Published by Sameer Kamat at

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GMAT Maths questions are considered (primarily by engineers) to be easier than GMAT verbal questions. But underestimating the difficulty levels of the GMAT Maths section is frequently the reason why many test takers score low on the quantitative section. GoGMAT continues the GMAT preparation series on MBA Crystal Ball, by taking up sample questions related to Remainders.

Remainder questions in the GMAT, like most Number Theory questions, are particularly hard to study for. No doubt about it, this is a space where mathematics enthusiasts have an edge. If you aren’t one, focus on becoming as familiar as possible with the questions and learning common resolution methods, useful shortcuts, and tricks. More likely than not, you will be able to draw on some of that preparation to relate the question to a problem you’ve done before or a particular property you studied, so it’s certainly worth practicing.

Remainders are closely related to the factors and multiples. A remainder is the quantity that is left following the division of two integers. Hence, 10 divided by 4 is both 2.5 and 2 remainder 2. Note that remainders, when nothing is said to the contrary, are usually assumed positive, such that the remainder of -42/8 isn’t -2 but rather +6.

To know that 4 is not a factor of 10, however, one doesn’t need to know much of anything. But what if the question was, “What is the remainder when 3725 is divided by 9?” or, “What is the remainder when (19)^^{16} is divided by 10?”

For some of these kinds of problems, doing the calculation is a viable solution method, for others not so much. Even when doing the calculation will get you the answer, employing shortcuts, properties, and logical thinking can save you time—always an important goal during the GMAT.

One shortcut in eliminating answer choices or even directly determining a remainder is to ascertain whether a given number is a factor of another. Here are some useful properties that will help you with this:

- Multiples of 3 can have any digit in the last position.
**Powers**of 3 can only end in 3, 9, 7, or 1 (the cycle repeating). Finally, 3 is a**factor**of n if the sum of the digits of n is divisible by 3. **Powers**of 4 can only end in 6 or 4. Moreover, 4 is a**factor**of n if n is divisible by 2 twice.- If n is divisible by both 2 and 3 (i.e. if n is even and the sum of its digits is a multiple of 3), then 6 is a
**factor**of n.**Powers**of 6 always end in 6. - A number is divisible by 8 if it can be evenly divided by 2 three times.
- If the sum of the digits of n is divisible by 9, than 9 is a
**factor**of n. Also,**powers**of 9 (or any number ending in 9) have a final digit that alternates between 1 and 9.

Using these properties, answering the two questions above becomes easier. In the first one, note that 3+7+2+5 = 17, which is not divisible by 9. But since 18 (or 9) is divisible by 9, we know that a number 1 larger (or 8 smaller) than 3725 will have a digits sum divisible by 9. The number 3726 (like 3717) is therefore a multiple of 9, and 3725 has remainder of 8. For the second question, simply recall the property that any power of 9 (or any number ending in 9) ends in either a 1 or a 9, depending on whether the exponent is even or odd, respectively. Since (19)^^{16} therefore must end in a 1, its remainder when divided by 10 will be 1.

Now look at some more realistic GMAT Maths related remainder questions:

GMAT Maths remainder questions often come in the form of Data Sufficiency problems, simply because otherwise these kinds of problems would be too easy to solve by simply substituting answer choices without actually “solving” the problem. Here is a data sufficiency remainder question.

When the positive integer n is divided by 6, the remainder is 3. What is the remainder when n is divided by 8?

- n is divisible by 11.
- n < 100.

A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.

B) Statement (2) ALONE is sufficient, but statement (1) 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.

Get a feel for any question like this by first picking a couple of low values for n (disregarding the statements) to see whether a logical solution or pattern exists. For this one, try n=9 and n=27. Both numbers have remainder 3 when divided by 6, but the first one has remainder 1 when divided by 8, while the latter has remainder 3. When you can’t see a pattern immediately, make a table with the first few possible values of n (especially useful when the question has a limiting parameter such as “n<100”)

| 3 | 9 | 15 | 21 | 27 | 33 | 39 | 45 |

| 3 | 1 | 7 | 5 | 3 | 1 | 7 | 5 |

By now you see a pattern: the sequence “3 1 7 5” repeats. Moreover, you can see that 33 is the only value of n in the table divisible by 11. Because the values of n increase in increments of 6 and 11 is a prime number, the next value of n that divisible by 11 will be 33 + (6*11) = 99.

But the remainder of n=99 could be the same as n=33, in which case the answer would be C (or A, if it remained the same for all succeeding multiples). It is easier to find the remainder of 99/8 by doing the actual calculation, but if that were too time-consuming (e.g. if 7000<n<7100), you could notice that when n=33 the remainder is 1, and the pattern repeats itself every four values of n.

Therefore, the remainder of the 12^{th} value of n after 33 will also have remainder 1, and the 11^{th} (the one we want) will have remainder 3. Since both n=33 and n=99 fit both conditions and have different remainders, the answer is E.

** **

For which of the following values of a and b will the integer (3)^(^{5a-3b)} have a remainder of 2 when divided by 5?

A) a=6, b=2

B) a=9, b=3

C) a=12, b=3

D) a=10, b=2

E) a=2, b=2

Remember that powers of 3 end in 3, 9, 7 or 1. A number will end with a 3 if the power is 1, 5, 9, etc; in a 9 if the power is 2, 6, 10, etc; in a 7 for powers 3, 7, 11, etc and in a 1 for powers 4, 8, 12, etc.

For any digit to have a remainder of 2 when divided by 5, it must end in either a 2 or a 7. Since no integer power of 3 ends in a 2, we are looking for a power that satisfies 3 + 4K for any integer K.

In other words, 5a-3b-3 must be a multiple of 4. In option A, 5a-3b-3 yields 30-6-3=21, not a multiple of 4; Option B yields 33, also not a multiple of 4; Option C yields 48 and is therefore the correct answer.

These two example questions would likely be towards the ‘Medium’ or ‘Hard’ end of the GMAT difficulty scale. As with any number theory problem, however, the assigned difficulty may not tell you much about whether you will find the solution easy or hard. This kind of problem really depends on whether your mind connects with it or not, and that is where practice and familiarity help.

Remember to learn the properties, substitute values when you’re out of ideas and logic doesn’t help, and don’t go too far down any one solution path. If you start using a method that’s taking up time and leading nowhere fast, try something different. Or, as always, just make an educated guess.

I hope you found this information and the examples useful. Happy GMAT Preparation!

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## 2 Comments

Thanks Sameer.

3/6 leaves remainder 3, had managed to forget that somehow.

The 2nd one knocked me out, 120secs!!…ll need more honing

Thank the GoGMAT team for the post, Shubh :-)

You can reach out to them on the GMAT Maths forum and ask them to share a few more difficult GMAT Maths questions.