How many factors does 36^2 have?

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ssy: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Sat Sep 29, 2007 7:17 pm; Thanked: 5 times; GMAT Score:660

by ssy » Thu Oct 25, 2007 7:14 am

How many factors does 36^2 have?
2
8
24
25
26

25

Can someone please explain how to work this out?

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Source: Beat The GMAT — Problem Solving | Thu Oct 25, 2007 7:14 am

samirpandeyit62: Master | Next Rank: 500 Posts; Posts: 460; Joined: Sun Mar 25, 2007 7:42 am; Thanked: 27 times

by samirpandeyit62 » Thu Oct 25, 2007 7:27 am

36^2 = 2^4*3^4

so factors = (4+1)(4+1) (power of each prime factor + 1)
=25

Regards
Samir

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jangojess: Master | Next Rank: 500 Posts; Posts: 175; Joined: Mon Jul 02, 2007 12:01 pmFollowed by: 1 member; Thanked: 15 times

by jangojess » Thu Oct 25, 2007 11:34 am

Rule - If n = a^x * b^y * c^z where a,b,c are prime numbers then number of factors that n have (including 1 and n) is (x+1)*(y+1)*(z+1)

Trying hard!!!

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ssy: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Sat Sep 29, 2007 7:17 pm; Thanked: 5 times; GMAT Score:660

by ssy » Thu Oct 25, 2007 10:26 pm

Thanks both of you, I get it now. =)

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ssy: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Sat Sep 29, 2007 7:17 pm; Thanked: 5 times; GMAT Score:660

by ssy » Thu Oct 25, 2007 10:27 pm

Thanks both of you, I get it now. =)

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Edthesock: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Mon Jan 19, 2009 12:11 am; Thanked: 3 times; GMAT Score:640

by Edthesock » Wed Jan 21, 2009 8:27 am

wow, so easy if you know that rule. I didn't...

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Wed Jan 21, 2009 9:36 am

And because of the rule jangojess cites above, perfect squares always have an *odd* number of divisors, because when you prime factorize a perfect square, the exponents are all even. When you add one to each power and multiply, you'll be multiplying odd numbers only, therefore getting an odd result. If you know that, you'll know without factorizing anything that the answer to the above question must be odd, and there's only one odd number among the answer choices.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Mozartain: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Tue Nov 11, 2008 8:38 pm; Thanked: 3 times; GMAT Score:740

by Mozartain » Thu Jan 29, 2009 10:58 pm

Hi Ian
could you please explain the rule Jangojess has given?

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awesomeusername: Master | Next Rank: 500 Posts; Posts: 226; Joined: Tue Jan 13, 2009 1:27 pmFollowed by: 1 member; Thanked: 23 times

by awesomeusername » Thu Jan 29, 2009 11:07 pm

Ian Stewart wrote:And because of the rule jangojess cites above, perfect squares always have an *odd* number of divisors, because when you prime factorize a perfect square, the exponents are all even. When you add one to each power and multiply, you'll be multiplying odd numbers only, therefore getting an odd result. If you know that, you'll know without factorizing anything that the answer to the above question must be odd, and there's only one odd number among the answer choices.

This is really cool Ian, thanks! So when I see something like:

How many factors does 81^99 have? I know that 81 is a perfect square and I know that an ODD number to the power of any positive integer is ODD, then the answer must be ODD. Since there was only one odd answer in the answer list, it looks like GMAC was doing this on purpose.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Fri Jan 30, 2009 12:34 am

Mozartain wrote:Hi Ian
could you please explain the rule Jangojess has given?

rather than explaining the rule with variables - which will get tedious, not to mention difficult to understand - i'll give an illustration with specific numbers.

let's say you want the total number of factors of the number 432, whose prime factorization is (2^4)(3^3).
here's the way to think about it: you have to choose (a) how many 2's, and, separately, (b) how many 3's are in the factorization. then you'll multiply these numbers of possibilities, in the same way as you'd multiply the numbers of possibilities for any other independent decision (such as 3 shirts x 4 pairs of pants = 12 outfits).

* because the factorization contains 2^4, there are 5 ways to choose the number of 2's: none, one, two, three, or four. note that this is 4 + 1 possibilities (= exponent + 1).
* because the factorization contains 3^3, there are 4 ways to choose the number of 3's: none, one, two, or three. note that this is 3 + 1 possibilities (= exponent + 1).
* then you'd multiply 5 x 4 = 20 possibilities total.

the same approach works in general, for any exponents that may happen to appear in a prime factorization. this is the genesis of the formula cited above.
of course, because TIME CONTRAINTS are a big deal on this exam, you should also just memorize the formula (i.e., you shouldn't have to derive it from scratch when you take the test, because that's a tremendous waste of time).

Ron has been teaching various standardized tests for 20 years.

