“How many odd factors does 210 have?”

“If n is the smallest integer such that 432 times n is the square of an integer, what is the value of y?”

“How many prime numbers are factors of 33150?”

If questions like these make you cringe, I’d like to convince you that only a few easy-to-understand concepts stand between you and doing these flawlessly.

## Idea #1: Prime Numbers

This is probably review, but just for a refresher: a prime number is any positive integer that is divisible by only 1 and itself. In other words, a prime number has only two factors: itself and 1. Numbers that have more than two factors are called composite. (By mathematical convention, 1 is the only positive integer considered neither prime nor composite.) The first few prime numbers are:

2 3 5 7 11 13 17 19 23 29

In preparation for the GMAT, it would be good to be familiar with this list. If you verify for yourself why each number from 2 to 30 is prime or composite, it will help you remember this list.

Occasionally, the GMAT will expect you know whether a larger two-digit number, like 67, is prime. Of course, if the number is even, it’s not prime. If it ends in a digit of 5, it’s divisible by five. For divisibility by 3, a good trick to know: if the sum of the digits is divisible by three, then the number is divisible by three. Here 6 + 7 = 13, not divisible by three, so 67 is not divisible by three.

To see whether a number less than 100 is prime, all we have to do is see whether it is divisible by one of the single digit prime numbers: 2, 3, 5, or 7. We’ve already checked 2, 3, and 5. The number 67 is not divisible by 7: 7 goes evenly into 63 and 70, not 67. That’s enough checking to establish irrevocably that 67 is prime.

## Idea #2: Prime Factorization

Every positive integer greater than 1 can be written in a unique way as a product of prime numbers; this is called its **prime factorization**. The prime factorization is analogous to the DNA of the number, the unique blueprint by which to construct the number. In other words, when you calculate the prime factorization of a number, you have some powerful information at your disposal.

How does one calculate the prime factorization of a number? In grade school, you may remember making “factor trees”: that’s the idea. To find the prime factorization of, for example, 48, we simply choose any two factors — say 6 and 8 — and then choose factors of those number, and then of those numbers, until we are left with nothing but primes.

Typically, once we are done, we sort the prime factors in numerical order:

Once we have the prime factorization, what can we do with it? See the next two items.

## Idea #3: The Number of Factors

Suppose the GMAT asks: how many factors does 1440 have? It would be quite tedious to count them all, but there’s a fast trick once you have the prime factorization. First of all, the prime factorization of 1440 is

Each prime factor has an exponent (the exponent of 5 is 1).

To find the total number of factors:

a) Find the list of exponents in the prime factorization — here {5, 2, 1}

b) Add one to each number on the list — here {6, 3, 2}

c) Multiply those together —

The number 1440 has thirty-six factors, including 1 and itself.

Suppose the GMAT asked the number of odd factors of 1440. We know that odd factors cannot contain any factor of 2 at all, so basically we repeat that procedure with all the factors except the factors of 2. Here: {2, 1} –> {3, 2} –> . The number 1440 has 6 odd factors, including 1. Just for verification, the odd factors of 1440 are

{1, 3, 5, 9, 15, and 45}

This also means it has 36 – 6 = 30 even factors.

## Idea #4: GCF and LCM

GCF = greatest common factor

LCM = least common multiple.

(Note: LCM and LCD are the same thing: a least common denominator, LCD, of two number is always their LCM.)

Suppose a GMAT Math question involves finding, say, the LCM (or LCD) of 30 and 48. There’s a very straightforward procedure to find the LCM.

- Find the prime factorizations of the two numbers: 30 = 2*3*5 and 48 = 2*2*2*2*3
- Find the factors they have in common – the product of these is the GCF. Here, the GCF = 2*3 = 6
- Express each number as the GCF*(other stuff): 30 = 6*5 and 48 = 6*8
- The LCM = GCF*(other stuff from first number)*(other stuff from the second number): LCM = 6*5*8 = 240

Let’s do one more, just for practice. Suppose, on a GMAT math problem, we need to find the LCM/LCD of 28 and 180

Step (a): 28 = 2*2*7, 180 = 2*2*3*3*5

Step (b): 28 = **2*2***7, 180 = **2*2***3*3*5; GCF = 2*2 = 4

Step (c) 28 = 4*7, 180 = 4*45

Step (d) LCM = 4*7*45 = 1260

## Practice Questions:

1) The number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 27 equal shipments by truck. What is the smallest number of boxes that could be in the warehouse?

