*This is the second installment in a new occasional series on our blog, authored by Karishma, one of our star GMAT prep instructors in Detroit, Michigan. From time to time Karishma will share some of her unique insights into how to maximize your potential on the GMAT. Enjoy!*

Let’s start with a very interesting and important topic that GMAT loves to test you on – Factors.

Factors are the divisors of a number. 1 is a factor of every number and the number itself is also a factor of that number. For the sake of giving the definition, let me say: A positive integer x is a factor of a positive integer N when there exists another positive integer y such that x × y = N. In other words, when N is divided by x, it doesn’t leave any remainder.

Now, let’s cut to the chase and go on to real business.

Tell me, how many factors does 315 have and what are they? (Isn’t that an ugly number! Unlike 36 or 72 or 81 – numbers easy to work with. But let’s work with what we have.)

First of all, let me break it down to smaller numbers: 315 = 63 × 5 = 7 × 9 × 5 (looks better now)

Now we know that 5, 7, 9, 3 are all factors of 315 because they divide 315 completely. But there are others too e.g. 15 (15 × 21 gives 315). So how do we ensure that we get ALL the factors of 315?

Let me digress here with a few questions:

Q: Is 2 a factor of 315?

A: No! 2 does not divide 315 completely.

Q: Is 9 a factor of 315?

A: Yes, we can see that it is. 9 × 35 gives us 315.

Q: So then 35 is also a factor of 315, isn’t it?

A: Yes, because 35 will divide 315 completely and give 9 as quotient.

Similarly, if 7 is a factor of 315, we should have a corresponding factor 45 such that, 7 × 45 = 315. Get the picture? (Let’s assume you nodded your head.)

Back to the topic at hand now. Let’s retain the Q&A format.

Q: 315 has lots of factors: 1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315. How do we get all of them?

A: For that, I first need to break it down to its prime factors 315 = 3^^{2} × 5 × 7. We get all the factors of 315 by combining the prime factors in as many different ways as we can. i.e. we take a 3 alone; it is a factor of 315. We take a 3^^{2} alone; it is also a factor of 315. We take a 5 alone; it is again a factor of 315. We take a 3 and a 5 and multiply them to get 15 as a factor and so on…

Q: In how many different ways can we combine the prime factors of 315?

A: In 3 × 2 × 2 ways

Q: Wait a sec! How did we get this?

A: We have to take a combination of 3, 3, 5 and 7 to make a factor. We can do this in any way we like. We can make a factor by taking a 3, a 5 and a 7. We get 105. Or we can make a factor by taking two 3s, no 5 and a 7. We get 63 as a factor. Therefore, we can choose 3 in three ways (take no 3, take one 3 or take two 3s), 5 in two ways (no 5 and one 5) and 7 in two ways (no 7 and one 7). So there are, in all, 3 × 2 × 2 = 12 ways to make a factor of 315. The diagram below illustrates this concept.

In general,** if a number can be written as N = x ^{^p}. y^^{q}.z^^{r}…..(where x, y, z are all distinct prime numbers), then total number of factors of N is (p + 1)(q + 1)(r + 1)…**

(the + 1 indicates the case in which you do not take that prime factor while making your factor)

Let me write down all the factors of 315 in increasing order:

1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315

Q: Do you notice a pattern? No?

Try again now

1, 3, 5, 7, 9, 15,

315, 105, 63, 45, 35, 21

Q: How about it?

A: Look at each column individually: 1 × 315 = 315; 3 × 105 = 315; 5 × 63 = 315; 7 × 45 = 315; 9 × 35 = 315; 15 × 21 = 315!

Q: Co-incidence?

A: I think not. Numbers equidistant from the left and the right multiply to give the original number. This is where the importance of the definition of factors given above comes in (You thought I was just wasting time by giving the definition, isn’t it?) Every factor x (let’s say 7) of a number (315 here) will have a factor y (45 here) such that x × y = N (7 × 45 = 315). The smaller the x, the greater the y to make up N. So if a factor is third smallest, it will multiply the third greatest to give you the number N.

So now, if we want to write down all the factors of 315, we just write down the small ones

1, 3, 5, 7, 9, 15

And then next to them write their corresponding big factors:

1, 3, 5, 7, 9, 15

315, 105, 63, 45, 35, 21

To ensure that we haven’t missed any of the small ones, we can use the formula and find out how many total factors does the number have. Life made easy!

