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)*