(A) 27

(B) 33

(C) 54

(D) 81

(E) 162

2) How many odd factors does 210 have?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8

3) If n is the smallest integer such that 432 times n is the square of an integer, what is the value of n?

(A) 2

(B) 3

(C) 6

(D) 12

(E) 24

4) How many distinct prime numbers are factors of 33150?

(A) Four

(B) Five

(C) Six

(D) Seven

(E) Eight

5) If n is a positive integer, then n(n + 1)(n – 1) is

(A) even only when n is even

(B) odd only when n is even

(C) odd only when n is odd

(D) always divisible by 3

(E) always one less than a prime number

## Answers

1) C

2) E

3) B

4) B

5) D

## Explanations

1) This tells us that the number of boxes is evenly divisible by both 6 and 27; in other words, it’s a common multiple of 6 and 27. The question says: what’s the smallest value it could have? In other words, what’s the LCM of 6 and 27? (This question is one example of a real-world set-up where the question is actually asking for the LCM.)

Step (a): 6 = 2*3 27 = 3*3*3

Step (b): 6 = 2*3 27 = 3*3*3 GCF = 3

Step (c): 6 = 3*2 27 = 3*9

Step (d) LCM = 3*2*9 = 54

Thus, 54 is the LCM of 6 and 27.

**Answer: C.**

2) Start with the prime factorization: 210 = 2*3*5*7

For odd factors, we put aside the factor of two, and look at the other prime factors.

set of exponents = {1, 1, 1}

plus 1 to each = {2, 2, 2}

product = 2*2*2 = 8

Therefore, there are 8 odd factors of 210. In case you are curious, they are {1, 3, 5, 7, 15, 21, 35, and 105}

**Answer: E.**

3) The prime factorization of a square has to have even powers of all its prime factors. If the original number has a factor, say of 7, then when it’s squared, the square will have a factor of 7^2. Another way to say that is: any positive integer all of whose prime factors have even powers must be a perfect square of some other integer. Look at the prime factorization of 432

432 = (2^4)*(3^3)

The factor of 2 already has an even power —- that’s all set. The factor of 3 currently has an odd power. If n = 3, then 432*n would have an even power of 2 and an even power of 3; therefore, it would be a perfect square. Thus, n = 3 is a choice that makes 432*n a perfect square.

**Answer: B.**

4) Start with the prime factorization:

33150 = 50*663 = (2*5*5)*3*221 = (2)*(3)*(5^2)*(13)*(17)

There are five distinct prime factors, {2, 3, 5, 13, and 17}

**Answer: B.**

5) Notice that (n – 1) and n and (n + 1) are three consecutive integers. This question is about the product of three consecutive integers.

If n is even, then this product will be (odd)*(even)*(odd) = even

If n is odd, this this product will be (even)*(odd)*(even) – even

No matter what, the product is even. Therefore, answers (A) & (B) & (C) are all out.

Let’s look at a couple examples, to get a feel for this

3*4*5 = 60

4*5*6 = 120

5*6*7 = 210

6*7*8 = 336

7*8*9 = 504

Notice that one of the three numbers always has to be a multiple of 3: when you take any three consecutive integers, one of them is always a multiple of 3. Therefore the product will always be divisible by 3.

Therefore, **Answer: D.**

BTW, for answer choice E of that question, you will notice that for some trios of positives integers, adding one to the product does result in a prime, but for others, it doesn’t.