Something to think about:

Q: What happens if the total number of factors is odd?

Hi, I just stumbled on your blog when I was searching for GMAT study materials. Good stuff.

Form your credentials, it appears that you have cracked GMAT quant.

I am not so sure about your verbal skills.

Read a sentence from one of your posts (December 6, 2010, filed in: GMAT, Quarter Wit, Quarter Wisdom ):

Quote You thought I was just wasting time by giving the definition, isn’t it? end-quote

Review this line carefully. Do you not see something odd in this?

Almost all Asians use such a construct. Why is that? Is it the result of transliteration from Indian to English?

Correct usage: You thought I was just wasting time by giving the definition, didn’t you?

I hope you will ackowledge this and try to rectify it in future.

Cheers- Mike

Mike..Your post is riddled with assumptions..I am not so sure about your critical reasoning skills…

You can offer constructive criticism without coming across as obnoxious or Bill O’Reilly rude –yes my own adverb.

You just enjoyed her article for free; a: “ thank you” would be a nice start..

This comment should be censored – it is very upsetting!! Just ruined my day…

Hi Mike,

How many languages do you speak, and probably you don’t speak the language of maths or intuition per-se very deep.

Imparting ideas have no language. Grammar is a construct. I am certain you inept jealous B…

If you look closely, I am sure you will find other errors. But I still claim that I can manage GMAT Verbal well. If you know your grammar, picking the correct option is easy. But when you write (especially to explain something to someone), you write the way it comes most naturally to you, the way you speak – with errors and all! Thankfully, GMAC (in AWA) and the business community at large forgive a few grammatical errors!

Hello Karishma,

I’ve read good reviews about you GMAT private tutoring. I am based in London and would like to take few private lessons.

Please let me know if it’s possible.

Regards,

Nika

Hey Nika,

We have some great instructors in London. In fact, I think Amit is there. He is a “Veritas Prep Instructor of the Year” award winner. I am sure you will be very happy under his tutelage.

You can call us at +44.208.099.7383

or mail us at info@veritasprep.com

for details.

Best,

Karishma

Hi

using this method ,factors of 100= 9

100= 2^2 * 5^2

so (2+1)*(2+1)= 9 which is correct .

what I cannot understand is how to use this information to find the factors of 100^2.

I could prime factor 10,000 directly and then use your method to get 25.

i.e = 10000=2^4 * 5^4

so ( 4+1) * (4+1)= 25

That is not the issue .

What I want to do is , use the factors of 100 to find the factors of 100^2 .This would save me some time , rather than having to prime factor 100^2. I would just find factors of 100 and then find the factors of 100^2 .

This somehow I am not able to do , or what is the relation ?

I just cannot add 9 factors of 100 to 9 factors of another 100 to give me a total of 18 factors of 100^2, which could be wrong.

nor can I multiply 9 factors of 100 to another 9 factors of 100 to give me 81 factors of 100^2 which could be ridiculous .

But since 10,000 = 100^2 , how do I use the factors of 100 to find the no. of factors

of 100^2.

Note I want to use your method , lets suppose I found 9 factors of 100 using your method, now what is the next step to get the factors of 100^2.

Is this the correct approach , or we cannot use this approach and we have to take prime factors of whole 10,000, please clarify .

You can write 100^2 as 10^4 and then prime factorize it as 2^4 * 5^4. Total number of factors = (4+1)*(4+1) = 25

or

100 = 2^2*5^2 so 100^2 = (2^2 * 5^2)^2 = 2^4 * 5^4 (proceed as above now)

Note: You cannot find the number of factors of 100 (which is (2+1)*(2+1) = 9) and then get the number of factors of 100^2 by adding/multiplying to the number of factors of 100. The reason is that when you square 100, every prime number power gets multiplied by 2 and then, 1 is ADDED to each term. You get (4+1)*(4+1) = 25.

Hi

Thank you , that definitely helps !

Karishma,

Have you discussed the below elsewhere in your blog?

“Q: What happens if the total number of factors is odd?”

This is the follow up post:

http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-factors-of-perfect-squares/

Thanks !

Very helpful article indeed, Thank you.

I just came across this post in GC

http://gmatclub.com/forum/how-many-even-different-factors-does-the-integer-p-have-132835.html#p1088587, which unfortunately was not backed up by an expert so I was not sure of its credibility.

