# All About Negative Remainders on the GMAT I could have sworn that I had discussed negative remainders on my blog but the other day I was looking for a post discussing it and much as I would try, I could not find one. I am a little surprised since this concept is quite useful and I should have taken it in detail while discussing divisibility. Though we did have a fleeting discussion of it here.

Since we did miss it, we will discuss it in detail today but you must review the link given above before we proceed.

Consider this: When n is divided by 3, it leaves a remainder 1.

This means that when we divide n balls in groups of 3 balls each, we are left with 1 ball. This also means that n is 1 MORE than a multiple of 3. Or, it also means that n is 2 less than the next multiple of 3, doesn’t it?

Say, n is 16. When you split 16 balls into groups of 3 balls each, you get 5 groups of 3 balls each and there is one ball leftover. n is 1 more than a multiple of 3 (the multiple being 15). But we can also say that it is 2 LESS than the next multiple of 3 (which is 18). Hence, the negative remainder in this case is -2 which is equivalent to a positive remainder of 1.

Generally speaking, if n is divided by m and it leaves a remainder r, the negative remainder in this case is -(m – r).

When n is divided by 7, it leaves a remainder of 4. This is equivalent to a remainder of -3.

n is 3 more than a multiple of m. It is also 2 less than the next multiple of m. This means m is 5.

This concept is very useful to us sometimes, especially when the divisor and the remainder are big numbers.

Let’s take a question to see how.

Question 1: What is the remainder when 1555 * 1557 * 1559 is divided by 13?

(A)   0

(B)   2

(C)   4

(D)   9

(E)    11

Solution: Since it is a GMAT question (a question for which we will have no calculator), multiplying the 3 numbers and then dividing by 13 is absolutely out of question! There has to be another method.

Say n = 1555 * 1557 * 1559

When we divide 1555 by 13, we get a quotient of 119 (irrelevant to our question) and remainder of 8. So the remainder when we divide 1557 by 13 will be 8+2 = 10 (since 1557 is 2 more than 1555) and when we divide 1559 by 13, the remainder will be 10+2 = 12 (since 1559 is 2 more than 1557).

So n = (13*119 + 8)*(13*119 + 10)*(13*119 + 12) (you can choose to ignore the quotient and just write it as ‘a’ since it is irrelevant to our discussion)

So we need to find the remainder when n is divided by 13.

Note that when we multiply these factors, all terms we obtain will have 13 in them except the last term which is obtained by multiplying the constants together i.e. 8*10*12.

Since all other terms are multiples of 13, we can say that n is 8*10*12 (= 960) more than a multiple of 13. There are many more groups of 13 balls that we can form out of 960.

960 divided by 13 gives a remainder of 11.

Hence n is actually 11 more than a multiple of 13.

We did not use the negative remainders concept here. Let’s see how using negative remainders makes our calculations easier here. The remainder of 8, 10 and 12 imply that the negative remainders are -5, -3 and -1 respectively.

Now n = (13a – 5) * (13a – 3) * (13a – 1)

The last term in this case is -5*-3*-1 = -15

This means that n is 15 less than a multiple of 13 i.e. actually 2 less than a multiple of 13 because when you go back 13 steps, you get another multiple of 13. This gives us a negative remainder of -2 which means the positive remainder in this case will be 11.

Here we avoided some big calculations.

I will leave you now with a question which you should try to solve using negative remainders.

Question 2: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Hint: I solved this question orally in a few secs using cyclicity and negative remainders. Don’t get lost in calculations!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!