We were taught in school to think of whole numbers in the context of two groups: Odds vs. Evens. I remember thinking it was like the black vs. white pieces on a chess board (I was kind of a nerd). As I’m sure you know, an Even number is simply a number divisible by 2, and an Odd number is any number that’s not even. But ask yourself this: what is the remainder when you divide an odd number by 2? Take a minute to think about this. Try out a few different odd numbers and see if you can identify a pattern.

The remainder will always be 1 when you divide an odd number by 2. Always. And when you divide an even number by 2? Well, by definition the even number is divisible by 2, so the remainder is therefore zero.

The GMAT loves taking this concept and testing how deep your understanding goes. Therefore, we must free ourselves of the simplistic odd vs. even framework that we were fed in school, and explore this concept to a much deeper level. That is exactly what I intend to do in this blog post.

I always joke with my students that if I were the number 3 on the number line, I would really hate my next door neighbor to the left (number 2). He thinks he’s so special because there’s a name for any multiple of him (“even”); and if that’s not enough to give him a big head, they also invented a name for any number that isn’t a multiple of him (“odd”). What do you call multiples of me? “multiple of 3″. What do you call numbers that are not multiples of me? “not a multiple of 3″. LAME

The point of the joke is to help students realize that just as we can think of the world in terms of odds vs. evens, we can also think of the world in terms of multiples of 3 vs. non multiples of 3 (or multiples of 7 vs. non multiples of 7, etc.)

This is where remainders come in. You see, there is more than one kind of non multiple of 3. If you’re a whole number that is not a multiple of 3, then either you have a remainder of 1 (meaning you’re sitting immediately to the right of a multiple of 3) or you have a remainder of 2 (meaning you’re sitting two spaces to the right of a multiple of 3). What happens if you’re three spaces to the right of a multiple of 3? Well, then you’re a multiple of 3 yourself, Mr. (or Ms.) Smarty Pants!

You can think of remainders as “positioning tags” that help you place the dividend in between two multiples of the divisor (see picture of long division for definitions).

In this picture, you can see that 16 is sitting immediately to the right of a multiple of 3 (because the remainder is 1).

#### Who cares??

Try solving this question. I made it up my self, but it could totally be a real GMAT problem (if GMAC folks are reading this, please don’t steal it):

When x is divided by 7, the remainder is 5. When y is divided by 7, the remainder is 4. If x and y are positive integers, what is the greatest integer that must divide (2x+y)?

(A)  1

(B)  2

(C)  4

(D) 5

(E)  7

I wasn’t kidding, try solving it on your own before reading on!!

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I don’t know where exactly x is on the number line, and I certainly don’t know where 2x is. The only thing I do know is that x is sitting 5 spaces to the right of a multiple of 7. I can infer that 2x is sitting (2×5=)10 spaces to the right of a multiple of 7 (which I could simplify to 3 spaces to the right of the following multiple of 7). You can also plug in numbers to figure this out: some possible numbers for x are 5, 12, 19, 26 and the corresponding 2x would be 10, 24, 38, 52, which are all 3 spaces to the right of a multiple of 7.

I also know that y is 4 spaces to the right of a multiple of 7.

When I add 2x+y I get some multiple of 7 (+3) and some multiple of 7 (+4). The (+3) and (+4) complete each other to one additional multiple of 7, and therefore the number 7 fits evenly into (2x+y). The correct answer is (E).

Takeaways:

1. Everything you’ve learned about evens vs. odds is applicable to multiples vs. non multiples of any number.
2. The remainder is a positioning tag that tells you where the dividend sits in between two multiples of the divisor (more specifically: how many spaces to the right of such a multiple).
3. If you memorized that Odd + Odd=Even, now you know why that is: try to apply my solution from above to prove this rule. Put your proof in the comments for extra credit : )

#### Avi Gutman

Avi Gutman earned his MBA at NYUâ€™s Stern School of Business. While at Stern, he took on the position of teaching fellow and taught his classmates several courses including Operations, Foundations of Finance, and New Venture Financing. Avi has been teaching test prep since 2004 and has worked for several different companies before joining Manhattan GMAT.

### 5 responses to Remainders – Who Needs Them Anyway??

1. Let m and n be 2 odd nos, such that m=2a+1 and n=2b+1 (a and b being 2 positive integers). From your discussion of positioning, if we divive m and n by 2, we get a remainder of 1 in each case, implying that the nos lie 1 place on the right of an even no. When we add m and n, we get some multiples of 2 and a remainder of 1 in each case, such that – m+n=2a(+1) + 2b(+1). 2a+2b gives an even no and the (+1)+(+1) of the 2 remainders point to the next consecutive even no (+2), hence making the some of 2 odd numbers even.

Consecutively: m+n=(2a+1) + (2b+1) = 2a+2b+2 = 2(a+b+1). Since the entire number is getting multiplied by 2, it is an even no.