distinct integers as factors". I guess you read it as "distinct digits as factors".
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05 Nov 2008, 12:31
Ahh there's my problem! Thanks.
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09 Jul 2010, 06:43
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To know the number of number of factors for a number , we need to split the number in the following way

Number = a^x * b^y * c^z .......so on

The number of factors would be (x+1)*(y+1)*(z+1)...so on

Applying the above to the current example

90 = 2^1 * 3^2 * 5^1

The number of factors = (1+1) * (2+1) * (1+1) = 2*3*2 = 12

Please send kudos if you like this
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10 Jul 2010, 03:17
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Possibly Correct, Could someone tell me for sure

First we factor 90 into primes 3*3*5*2

I thought of this as a combination problem with a duplicate choice 4! / 2 !

4! because I have to put the numbers in order A*B*C*D

2! because I have a duplicate choice (3)

Can anyone tell me if this will work for other situations?
or did I just confuse myself and everyone else?

Alternatively, I write these questions out starting with the largest and smallest factor
(I am using the .... to show that I write them a good distance apart on my scratch paper.)
1 .............................................................................................................. 90
next
1, 2,...........................................................................45, 90
1, 2, 3 30, 45, 90
1, 2, 3, 5, 18, 30, 45, 90
If find that doing things this way prevents me from having duplicates and quickly gets me to a situation where I know I have covered all the possibilities.
This method is really only good for number less than 100. For example if you had 87452, it might take too much time to do this.
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10 Jul 2010, 05:48
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E

I found a very useful formula (from GMAT Club forum) number of factors of a^x*b^y*c^z = (x+1)(y+1)(z+1).

Do hope it helps.
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16 Jul 2010, 02:43
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90, when prime-factorized, can be expressed as: 2*(3^2)*5
i.e (2^1)(3^2)(5^1).
generally, distinct factors [of N = (a^x)(b^y)(c^z)]
is (x+1)(y+1)(z+1)

so, for 90, the distinct prime factors = 2*3*2 = 12
OA = E.
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Status: Time to step up the tempo
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16 Jul 2010, 18:32
I have been going through some of the materials from MGMAT. I would go about solving this problem this way....

90 -- 3^2*2^1*5^1

Number of factors would be (2+1)(1+1)(1+1) = 12.

I believe that this is the quick way of solving these problems....

Are there any other quick ways to solve these kinds of problems ?????
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Joined: 23 Jun 2010
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02 Aug 2010, 15:07
tiruraju wrote:
To know the number of number of factors for a number , we need to split the number in the following way

Number = a^x * b^y * c^z .......so on

The number of factors would be (x+1)*(y+1)*(z+1)...so on

Applying the above to the current example

90 = 2^1 * 3^2 * 5^1

The number of factors = (1+1) * (2+1) * (1+1) = 2*3*2 = 12

Please send kudos if you like this

This is definitely a more methodical approach to the question. But it requires double the work. Since you anyways have to prime factor number N, I found it easier to factorize the non-prime factor further until it is represented as prime. For example, 90 = 45 x 2. Then 45 = 15 x 3. 15 = 5 x 3 and so on. At the end it is a simple math of counting the factors!
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11 Aug 2010, 23:47
tiruraju wrote:
To know the number of number of factors for a number , we need to split the number in the following way

Number = a^x * b^y * c^z .......so on

The number of factors would be (x+1)*(y+1)*(z+1)...so on

Applying the above to the current example

90 = 2^1 * 3^2 * 5^1

The number of factors = (1+1) * (2+1) * (1+1) = 2*3*2 = 12

Please send kudos if you like this

This is very useful if you have to determine the number of factors of very large numbers (kudos!). Finding each factor by trial-and-error would cost way to much time. Try 2496: prime factorization is much easier (2^5)*3*13 -> (5+1)*(1+1)*(1+1)=24
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13 Jul 2011, 05:52
I wrote all the factors out and got 10. I forgot to add 1 and 90 as factors. Grrrr!!
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13 Jul 2011, 11:43
I got the correct answer, (E): 12 however I do have one question, do negative integers not qualify as "distinct" factors? Meaning the answer would be 12 positive distinct factors + 12 negative distinct factors = 24 distinct factors in total?
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17 Jul 2013, 05:55
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Expert's post
Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

Back to the original question:

How many distinct integers are factors of 90?

A. 6
B. 8
C. 9
D. 10
E. 12

$$90=2*3^2*5$$, which means that the number of factors of 90 is: $$(1+1)(2+1)(1+1)=12$$.

Similar questions to practice:
how-many-odd-positive-divisors-does-540-have-106082.html
how-many-factors-does-36-2-have-126422.html
how-many-different-positive-integers-are-factor-of-130628.html
how-many-distinct-positive-factors-does-30-030-have-144326.html

Hope it helps.
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Re: m02#24   [#permalink] 17 Jul 2013, 05:55
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# m02#24

Moderator: Bunuel

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