Making remainder questions easier

At times, it will be helpful to know the kind of terminology we’re taught in grade school, while at other times, we’ll simply want to select simple numbers that satisfy the parameters of a Data Sufficiency statement. So let’s explore each of these scenarios in a little more detail. A simple example can illustrate the terminology: if we divide 7 by 4, we’ll have 7/4 = 1 + ¾. 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. In the abstract, the equation is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get: Dividend = Quotient*Divisor + Remainder. Simply knowing this terminology will be sufficient to answer the following official 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?  A) ST B) S + V C) ST + V D) T(S+V) E) T(S – V)  In this 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 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 equation, 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. Take the following example: 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 in our x values: x = 4 or 9 or 14… In essence, the first allowable value of x is the remainder. Afterwards, we’re simply adding the divisor, 5, over and over. 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 In this problem, Statement 1 gives us potential values for x – y. 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 alone. Statement 2 gives us potential values for x + y. 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 alone. Now 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. We need to see if this will ever change, so 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 what we select, this will be the case – we know definitively that the remainder is 0. Together the statements are sufficient, so the answer is C. Takeaway: You’re virtually guaranteed to see remainder questions on the CAT, so you want to make sure you have this concept mastered. First, make sure you feel comfortable with the following equation: 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.