If k is a positive integer and n=k(k+7), is n divisible by 6

[AceTheGM@]

I found this question in in the GMAT Prep #4 Cat Exam.

If k is a positive integer and n=k(k+7), is n divisible by 6?
(1) k is odd.
(2) When k is divided by 3, the remainder is 2.
The answer is B


I met with an MGMAT private tutor about this question, who unfortunately could only explain this question to me using the not-so-airtight method of picking numbers. I'd like to know how to solve this using Picking Numbers, Even/Odds, and the Remainder Equation.

Picking Numbers
For (1), k=1 results in "no" and k=3 results in "yes" so A is insufficient.
For (2), k=2 or 5 or 8 all result in "yes" which I guess is enough to go with B. But this isn ot reassuring in case there's a bigger number that would dispove Statement 2! How do I know if that's likely to happen?

Odds/Evens
(1) will create an scenario of (odd)(even)=even. We don't know if the even is divisible by 6, as some evens are divisible by 6 and others aren't. Cross off AD. My tutor said this was valid.
(2) You end up with (odd)(even) or (even)(even), so either way, even. But that number could be, say, 4 (no) or 24 (yes), so why isn't this considered ambiguous?

Remainder Equation
I believe there's some way to solve this using X=Qy+R. My tutor said this was the best way to go, but I don't have anything in my notes about how to apply that formula. How would you solve this using the Remainder Equation?

Thanks.

[mondegreen]

I found this question in in the GMAT Prep #4 Cat Exam.

If k is a positive integer and n=k(k+7), is n divisible by 6?
(1) k is odd.
(2) When k is divided by 3, the remainder is 2.
The answer is B

Picking Numbers
For (1), k=1 results in "no" and k=3 results in "yes" so A is insufficient.
For (2), k=2 or 5 or 8 all result in "yes" which I guess is enough to go with B. But this isn ot reassuring in case there's a bigger number that would dispove Statement 2! How do I know if that's likely to happen?

You are clear with statement 1. So let's jump to statement 2.

Given that k = 3p+2, where p is an integer.

Substituting this to the equation, we have n= (3p+2)(3p+2+7)
--> n = (3p+2)(3p+9)--> 9p^2+33p+18 --> 3p(3p+11)+18
Now, 18 is always divisible by 6. So let's just concentrate on 3p(3p+11)

Case I : p is odd --> 3*odd*(odd+odd) = 3*odd*even = 3*odd*(some multiple of 2) = 6*some integer. Sufficient.

Case II : p is even --> 3*even*(even+odd) = 3*even*odd = 3*(some multiple of 2)*odd = 6*some ineteger. Sufficient.

B.

[RonPurewal]

using the not-so-airtight method of picking numbers

Au contraire, number-picking IS "airtight" for remainders and digits.

Remainders and digits are, by definition, repeating patterns. When you plug in numbers, you'll quickly see ... repeating patterns.
When patterns emerge on a remainder/digit problem, you can trust them.

What's more, they'll emerge FAST. Very fast, since GMAT remainder problems exclusively involve division by small numbers.
If you were dividing by, say, 109, then it would take a long, long time for patterns to emerge. But the official problems are specifically written to be friendly to the number-picking method, so that's not going to happen.

[RonPurewal]

See that algebra in the post above? Quite well done.
However, GMAC will bend over backward to make sure that you don't have to do that kind of thing to solve the problems!
They don't want just "mathematicians" to do well on this thing, so they'll engineer the problems to allow things like number picking as well. Yes, they will.

In fact, it's even more than that. When DS remainder/digit problems exist, they are almost always written so that number-picking is the easiest and most efficient way to solve them.
Yes you can do algebra/theory, but it's usually somewhat annoying (as you can see above). That's actually the point of these problems--to DISadvantage the use of algebra/theory, and to advantage the use of number picking. By calling number picking "not airtight", you're pretty much barking up the wrong tree here.

Try it"”try going through all of the official remainder problems you can get your hands on (especially DS). You should find that number-picking works on every single one of them.
Every single one.
That's pretty airtight.

[RonPurewal]

By the way, if you're prone to making mistakes with algebra (like me), then you can reason your way through the possibilities instead.

First, note that, to have a multiple of 6, you need a 2 and a 3 (= the prime factors of 6) in there somewhere.

Statement 2 tells you that k is two more than a multiple of 3.
That means it's one less than the NEXT multiple of 3. (If you don't understand this right away, list out a few multiples of 3, add 2 to them, and see.)
In other words, k + 1 is a multiple of 3.
So is k + 4, and likewise k + 7.
So, the (k + 7) in your product is a multiple of 3.

The only thing that's left to worry about is the "2".
But you've got k and k + 7.
One of those is odd; the other is even. So, one of them--whichever is even--has the "2" that you need.

So you always have your 3 and your 2, so you're always a multiple of 6. Sufficient.

But much, much, MUCH more work than picking numbers.
(:

Ironically, to me personally, these theory-based methods actually seem less airtight than picking numbers.
With picking numbers, the patterns are clear, present, immediate, and obvious.
With these kinds of approaches, on the other hand, there's always that nagging thought in the back of my mind, "Wait, what if I missed a case?" With picking numbers, this is not an issue at all.

[nelosnp503]

nice post.

[tim]

:)

[RonPurewal]

nice post.

thanks

[AceTheGM@]

This helps- thanks! But I still have a question about the "not-so-airtight" aspect of picking numbers. See this part of my original post:

For (2), k=2 or 5 or 8 all result in "yes" which I guess is enough to go with B. But this isn ot reassuring in case there's a bigger number that would dispove Statement 2! How do I know if that's likely to happen?

I agree that picking numbers leaves, if done correctly, leaves no room for error. I think my anxiety comes from the fact that I am never sure if I'm forgetting to test an important a number. So in my statement above, I had time to try 2, 5, and 8. However, on other similar questions, I've only had time to try 2 numbers, both of which give a consistent answer. So then normally, I think I'm all set. The problem is when there's a much weirder number that I haven't tried plugging in, which would create an insufficiency. I usually try to choose wisely, but have run into errors in the past where picking 2 numbers appears to be sufficient but really isn't. How do I avoid that?

[RonPurewal]

I agree that picking numbers leaves, if done correctly, leaves no room for error. I think my anxiety comes from the fact that I am never sure if I'm forgetting to test an important a number. So in my statement above, I had time to try 2, 5, and 8. However, on other similar questions, I've only had time to try 2 numbers, both of which give a consistent answer.

Maybe you aren't testing the cases efficiently.

In particular, you don't have to multiply out the numbers"”you just have to see whether the factors contain a 6, or else a 2 and a 3.
For instance, in your case k = 8, you get 8 x 15. The 8 contains a 2, and the 15 contains a 3, so 8 x 15 is a multiple of 6.

With divisibility in general, you can follow such an approach. No need to multiply things out, in case that's what you were doing. (I'm guessing that's what you were doing, if you only had the time to test 3 cases.)

[RonPurewal]

The problem is when there's a much weirder number that I haven't tried plugging in, which would create an insufficiency.

Won't happen, for 2 reasons.
1/ Remainders create predictable patterns. Remainders ARE patterns.
2/ The GMAT never includes any "trick questions" of any kind. They don't stick the proverbial foot out and try to trip you up.

In any case, you should just test cases until you see a pattern that's clearly a pattern.
If this takes a bit longer than you had originally anticipated, then it's time well spent. (If you KNOW you are on your way to a legitimate solution"”"”i.e., you are 100.0000% sure, so much that you'd continue doing the work with a gun to the back of your head"”"”then you should not quit because of some arbitrary self-imposed time limit.)