3*4*5 + 1 = 61 = prime

4*5*6 + 1 = 121 = 11^2 (not prime)

5*6*7 + 1 = 211 = prime

6*7*8 + 1 = 337 = prime

7*8*9 + 1 = 505 = 5*101 (not prime)

This is a mathematical idea far far more advanced than anything on the GMAT, but it is mathematically impossible to create an easy rule or formula that will always result in prime numbers. The prime numbers follow an astonishingly complicated pattern, which is the subject of the single hardest unanswered question in modern mathematics: the Riemann Hypothesis. Fascinating stuff for leisure reading, but absolutely 100% not needed for the GMAT :).

What will be the difference in the number of factors of N^2013 and N^2015 ?

Dear Devyani,

Normally, we don’t answer outside questions in blog comments, but I’ll address this, because it’s quick. My friend, the question as you pose it is not well-defined. The number of factors would depend very much on N. If N is a prime number, then N^2015 has only two more factors — N^2014 and N^2015. By contrast, if N were a number loaded with factors, then the number of new factors in N^2015 would increase considerably. Even if N has just two prime factors (e.g. N = 6 or N = 15), the number of new factors would be in the 1000s. The question really doesn’t make sense without specifying N.

Does all this make sense?

Mike

Hello,

Thanks a lot for your very helpful explanations. This is pertaining to a question on the GMAT review number 13 on the diagnostic test. Although I have read the topic on remainder and tried the practice questions therein I still was unable to apply the format used to solve this particular question: if s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t?

Please can you help me with this.

Regards,

Rachael

Dear Rachael,

Normally, we do not answer outside questions on the blog, even official questions, but your question is so directly pertinent to this blog article that I decided to answer it here.

Let R be the remainder. R/t = 0.12 = 12/100 = 3/25. So, it could be that R = 3 and t = 25, or they both could be multiples of those numbers. (3k)/(25k). R would absolutely have to be a multiple of 3. The only multiple of 3 is

(E)45.Does this make sense?

Mike

Hey Mike,

From Practice Question No. 5’s explanation, is it safe to assume that for a set of N consecutive integers, their product will ALWAYS be divisible by N?

Thanks!

Dear Karan,

Yes, you are perfectly correct. Even if the set of N consecutive integers includes 0, the product will be zero, and zero is divisible by every integer. You are 100% correct.

Mike

There’s a typo in Idea #3: The Number of Factors

the number should be 1440 and not 1400

Best

Dear Leszek,

Thanks. I just fixed that! Best of luck to you, my friend.

Mike

Hey Mike!

Thanks for the post! It’s really helpful!

I have an issue with Question #5.

If n=1, the result would be 1*2*0 = 0, which is not divisible by 3.

Then there would ne no correct answer.

Maybe shoud mention n is positive integer >1?

Am I wrong?

Thanks a lot!

William

Dear William,

Zero is divisible by every positive integer, precisely because zero is a multiple of every positive integer.

Mike

Hi. My problem with LCM and GCF (word problems) is that I don’t know when to use each.how can I tell if they are asking me to calculate the LCM or the GCF? any language tricks????!!!! plz clarify.Thanks.

Dear Sanjoy,

I’m happy to respond. It’s true, these two ideas can show up in many different ways. I guess the only think I would say is: if the problem is collecting smaller groups to make a larger group, and asks what’s the smallest larger group that would do X — chances are good that’s a LCM problem. If we are adjusting the size of the smaller group, trying to make it as large as possible to make up some larger group, that’s a GCF. Does this make sense?

Mike

Hi,

I have a question regarding idea #3, I am having hard time understanding the concept here when you said that ” each prime factor has an exponent”. I am not seeing where these numbers coming from. can you help me out here? thanks,

Vivian,

I’m happy to help.

The number 432 is even, so let’s start by dividing it by 2.

432/2 = 216

Well, here, it would be helpful to have memorized that 6^3 = 216. I think it’s good to have memorized the perfect cubes of the first ten positive integers. But suppose you don’t have that memorized. Well, 216 is still even, so keep dividing by 2

216/2 = 108

108/2 = 54

54/2 = 27

That was four divisions by two, so the original number has four factors of two:

432 = (2^4)*27

Now, 27 is just 3^3 — that’s one you definitely have to have memorized! That gives us

432 = (2^4)(3^3)

All of those calculations you should be able to do without a calculator. Once we have it in this form, we can see the exponents, and see which ones are even or odd.

