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# how many integers can be factors of 78? (do it without

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CEO
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how many integers can be factors of 78? (do it without [#permalink]

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08 Nov 2007, 02:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

how many integers can be factors of 78? (do it without tediously listing the factors)
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08 Nov 2007, 05:53
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

4
I'm not listing the factors
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08 Nov 2007, 06:23
1,2,3 and 13!!

Is there some "hidden" trick or "short" method behind it, if Yes, could you please explain?
Director
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08 Nov 2007, 10:24
sidbidus wrote:
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

4
I'm not listing the factors

Are you guys doping prime factorization? it is asking only factors not different prime factors. so i guess it should be 8.

1, 2, 3, 6, 13, 26, 39, 78.
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08 Nov 2007, 10:38
LM wrote:
1,2,3 and 13!!

Is there some "hidden" trick or "short" method behind it, if Yes, could you please explain?

this is a variation of one of the Challenges.

we get the prime factors of 78:

2 *39
2* 3* 13

Now here's where it gets nice.
we look at the exponents of each of the factors and add one.
2, 2, 2

we multiply all the modified exponents together= 2*2*2=8 factors in all
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08 Nov 2007, 11:56
factors of 78

78 = 2 * 39

so (1+1)* (1+1) = 2*2 = 4
total four factors.
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09 Nov 2007, 18:31
LM wrote:
1,2,3 and 13!!

Is there some "hidden" trick or "short" method behind it, if Yes, could you please explain?

http://www.gmatclub.com/forum/t55022
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09 Nov 2007, 19:35
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

something i have noticed, unless it is a perfect square (ex. 4 or 25) integer must have EVEN # of factors

i get 8:
1,2,3,6,13,26,39,78

using the method that bkk145's shown me:

prime factorization
2*3*13
(1+1)(1+1)(1+1) = 8

either way its 8
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10 Nov 2007, 01:13
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

If the prime factorization of an integer is a^n * b^m * c^p, then the number of factors that integer has = (n+1)(m+1)(p+1).

In this case that gives us 8 factors.
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10 Nov 2007, 01:18
GK_Gmat wrote:
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

If the prime factorization of an integer is a^n * b^m * c^p, then the number of factors that integer has = (n+1)(m+1)(p+1).

In this case that gives us 8 factors.

Perfect !

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17 Nov 2007, 20:20
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

Sorry, but I'm a little confused. I came across a similar problem and had trouble with your method. The question is how many positive intergers are factors of 441?
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17 Nov 2007, 22:43
Ant wrote:
bmwhype2 wrote:
how many integers can be factors of 78? (do it without tediously listing the factors)

Sorry, but I'm a little confused. I came across a similar problem and had trouble with your method. The question is how many positive intergers are factors of 441?

factor 441

441 | 7

63 | 3

21 | 7

3

3^2*7^2 = (2+1)*(2+1) = 9

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18 Nov 2007, 00:21
can someone explain this method with apples and oranges please?
I get lost after finding the prime factors.
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18 Nov 2007, 00:56
1
KUDOS
Expert's post
asdert wrote:
can someone explain this method with apples and oranges please?
I get lost after finding the prime factors.

it is very good that
a^n * b^m * c^p => Nf = (n+1)(m+1)(p+1). (a,b,c -prime numbers)

why is it correct?

one has such cases: {a^0,a^1,.......a^n}*{b^0,b^1,......b^m}*{c^0,c^1,......c^p}
therefore, for a - n+1 (from 1=a^0 to a^n)
b - m+1
c - p+1

Nf=(n+1)(m+1)(p+1)
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18 Nov 2007, 10:46
So, is this correct then?

Factors of 60:

60/2
30/2
15/3
5

2² * 3 * 5, therefore (2+1)(1+1(1+1) = 12

Outcome number includes 1 and itself, correct? awesome! thanks for the tip.
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18 Nov 2007, 11:01
asdert wrote:
So, is this correct then?

Factors of 60:

60/2
30/2
15/3
5

2² * 3 * 5, therefore (2+1)(1+1(1+1) = 12

Outcome number includes 1 and itself, correct? awesome! thanks for the tip.

Yes {60,30,20,15,12,10,6,5,4,3,2,1} = 12

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18 Nov 2007, 12:45
Does this holds tru for any number or are there any exceptions? I've tested it on a bunch and all seem to be ok.

How come this stuff is not on any study guide?

I thought Mgmat was going to cover all this important tricks. I guess that's why the forum rocks.

Is there a thread somewhere with tips like this? (not necessarily factors related)

Thanks!
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18 Nov 2007, 13:44
asdert wrote:
Does this holds tru for any number or are there any exceptions? I've tested it on a bunch and all seem to be ok.

How come this stuff is not on any study guide?

I thought Mgmat was going to cover all this important tricks. I guess that's why the forum rocks.

Is there a thread somewhere with tips like this? (not necessarily factors related)

Thanks!

I think that walker gave a very good proof for this.

18 Nov 2007, 13:44
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