### Patterns in Divisibility Problems

Patterns in Divisibility Problems

Today we’re going to tackle a couple of tough divisibility problems from GMATPrep. The two problems I’ve chosen share some interesting characteristics. Here’s your first one; set your timer for 2 minutes and go!

If n is a multiple of 5 and n = p 2 q , where p and q are prime numbers, which of the following must be a multiple of 25?”
(A)  p 2

(B)  q 2

(C) p q

(D) p2q2

(E) p3q

Hmm. So p and q are primes. They could be 2, 3, 5, 7 or so on. It doesn’t say that p and q are different prime numbers, so they could also be the same number. And I’m going to use some theory here: if p2q equals n and n is a multiple of 5, then that 5 must be contained in either p or q. And since those two numbers are primes, either p or q is 5. (Or maybe both are!)

How did I figure that out? Look at this (totally different) example:

56 = 8 Ã— 7

56 is a multiple of 4. Do either of the two numbers on the right-hand side contain a 4? Yes, 8 contains a 4.

Come up with an equation like the one above in which the number on the left is a multiple of some number, but you CANNOT find that number in one of the numbers on the right.

56 = 7 Ã— [something that does NOT contain a 4]

We can’t do it. If the two sides are equal, they must simplify down to the same primes:

56 = 2 Ã— 2 Ã— 2 Ã— 7

8 Ã— 7 = 2 Ã— 2 Ã— 2 Ã— 7

Okay, so that’s how I know that if the n, on the left side, contains a 5 as a factor, then the p2q on the right side also has to contain a 5 somewhere. Since we were told that p and q are both primes, we know that at least one is equal to 5.

Which one? We have no idea. Now, the problem is asking us how to get 25. Well, if p is 5, then we’d need p2 to get 25. And if q is 5, then we’d need q2 to get 25.

So how can we guarantee that we get 25? Have both of those! With p2q2, it doesn’t matter whether the p is 5 or the q is 5 “ either way, we square whichever one is 5 to get 25!

The key takeaway here is this: when we’re told two expressions equal each other, then those two expressions must simplify down to the same pieces (on each side of the equation). Remember that.

Ready for the next one? You know the drill (2 minutes!).

If n and y are positive integers and 450y = n3, which of the following must be an integer?”

I.

II.

III.

(A) None
(B) I only
(C) II only
(D) III only
(E) I, II, and III

Oh, yay. A roman numeral question. (Yes, that’s sarcasm.)

We’re told that we have positive integers and there’s an equation. Now, what’s the question asking? Which of the following must be an integer okay, and then each of the roman numerals contains y divided by some primes oh, I get it! If doing the division results in an integer, then that means y is divisible by those primes. This is actually a divisibility question, too, even though they don’t say that word in the problem! Also, this is a MUST be true question, just like the last one.

So, what do we do with the equation, again, on divisibility questions? Ah yes “ that means the left-hand side must ultimately break down to the same thing on the right-hand side. Let’s see if that helps us at all on this one.

450 = 45*10 = 3 Ã— 3 Ã— 5 Ã— 2 Ã— 5

n3 = n Ã— n Ã— n

Hmm. So, here’s what we’ve got (I’m going to rearrange the numbers that make up 450 to put them in order):

2 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— y = n Ã— n Ã— n

Now what? Hmm, what’s the significance of having three n‘s on the right-hand side? Each n is the same number and the n‘s on the right have to contain all of the same pieces as the stuff on the left

That’s interesting. Each n has to contain the exact same thing, and I can’t break primes down any further. So if there’s one 2 on the left, there actually has to be at least three 2’s, because each n needs to get its own 2.

Too theoretical? Try a simpler, real-numbers-only example.

2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5 = b Ã— b Ã— b

What is the value of just one single b? We just split each set of 3 numbers on the right “ one 2 for each n and one 5 for each n, or n = 2 Ã— 5.

In our harder problem, with the y, we don’t have all of the numbers on the left because we’ve got this variable y. So we basically have to infer that, if there’s one 2, there must be at least three 2’s (so that each n can have its own 2). By the same token, if there are two 3’s, there actually must be at least three 3’s, and ditto for the 5’s.

Where are those other numbers? They have to be part of the y variable. So the y has to contain at least two 2’s, one 3, and one 5. (Technically, it could contain even more “ maybe it contains three 17’s as well! But the question asks us what must be true, so we don’t have to speculate about what else we might have.)

If the y must contain two 2’s, one 3, and one 5, then how can we use that knowledge to evaluate the three roman numerals? First, let’s represent y mathematically:

y = 2 Ã— 2 Ã— 3 Ã— 5 Ã— ?

We know we have those 4 numbers plus there might be other stuff, so we’ll include the question mark, just in case.

If I take that and plug it into the first roman numeral, am I going to get an integer?

Let’s see:

I.

Start crossing stuff off. Hey, I can definitely eliminate everything on the bottom! That means I don’t have anything left on the denominator except for the number 1, so this must simplify to an integer. Yay! Roman numeral 1 works.

Eliminate answers A, C, and D. Excellent! Notice that we only have to test one more roman numeral, either II or III. That’ll be enough to get the answer, because it can only be either B or E at this point. If one of those roman numerals looks a lot easier to you than the other, make sure to do the easier one next. (In my opinion, these two look pretty similar, but sometimes there is a big difference in difficulty level.)

Let’s try roman numeral II.

Hmm. I can cross off the 2 and the 5. I can also cross off one of the 3s. Can I cross off the other 3? Maybe “ maybe the ? contains another 3 “ but I don’t know for sure. So there’s a possibility that I’m left with a 3 on the bottom of this fraction which means I won’t get an integer. This one isn’t a MUST be true statement, so I can eliminate answer E.

Let’s look at roman numeral III to be thorough (though on the real test, don’t do more work than you have to do!).

This one is like roman numeral II. I know I can definitely divide out the 2, the 3, and one of the 5’s on the denominator, but I don’t know that I can definitely divide out both 5s. This one is also not a MUST be true statement.

### Key Takeaways for Divisibility Problems:

(1) If the problem is about divisibility and we’re given an equation, then we’re probably going to need to break both sides of the equations down to the basic components and compare.

(2) A divisibility problem might not use normal divisibility language (divisible, multiple, factor, ). For example, it may just ask us whether we get an integer for some math setup that includes division “ we have to recognize that they really are asking about divisibility and factors.

(3) On MUST be true questions, we have to find something that is always true, not just sometimes true. There will be traps revolving around things that could be true but don’t absolutely have to be true “ watch out for those traps!

* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

1. Molly December 26, 2014 at 6:49 pm

Ã— is showing up instead of the multiplication symbol. Makes the explanation extremely confusing…Is probably a bug in your website. Please fix.

2. Business School Admissions Blog | MBA Admission Blog | Blog Archive » GMAT Impact: Number Properties on the GMAT April 11, 2013 at 4:27 pm

[…] the article “Patterns in Divisibility Problems,” we examine two GMATPrep problems that share some interesting characteristics. In this […]

3. cancer April 9, 2013 at 10:56 am

Will you be capable of manual us on your webmaster or the guy that takes care of your site, I have to know if it could be possible to be considered a guest poster.

4. Stacey Koprince November 28, 2012 at 12:40 pm

that’s a formatting typo – it should read n^3. I’ll have someone fix it!

5. Roberto November 27, 2012 at 9:07 pm

The n3 comes out like n3.

6. Yajur March 2, 2012 at 11:48 am

In the second problem. It says 3n in the question but somehow switches to n^3 halfway. Can you please clarify which one is the right answer?

Thanks,

Yajur