Does all this make sense?

Mike

Mike – great blog post as always. Quick question – how come there is a good trick to finding the number of odd factors in an equation, but not even factors? In the example at the beginning of the post, there were 3×2=6 odd factors but not 6 even factors (5+1). Instead you had to find the total number of factors and subtract the number of odd factors to find even factors. Thank you!!

Misty,

It’s easy to count all the factors. It’s also easy to count all the odd factors, because we simply exclude the 2’s and count everything else. For even factors, we would have to include 2 (or powers of 2) and count everything that includes it: it just a more complicated procedure, especially if there are several powers of two involves. The subtraction trick is easy to explain and works 100% of the time.

Technically, if there are n powers of 2 in the prime factorization of the number, then 1/(n + 1) of the total factors will be odd, and n/(n + 1) of the total factors will be even. Now, would you rather remember that, or just the simple subtraction trick?

Does all this make sense?

Mike

Makes sense to me! I wouldn’t have realized it unless you pointed out the subtraction method at the beginning of the post, which I am glad you did. Perhaps it’s something worth highlighting in future posts/discussions on this topic. Thanks again Mike and keep up the great work!

Misty,

You are quite welcome. Best of luck to you, my friend.

Mike

Is problem 5, what if value of n is 1. Then the whole equation turns out to be zero. The answer option D does not seem to be correct in all situations.

Dear Kumar,

This is a little appreciated mathematical fact, but zero is a multiple of every positive integer, and hence, zero is divisible by every positive integer. When we divide 0 by 3, we get a quotient of 0 with no remainder — if we can divide one number into another with no remainder, that’s the definition of divisibility. Does all this make sense?

Mike

Hello Mike. You state that 1 is not prime. You also state that all positive integers can be written as a product of prime numbers. You did not intend to include 1, in the latter statement, did you?

Allan,

Good catch. That’s a niggling little exception that I forget to mention. I edited the text above to include this. The idea of prime factorizations and the Fundamental Theorem of Arithmetic apply to every positive integer *greater than 1*.

Mike

Hi Mike. Could you please explain how you arrived at the answer for #3 in further detail? I do not understand what you mean when you say that you “have to add additional factors for anything with odd exponents.”

Allison,

I’m happy to explain this.

As I say in the solution to #3 above, every prime factor in a perfect square must have an *even* exponent. If *any* of the prime factors has an odd exponents, then automatically the number cannot possibly be a perfect square. That’s the really BIG idea.

Given that big idea, suppose we have a problem such as #3, — given some big number, say, N = 150, by what factor do we have to multiply this number, so that the product is a perfect square?

Well, first we find the prime factorization of 150 –> (2^1)*(3^1)*(5^2). That has some odd exponents and some even exponents. The even exponents — they’re fine. The odd exponents are the problem — that’s precisely what prevents this number from being a perfect square right now. If we want to change this number, 150, into a perfect square, we have to multiply it by something that changes those odd exponents into even exponents. That’s precisely what I mean when I say that we “have to include additional factors for anything with odd exponents.” Here, the factor of 5, with an even exponent, is already all set. It’s those factors of 2 & 3, each with an odd exponent, that pose a problem and need reconjiggering. To turn this into a perfect square, we would have to bump each one of those exponents of 1 up to 2, the next biggest even number. This means, we need to multiply the original number, 150, by one factor of 2 and by one factor of 3, that is, by a factor of 6 altogether. Sure enough, if we multiply 150 by 6, we get 900, which is a perfect square, 30^2.

Any number in which some of the exponents of prime factors are odd and some are even is thereby not a perfect square. If we want to multiply it by something to turn it into a perfect square, then for each factor with an odd exponent, we need to multiply once by that factor, to bump that exponent up from odd to even. In other words, we “have to include additional factors for anything with odd exponents.”

Does all this make sense?

