Patterns in Divisibility Problems
Today were going to tackle a couple of tough divisibility problems from GMATPrep. The two problems Ive chosen share some interesting characteristics. Heres your first one; set your timer for 2 minutes and go!
If n is a multiple of 5 and n = [pmath]p^2[/pmath]q, where p and q are prime numbers, which of the following must be a multiple of 25?(A) [pmath]p^2[/pmath]
(B) [pmath]q^2[/pmath]
(C) pq
(D)[pmath]p^2*q^2[/pmath]
(E) [pmath]p^3[/pmath]q
Hmm. So p and q are primes. They could be 2, 3, 5, 7 or so on. It doesnt say that p and q are different prime numbers, so they could also be the same number. And Im going to use some theory here: if[pmath]p^2[/pmath]q 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 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 <something that does NOT contain a 4>
We cant do it. If the two sides are equal, they must simplify down to the same primes:
56 = 2 2 2 7
8 7 = 2 2 2 7
Okay, so thats how I know that if the n, on the left side, contains a 5 as a factor, then the [pmath]p^2[/pmath]q 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 wed need[pmath]p^2[/pmath] to get 25. And if q is 5, then wed need[pmath]q^2[/pmath] to get 25.
So how can we guarantee that we get 25? Have both of those! With[pmath]p^2*q^2[/pmath], it doesnt matter whether the p is 5 or the q is 5 either way, we square whichever one is 5 to get 25!
The correct answer is D.
The key takeaway here is this: when were 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 =[pmath]n^3[/pmath], which of the following must be an integer?I. [pmath]y/{(3)(2^2)(5)}[/pmath]
II. [pmath]y/{(3^2)(2)(5)}[/pmath]
III. [pmath]y/{(3)(2)(5^2)}[/pmath]
(A) None
(B) I only
(C) II only
(D) III only
(E) I, II, and III
Oh, yay. A roman numeral question. (Yes, thats sarcasm.)
Were told that we have positive integers and theres an equation. Now, whats 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 dont 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. Lets see if that helps us at all on this one.
450 = 45*10 = 3 3 5 2 5
[pmath]n^3[/pmath] = n n n
Hmm. So, heres what weve got (Im going to rearrange the numbers that make up 450 to put them in order):
2 3 3 5 5 y = n n n
Now what? Hmm, whats the significance of having three ns on the right-hand side? Each n is the same number and the ns on the right have to contain all of the same pieces as the stuff on the left
Thats interesting. Each n has to contain the exact same thing, and I cant break primes down any further. So if theres one 2 on the left, there actually has to be at least three 2s, because each n needs to get its own 2.
Too theoretical? Try a simpler, real-numbers-only example.
2 2 2 5 5 5 = b b 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 5.
In our harder problem, with the y, we dont have all of the numbers on the left because weve got this variable y. So we basically have to infer that, if theres one 2, there must be at least three 2s (so that each n can have its own 2). By the same token, if there are two 3s, there actually must be at least three 3s, and ditto for the 5s.
Where are those other numbers? They have to be part of the y variable. So the y has to contain at least two 2s, one 3, and one 5. (Technically, it could contain even more maybe it contains three 17s as well! But the question asks us what must be true, so we dont have to speculate about what else we might have.)
If the y must contain two 2s, one 3, and one 5, then how can we use that knowledge to evaluate the three roman numerals? First, lets represent y mathematically:
y = 2 2 3 5 ?
We know we have those 4 numbers plus there might be other stuff, so well 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?
Lets see:
I. [pmath]{2*2*3*5*?}/{(3)(2^2)(5)}[/pmath]
Start crossing stuff off. Hey, I can definitely eliminate everything on the bottom! That means I dont 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. Thatll 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.)
Lets try roman numeral II.
II. [pmath]{2*2*3*5*?}/{(3^2)(2)(5)}[/pmath]
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 dont know for sure. So theres a possibility that Im left with a 3 on the bottom of this fraction which means I wont get an integer. This one isnt a MUST be true statement, so I can eliminate answer E.
The correct answer is B.
Lets look at roman numeral III to be thorough (though on the real test, dont do more work than you have to do!).
III. [pmath]{2*2*3*5*?}/{(3)(2)(5^2)}[/pmath]
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 dont 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 were given an equation, then were 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 dont 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.