Several months ago, I wrote an article about remainders. Because the concepts of division, quotients, and remainders show up so often on the GMAT, I thought it would be useful to revisit this frequently-tested topic. (After all, there always remains at least something unsaid about remainders, right?) When you encounter quotient/remainder problems on the GMAT, at times, it will be helpful to know the kind of division terminology we’re taught in grade school – in particular the quotient + remainder formula you’ll see detailed below – while at other times, you will simply want to select simple numbers that satisfy the parameters of a Data Sufficiency statement.
To ensure that you’re prepared for all types of quotient/remainder problems on the GMAT, let’s explore each of these division-related scenarios in a little more detail. A simple example can illustrate the important division terminology: if we divide 7 by 4, we’ll have 7/4 = 1 + 3/4.
7, the term we’re dividing by something else, is called the dividend. 4, which is doing the dividing, is called the divisor. 1, the whole number component of the mixed fraction, is the quotient. And 3 is the remainder. This probably feels familiar even if the terminology takes a little reminding to come back to you.
In the abstract, the classic remainder formula is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get another helpful variant of the remainder formula: Dividend = Quotient*Divisor + Remainder.
Simply knowing this terminology and the remainder equation will be sufficient to answer the following official GMAT question:
When N is divided by T, the quotient is S and the remainder is V. Which of the following expressions is equal to N?
B) S + V
C) ST + V
E) T(S – V)
In this division problem, N – which is getting divided by something else – is our dividend, T is the divisor, S is the quotient, and V is the remainder. Plugging the variables into our remainder equation of Dividend = Quotient*Divisor + Remainder, we get N = ST + V… and we’re done! The answer is C.
(Note that if you forgot the remainder formula, you could also pick simple numbers to solve this problem. Say N = 7 and T = 3. 7/3 = 2 + 1/3. The Quotient is 2, and the remainder is 1, so V = 1. Now, if we plug in 3 for T, 2 for S, and 1 for V, we’ll want an N of 7. Answer choice C will give us an N of 7, 2*3 + 1 = 7, so this is correct.)
When we need to generate a list of potential values to test in a data sufficiency question, often a statement will give us information about the dividend in terms of the divisor and the remainder.
Consider, for example, the following question: when x is divided by 5, the remainder is 4. Here, the dividend is x, the divisor is 5, and the remainder is 4. We don’t know the quotient, so we’ll just call it q. In equation form, it will look like this: x = 5q + 4. Now we can generate values for x by picking values for q, bearing in mind that the quotient must be a non-negative integer.
If q = 0, x = 4. If q = 1, x = 9. If q=2, x = 14. Notice the pattern that emerges with our x values: x = 4 or 9 or 14, or 19… In essence, the first allowable value of x is the remainder. Afterwards, we’re simply adding the divisor, 5, over and over. Without much math at all, you could continue this cycle indefinitely: 4, 9, 14, 19, 24, 29, etc. This is a handy shortcut to use in complicated data sufficiency problems, such as the following:
If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?
1) When x – y is divided by 5, the remainder is 1
2) When x + y is divided by 5, the remainder is 2
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
(C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
(D) EACH statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient
In this problem, Statement 1 gives us potential values for x – y. Remember the pattern discussed above: x – y must be 1 greater than a multiple of 5. If we begin with the remainder (1) and continually add the divisor (5), we know that x – y = 1 or 6 or 11, etc. If x – y = 1, we can say that x = 1 and y = 0. In this case, x^2 + y^2 = 1 + 0 = 1, and the remainder when 1 is divided by 5 is 1. If x – y = 6, then we can say that x = 7 and y = 1. Now x^2 + y^2 = 49 + 1 = 50, and the remainder when 50 is divided by 5 is 0. Because the remainder changes from one scenario to another, Statement 1 is not sufficient by itself.
Statement 2 gives us potential values for x + y. Once again, let’s use that pattern: x + y must be 2 greater than a multiple of 5. If we begin with the remainder (2) and continually add the divisor (5), we know that x + y = 2 or 7 or 12, etc. If x + y = 2, we can say that x = 1 and y = 1. In this case, x^2 + y^2 = 1 + 1 = 2, and the remainder when 2 is divided by 5 is 2. If x + y = 7, then we can say that x = 7 and y = 0. Now x^2 + y^2 = 49 + 0 = 49, and the remainder when 49 is divided by 5 is 4. Because the remainder changes from one scenario to another, Statement 2 is also not sufficient on its own.
Now that we find ourselves in the classic C or E scenario, let’s test them together – simply select one scenario from Statement 1 and one scenario from Statement 2 and see what happens. Say x – y = 1 and x + y = 7. Adding these equations, we get 2x = 8, or x = 4. If x = 4, y = 3. Now x^2 + y^2 = 16 + 9 = 25, and the remainder when 25 is divided by 5 is 0.
Now remember this: for us to pick a non-E answer on Data Sufficiency, we must know that the value will stay the same in any scenario allowed by the question and statements. To be safe, let’s try another scenario. Say x – y = 6 and x + y = 12. Adding the equations, we get 2x = 18, or x = 9. If x = 9, y = 3, and x^2 + y^2 = 81 + 9 = 90. The remainder when 90 is divided by 5 is, again, 0. No matter which values we select, this will be the case – we can prove definitively that the remainder is 0. Together, the statements are sufficient, so the correct answer is C.
Now let’s summarize some important takeaways regarding GMAT quotient/remainder problems and the ever-important remainder formula. You’re virtually guaranteed to see remainder questions on the GMAT, so you want to make sure you have this concept mastered. First, make sure you feel comfortable with the remainder formula: Dividend = Divisor*Quotient + Remainder. Second, if you need to select values, you can simply start with the remainder and then add the divisor over and over again. If you internalize these two ideas, remainder questions will become considerably less daunting.
Remember, also, that division is something you were once quite good at as a primary school student, so do not let the terminology intimidate you as an adult! The GMAT thrives on abstraction in these problems, so if you find yourself distracted by terminology or abstraction, simply try using small numbers to remind yourself how the operation works. The remainder equation Dividend = Divisor*Quotient + Remainder is an important one, but if you blank on it you can reconstruct it. Try, as we did at the beginning, 7 divided by 4. The result of that is 1, remainder 3. And the quotient is 1 because 4 goes into 7 one time, leaving 3 left over. So you can reconstruct the equation: to get back to 7, multiply the divisor (4) by the 1 time it went in to 7, and then add back the remaining 3: 7 = 4(1) + 3. Once you’ve stripped away the abstraction in quotient/remainder problems, the remainder is a concept you’ve been quite adept with your whole life!
Interested in more practice with GMAT division problems and the remainder equation? Check out some of our other articles on this frequently-tested GMAT topic, or try your hand at some practice questions via the Veritas Prep GMAT Question Bank or practice tests.
*GMATPrep questions courtesy of the Graduate Management Admissions Council.