Mike

Thanks so much Mike, that clears it up

Allison,

You are quite welcome, my friend. Best of luck to you!

Mike

Hi Mike,

I wanted to ask a silly question. At question 4, when prime factoring 33,150, I got stuck when coming to 2 * 5 * 5 * 3 * 221.

221 can be factored into 13 * 17 which takes some work. Is there any quick way to go about these kind of numbers ???

Naifz,

My friend, that is not at all a silly question. I will admit, I kinda cheated here — I picked a number for which I happen to know the factors, but which would probably be particularly difficult for most folks to factor on the fly without a calculator. In other words, I was being a little mean here — the GMAT will not be that mean to you.

BTW, in case you’re curious, here’s a slick factoring trick, if you happen to notice it. If you happen to know that 15^2 = 225, and notice that 221 is exactly 4 less than 225 — of course, 4 = 2^2 — then we can express 221 as a Difference of Two Squares.

a^2 – b^2 = (a + b)(a – b)

221 = 225 – 4 = 15^2 – 2^2 = (15 + 2)(15 – 2) = 17*13

I discuss these factoring techniques more in this post:

http://magoosh.com/gmat/2012/advanced-non-calculator-factoring-on-the-gmat/

Again, all this is at the very outer edge of what you might have to understand if you get absolutely everything else correct on the quant section and the CAT is throwing the hardest possible questions at you. Most folks will not have to worry about this at all.

Does all this make sense?

Mike

Hi Mike,

Many thanx for your detailed explanation :). Do you think these kind of questions can come at GRE ? I was actually watching the magoosh video for counting divisors of large numbers, from there I came to your related post here.

And what is the difference between GRE and GMAT quant questions qualitatively ? I have heard GMAT math is more hard, hence, do you think practicing GMAT type questions will improve my quant score at GRE ?

Thanx again.

Dear Nafiz,

This is a notch harder than anything the GRE is likely to throw at you, except perhaps in a very very hard challenge problem if one happens to be included in your Quant section.

Yes, GMAT math is a tad harder than GRE math, and on the GMAT, you are not allowed to use a calculator, so it forces you to use all the mental math tricks and shortcuts which, if you master them, can save you a ton of time on the GRE. Many GRE students think the calculator to which they have access will help them on most problems, whereas actually, ETS designs many many problems to *punish* all who reflexively reach for the calculator rather than doing critical thinking first. Practicing GMAT math is a good way of avoiding all the “calculator traps.”

Does all this make sense?

Mike

I took more than 15 secs to realize that 221 is composite. But the derivation you provided using a^2-b^2 was wonderful. I’ll try to keep it in my mind next time I reach such a stage. Thanks

Dear Subh,

I’m glad you found that helpful. Best of luck to you.

Mike

Hiya! I just would like to give a huge thumbs up for the great information you have right here on this post. I shall be coming again to your blog for more soon.

Thank you for your compliments.

Mike

Hi Mike

In question #3, u checked straight away for 432×3 because it was odd or should we check with 2 first n then proceed?am i missing out something?because checking with 2 will kill time?is it a shortcut or something?

Taru

Dear Taru,

What I checked first were the EXPONENTS of the factors. Anything with an even exponent could be ignored, and we had to add additional factors for anything with an odd exponents. That question is all about the exponents of the prime factors. It has nothing to do with whether the factors, 2 or 3, are odd. It has everything to do with whether the exponents of the prime factors are odd.

Does that make sense?

Mike

If n is the smallest integer such that 432 times n is the square of an integer, what is the value of y?

(A) 2

(B) 3

(C) 6

(D) 12

(E) 24

I think there is a typo in this question, shouldn’t it read what is the value of N? (instead of Y)

Ooops! That’s a silly mistake I made there. I just corrected it. Thank you very much for the heads up.

Mike

“The prime factorization is analogous to the DNA of the number, the unique blueprint by which to construct the number” – Its a great analogy mike.

The examples on number of prime factors and odd factors are good.

Thank you. I’m glad you found it helpful. These are powerful strategies if you master them. Best of luck to you!

Mike