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Mozartain: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Tue Nov 11, 2008 8:38 pm; Thanked: 3 times; GMAT Score:740

by Mozartain » Fri Jan 30, 2009 11:16 am

Hi Ron

Thanks so much for the very clear explanation. I much appreciate the time you took for my question.

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gmatpill: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Thu Apr 16, 2009 11:44 amFollowed by: 9 members; Thanked: 8 times

by gmatpill » Tue Jul 13, 2010 1:56 pm

How many factors does 36^2 have?
a) 2
b) 8
c) 24
d) 25
e) 26

Step 1:
Recognize this question is asking for all the numbers that go into the big number. Breakdown the big number into it's prime factorization - put into bases of 2, 3, 5, 7, etc...and assign exponents

Step 2
36^2 =
36 * 36
= 6 * 6 * 6 * 6
= 6^4
= (2*3)^4
= 2^4 * 3^4

Step 3:
Once we have the prime factorization (put into bases of 2, 3, 5, 7, etc), we can figure out the number of factors.

It turns out that the exponents can tell us how many factors are in that number. On a smaller scale, if we had the number 8, its factors are 1, 2, 4, 8.

8 = 2^3
All lower exponents of 2^3 are also factors, such as 2^2, 2^1 and 2^0. Thus 2^3 has 3 factors + "1" which is also a factor so 2^3 has 4 factors. In other words, you just take the exponent and add 1. Pretty simple, right?

Now that was for 8, what about if we multiplied by a different base like 3? so 8*3 = 24, which breaks down to 2^3 * 3. How many factors does 24 have?

Well 24 = 2* 2* 2* 3; the factors are basically any combination of the prime numbers. By adding the 3, we added more ways to combine the prime numbers. Now you can add all previous factors by 3 and create a new set of factors: 2^0 * 3, 2^1 *3, 2^2 * 3, 2^3 * 3 = {3, 6, 12, 24}
in addition to {1, 2, 4, 8}; essentially multiplying the number of factors by 2. 4*2 = 8 factors now

So this is what it looks like mathematically

24 = (2^3)*(3^1)
= (set of 4 factors) * (set of 2 factors) --we multiply because we want to find the number of combinations of 2's and 3's that form new factors

= 8 factors

Step 4)
Now, going back to our big number:
36^2 = 2^4 * 3^4

The number of factors are combinations of factors of 2^4 with factors of 3^4. So, the # of factors for 2^4 multipled by # of factors of 3^4 will give you the answer

2^4 = 16; has factors 2^0, 2^1, 2^2, 2^3, 2^4 = 1, 2, 4, 8, 16 (or 5 factors)

3^4 = 81; has factors 3^0, 3^1, 3^2, 3^3, 3^4 = 1, 3, 9, 27, 81 (or 5 factors)

Now we can combine any of the first set of factors with the second set of factors = 5*5 = 25
You can take {1, 3, 9, 27, 81} and multiply by 1.

Or you can multiply by 2. {3, 6, 18, 54, 162} are all factors of 36^2.

Or you can multiply by 4. By 8, By 16.

The largest number you get is 16*81 = 1296. And you will see this is equal to 36^2, so you know as a check that these two numbers multiplied still give you a factor. That factor being 1296, does go into 36^2.

Total there are 5*5 = 25 combinations of different factors generated.

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maxm: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Wed May 30, 2012 11:16 am

by maxm » Tue Jun 26, 2012 7:37 am

The statement says: How many factors does 36^2 have?
So, it's asking for the total number of factors, aren't we me missing the negative facors?

So, it would be 25*2 = 50 factors total.
Isn't 25 the number of positive factors?

I'm kind of lost. Thanks for clarifying.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Jun 26, 2012 9:48 am

maxm wrote:The statement says: How many factors does 36^2 have?
So, it's asking for the total number of factors, aren't we me missing the negative facors?

So, it would be 25*2 = 50 factors total.
Isn't 25 the number of positive factors?

I'm kind of lost. Thanks for clarifying.

Good question. Normally, GMAT questions will restrict the values to consider only positive factors. In general, however, the GMAT allows for negative factors (divisors), as seen in this excerpt from the Official Guide:

An integer is any number in the set {. . . -3, -2, -1, 0, 1, 2, 3, . . .}. If x and y are integers and x " 0, then x is a divisor (factor) of y provided that y = xn for some integer n. In this case, y is also said to be divisible by x or to be a multiple of x. For example, 7 is a divisor or factor of 28 since 28 = (7)(4), but 8 is not a divisor of 28 since there is no integer n such that 28 = 8n.

Since the original question does not restrict the factors to only positive factors, the correct answer here is 50.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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se_abhi: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Wed Sep 25, 2013 6:39 am

by se_abhi » Mon Sep 30, 2013 6:52 am

Thanks.

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