Welcome back to our series on number properties. In this article, we deal with the fallout of failed divisibility: remainders. Remainders exist whenever an integer is divided by another integer that is not one of its factors. When integer n is divided by integer x, the remainder is the difference between n and the nearest multiple of x that is less than n.

If the quotient of division is represented as a mixed fraction, the numerator of the fraction is the remainder.

24 / 5 = 4 ⅘ → remainder 4

38 / 7 = 5 ⅜ → remainder 3

61 / 6 = 10 ⅙ → remainder 1

If we “multiply back” any of these examples, we can see that the remainder is what is “missing” from n.

24 / 5  →  5 * 4 = 20  →  24 – 20 = 4 (remainder)

38 / 7  →  7 * 5 = 35  →  38 – 35 = 3 (remainder)

61 / 6  →  6 * 10 = 60  →  61 – 60 = 1 (remainder)

A remainder must always be less than the divisor in the operation. To generalize this, in the division n/x, where n and x are integers, the remainder is an integer less than x. If x is a factor of n (that is, n is divisible by x), there is no remainder or the remainder equals 0.

GMAT quant problems incorporate remainders in a variety of ways. Here’s an official problem for practice:

The dial shown above is divided into equal-sized intervals. At which of the following letters will the pointer stop if it is rotated clockwise from S through 1,174 intervals?

(A) A

(B) B

(C) C

(D) D

(E) E

Yes, this is the problem I meant to show you. Many GMAT remainder problems don’t use the term “remainder” or even tell you that division is happening. But the remainder is nonetheless the controlling concept for this problem.

The dial in the problem is divided into 8 intervals. It doesn’t matter how many times the pointer goes around the dial; it only matters where it stops on its last “lap.”

Since there are 8 intervals, every multiple of 8 represents the pointer back at its starting point of S. What matters, then, is the difference between 1,174 and the highest multiple of 8 before 1,174. In other words, we need the remainder of 1,1174/8.

Since 1000 is divisible by 8 (1000 / 8 = 250), we need only consider 174. An easy way to go about it is to consider that 160 is a multiple of 8 and then count by 8s up to 174.

So a multiple of 8 occurs at 1,160, another occurs at 1,168, and another at 1,176. 1,176 is greater than 1,174, so the remainder we’re looking for is 1,174 – 1,168 = 6. On its final “lap,” the pointer rotates clockwise through 6 intervals, stopping at E.

Here’s another official problem to try:

Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than members?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Another remainder problem wearing a wig and a fake mustache. Let’s remove the disguise:

10 < x < 40

4 members per table with 3 at one table  →  x/4 has remainder 3

5 members per table with 3 at one table  →  x/5 has remainder 3

What is the remainder of x/6?

Interestingly, x/4 and x/5 have an equal remainder of 3. This means that x is 3 greater than some multiple of 4, and x is also 3 greater than some multiple of 5. In other words, the integer (x – 3) is a multiple of 4 and a multiple of 5. The only number in the given range that satisfies this condition is 20.

So Club X has 23 members. Sometimes the first 20 members are seated at 5 tables, 4 per table. And sometimes they are seated at 4 tables, 5 per table. Either way, the remaining 3 members get their own table.

The final step for this problem is to find the remainder of x/6. If the 23 members of Club X sit 6 per table, they will be able to seat 18 members at 3 tables (6 * 3 = 18). And there will be 5 members remaining (23 – 18 = 5) to sit at the last table.

Here’s a final, similar problem to close out the article:

There are between 100 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection?

(A) 101

(B) 103

(C) 106

(E) 107

(D) 109

By now, you should be seeing right through these remainder problems. Let’s translate it like we did the last one:

100 < x < 110

x/3 has remainder 2 (x is 2 greater than the last multiple of 3)

x/4 has remainder 1 (x is 1 greater than the last multiple of 4)

Let’s map out the multiples of 3 and 4 in the given range:

Multiples of 3: 99, 102, 105, 108

Multiples of 4: 100, 104, 108

I’ve included 99 and 100 because these multiples would still allow for an x value between 100 and 110. In fact, 99 and 100 are the multiples of 3 and 4, respectively that “line up” the way we need them to. x must be 2 greater than a multiple of 3 and 1 greater than a multiple of 4. This means that x is the third of three consecutive integers where the first is a multiple of 3 and the second is a multiple of 4. 99 and 100 are the only consecutive pair in our range (we also have 105 as a multiple of 3 and 104 as a multiple of 4, but these are “backwards”), meaning x = 101. The correct answer choice is A.

This concludes our introduction of remainders and remainder problems on GMAT quant. In the next article, we’ll look at a certain kind of remainder problem involving days of the week.

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Contributor: Elijah Mize (Apex GMAT Instructor)

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