How many factors of 36^2?

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maxim730: Senior | Next Rank: 100 Posts; Posts: 76; Joined: Tue Dec 26, 2006 11:26 amFollowed by: 3 members

by maxim730 » Fri Mar 09, 2007 8:35 am

How many factors does 36^2 have?

A)2
B)8
C)24
D)25
E)26

Answer in a bit.

My solution:

factors of 36 = 2^2 * 3 ^2 = 9 factors when you write it out:
1,2,3,4,,9,12,18,6, and 36

Therefore, 36^2 = 9 * 9 factors = 81 factors.. but 81 isn't a choice..

Thanks!!!

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Source: Beat The GMAT — Problem Solving | Fri Mar 09, 2007 8:35 am

Vasudha: Senior | Next Rank: 100 Posts; Posts: 46; Joined: Fri Jan 12, 2007 3:24 am; Location: Notre Dame; Thanked: 1 times

Re: How many factors of 36^2?

by Vasudha » Fri Mar 09, 2007 8:54 am

See attached..

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pmFollowed by: 6 members

by BTGmoderatorRO » Sun Sep 03, 2017 11:51 am

let us take the our N=36
N^2=36^2=1296
finding the prime factors of 1296 gives 4*4*9*9
=(2^2) * (2^2) *(3^2) *(3^2)
solving using indices
=(2^4) * (3^4)
using the factor formula f= (p+1)(r+1)
where p and r are the power factors of the prime number
f= (4+1) * (4+1)
=5 * 5 =25

option D is the correct option

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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 » Sun Sep 03, 2017 12:51 pm

maxim730 wrote:How many factors does 36^2 have?
A)2
B)8
C)24
D)25
E)26

-------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
---NOW ONTO THE QUESTION-----------------

36 = (2)(2)(3)(3)
So, 36Â² = (36)(36)
= (2)(2)(3)(3)(2)(2)(3)(3)
= (2^4)(3^4)
So, the number of positive divisors of 36Â² = (4+1)(4+1)
= (5)(5)
= 25

Answer: D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pmFollowed by: 247 members; Thanked: 1443 times

factoring

by ceilidh.erickson » Mon Sep 04, 2017 8:24 am

Roland2rule wrote:let us take the our N=36
N^2=36^2=1296
finding the prime factors of 1296 gives 4*4*9*9
=(2^2) * (2^2) *(3^2) *(3^2)
solving using indices
=(2^4) * (3^4)
using the factor formula f= (p+1)(r+1)
where p and r are the power factors of the prime number
f= (4+1) * (4+1)
=5 * 5 =25

option D is the correct option

You will not have a calculator on the GMAT Quant section. As such, it would be pointless to multiple out 36*36 to get 1296, and then factor it again. This would be a serious waste of time.

As a general rule when you don't have a calculator, you should never make numbers bigger. Always make them smaller!

Break 36*36 down into (4*9)(4*9) and then 2*2*2*2*3*3*3*3 without taking the intermediate step of multiplying it into a larger number.

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pmFollowed by: 247 members; Thanked: 1443 times

by ceilidh.erickson » Mon Sep 04, 2017 8:29 am

There is also a shortcut to this particular problem:

When we're looking for DISTINCT FACTORS of an integer, all positive integers have an EVEN number of distinct factors... except for perfect squares! This is because we can count distinct factors in pairs:

5:
1 x 5
primes will always have exactly 2 distinct factors - even number

18:
1 x 18
2 x 9
3 x 6
6 distinct factors - even number

36:
1 x 36
2 x 18
3 x 12
4 x 9
6 x 6
9 distinct factors - odd number

Since we're only looking for DISTINCT factors, we only want to count one of the 6's for 36. Perfect squares will always have an ODD number of distinct factors.

In the given question, we know that 36^2 must also have an odd number of distinct factors. Since D is the only odd answer choice, that must be the answer.

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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Admin1: Site Admin; Posts: 28; Joined: Wed Aug 30, 2017 10:11 pm

by Admin1 » Mon Sep 04, 2017 11:46 am

36^2 can be written as a product of prime factors as,
(2 x 2 x 3 x 3)^2
= 2^2 x 2^2 x 3^2 x 3^2
= 2^(2+2) x 3^(2+2)
= 2^4 x 3^4

To find the number of factors, add 1 to each exponent and multiply.
Number of factors = (4 + 1)(4 + 1)
Number of factors = (5)(5) = 25

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Admin1: Site Admin; Posts: 28; Joined: Wed Aug 30, 2017 10:11 pm

by Admin1 » Tue Sep 05, 2017 5:44 am

Solution:
36^2 = (6^2)^2 = 6^4 = (2 x 3)^4 = (2^4)(3^4)
So, if n is a factor of 36^2, then
n = (2^a)(3^b),
where
0 <= a <= 4, 0 <= b <= 4.
For a, there are five choices: 0, 1, 2, 3, 4. Similarly there are five choices for b. For each choice of a, b, there is a different factor for 36^2.
Thus, there are
5 x 5 = 25 factors in 36^2.

Answer: D

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: How many factors of 36^2?

by Scott@TargetTestPrep » Sun Feb 09, 2020 4:38 am

maxim730 wrote: ↑
Fri Mar 09, 2007 8:35 am
How many factors does 36^2 have?

A)2
B)8
C)24
D)25
E)26

Answer in a bit.

My solution:

factors of 36 = 2^2 * 3 ^2 = 9 factors when you write it out:
1,2,3,4,,9,12,18,6, and 36

Therefore, 36^2 = 9 * 9 factors = 81 factors.. but 81 isn't a choice..

Thanks!!!

Solution:

Since 36^2 = (9 x 4)^2 = (3^2 x 2^2)^2 = 3^4 x 2^4, then 36^2 has (4 + 1)(4 + 1) = 5 x 5 = 25 factors.

Alternate Solution:

We see that 36^2 is a perfect square and recall that a perfect square always has an odd number of factors. Therefore, the only answer choice that can be correct is D.

Answer: D

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