So here goes:

What if the required factors were more specific in nature?

Lets take 450 as an example, prime factorization 2^1 3^2 5^2

1) For even factors = 1(2+1)(2+1) = 9

2) Odd factors = (2+1)(2+1)= 9

3) How to find how many factors that contain 5? The method given in the above link describes something as this 450 = 5(2^1 3^2 5^1) -> (1+1)(2+1)(1+1)=12 which is correct

4) How many factors contain 3? Again 3(2^1 3^1 5^2)

(1+1)(1+1)(2+1)= 12, which seems correct too.

5) How many factors contain 6? 3 * 2 (3^1 5^2) -> (1+1) (2+1) = 6 which looks correct also.

So is this the proper method or is it just a coincidence?

If the above method works then to find the all the even factors I could also just find all the factors that contain 2 , could I not?

2(3^2 5^2) -> (2+1)(2+1) = 9

So just wanted to know if all the factors that contain 3 or 5 or 6 or 10 are required for a given number (here 450) is there a method for that? This method given in the link above seems to work.

Thank you

Here is the logic you can use for any number:

To get even factors, you need at least one 2.

So given N = 2^1 * 3^2 * 5^2, you have one 2, two 3s and two 5s.

To get even factors, you must pick one the one 2 you have (1 way) and you can pick a 3 and a 5 in 3 ways each (pick one no 3, pick one 3 or pick two 3s etc)

To get the number of factors that have 6, you must pick a 2 and a 3 to make 6. So you can pick 2 in one way (you must pick the 2 you have), you can pick a 3 in two ways (either pick one 3 or two 3s) and you can pick a 5 in three ways (pick no 5, one 5 or two 5s). Hence you get 1*2*3 = 6 ways

Say N = 2^3 * 5^2 * 7^2

How many factors are even? For an even factor, you need at least one 2.

So you can pick a 2 in three ways (pick one 2, two 2s or three 2s), a 5 in three ways (no 5, one 5 and two 5s) and a 7 in three ways too. Hence you get 3*3*3 = 27 even factors.

Hope this is clear.

Say N = 2^3 * 5^2 * 7^2

so to find the factors of 10, I must have one 2 and one 5

One 2 – 3 ways

One 5 – 2 ways

7 are not compulsory so I can have one 7 or 2 seven or no seven too! So there are 3 ways to pick a 7

Hence 3*2*3= 18 factors of 10.

How many factors of 7

One seven is compulsory – 2 ways to pick seven (we cannot have zero 7 case here)

2 not compulsory – So apart from the three 2′s we have we can also have a NO 2 or zero 2 case- so 4 ways ( zero 2, one 2 , two 2 and three 2′s )

5 not compulsory – So we can have zero 5′s, one 5 , or two 5′s -> that’s 3 ways

So 2*4*3= 24

Let’s push a little more

How many factors contain 35?

Got to have a 5 and a 7

5′s – > 2 ways only, cannot have zero 5′s for factors of 35, can we?

7′s-> 2 ways only, cannot have zero 7′s for factors of 35, can we?

2′s -> not compulsory, do not need a 2 for factors of 35 , but 2 can we there also , so zero 2,one 2, two 2, and three 2′s, we cannot have four 2′s as max power of 2 is 3 .So total 4 ways of selecting the 2

So – 2*2*4= 16 so the number N = 2^3 * 5^2 * 7^2 has 16 factors that contain 35.

Well if these are correct then I have understood your explanation.

Incidentally the method in the Link that I posted initially does the same thing, although the meaning may not be clear in that method.

So what are your views?

a) Have I understood your technique, the proper technique?

b) Can we use the method in the GC link as a short cut method; it does seem to give the same result as the proper method every time.

Thank you.

Yes, all the solutions are correct.

Also, the GC link gives the same method with the same thought process. Just how you denote it is very slightly different.

N = 2^3 * 5^2 * 7^2

Now if you need even factors, you can either say that I must have a 2 so 3 ways for 2 and 3 ways each for 5 and 7 so 3*3*3 = 27 ways

or you can say that keep a 2 separately (since we must have it)

2 ( 2^2 * 5^2 * 7^2) and find the factors of the leftover 3 ways for 2, 3 ways for 5 and 3 ways for 7 = 3*3*3 = 27

When you use it in the actual exam, you should not need to write down anything other than 3*3*3 so which method you use is immaterial.