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

Question

how many integers can be factors of 78? (do it without tediously listing the factors)

sidbidus · Answer

4
I'm not listing the factors

LM · Answer

1,2,3 and 13!!

Is there some "hidden" trick or "short" method behind it, if Yes, could you please explain?

Fistail · Answer

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.

bmwhype2 · Answer

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

botplant · Answer

factors of 78

78 = 2 * 39

so (1+1)* (1+1) = 2*2 = 4
total four factors.

bkk145 · Answer

https://www.gmatclub.com/forum/t55022

beckee529 · Answer

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

GK_Gmat · Answer

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.

KillerSquirrel · Answer

Perfect !

 <img src="{SMILIES_PATH}/icon_smile.gif" alt=":)" title="Smile" />

Ant · Answer

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?

KillerSquirrel · Answer

factor 441 441 | 7 63 | 3 21 | 7 3 3^2*7^2 = (2+1)*(2+1) = 9

asdert · Answer

can someone explain this method with apples and oranges please? I get lost after finding the prime factors. :oops:

walker · Answer

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)

asdert · Answer

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.

KillerSquirrel · Answer

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

asdert · Answer

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!

KillerSquirrel · Answer

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

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

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