Mathematical Operations on Terms with Remainders | Number Properties

Our Divisibility and Remainders modules discusses all the concepts of this topic in detail. We have seen how to handle remainders with mathematical operations on terms. Let’s take a look at an application of that today.

Say “ $x$ ” gives you a remainder of 2 when divided by 6. What will be the remainder when $x + 1$ is divided by 6?

Go back to the divisibility concepts discussed. When $x$ balls are split into groups of 6, we will have 2 balls leftover. If we are given 1 more ball, it will join the 2 balls and now we will have 3 balls leftover. The remainder will be 3.

What happens in the case of $x + 6$ – what will be the remainder when this is divided by 6? This additional 6 balls will just make an extra group of 6, so we will still have 2 balls leftover.

What about the case of $x + 9$ ? Now, of the extra 9 balls, we will make one group of 6 and will have 3 balls leftover. These 3 balls will join the 2 balls leftover from $x$ , giving us a remainder of 5.

Now, what about the case of $2x$ ? Recall that $2x = x + x$ . The number of groups will double and so will the remainder, so $2x$ will give us a remainder of $2\times2 = 4$ .

On the other hand, if $x$ gives us a remainder of 4 when divided by 6, then $2x$ divided by 6 will have a remainder of $2\times4 = 8$ , which gives us a remainder of 2 (since another group of 6 will be formed from the 8 balls).

Let’s consider the tricky case of $x^2$ now. If $x$ gives us a remainder of 2 when it is divided by 6, it means:

$x = 6Q + 2$

$x^2 = (6Q + 2) \times (6Q + 2) = 36Q^2 + 24Q + 4$

Note here that the first and the second terms are divisible by 6. The remainder when you divide this by 6 will be 4.

We hope you understand how to deal with these various cases of remainders. Let’s take a look at a GMAT sample question now:

Question: If ${z}$ is a positive integer and ${r}$ is the remainder when ${z^2 + 2z + 4}$ is divided by 8, what is the value of ${r}$ ?

Statement 1: When ${(z-3)^2}$ is divided by 8, the remainder is 4.

Statement 2: When ${2z}$ is divided by 8, the remainder is 2.

This is not our typical, “When $z$ is divided by 8, $r$ is the remainder” type of question. Instead, we are given a quadratic equation in the form of $z$ that, when divided by 8, gives us a remainder of $r$ . We need to find $r$ . This question might feel complicated, but look at the statements – at least one of them gives us data on a quadratic! Looks promising!

Statement 1: When $(z-3)^2$ is divided by 8, the remainder is 4

$(z - 3)^2 = z^2 - 6z + 9$

We know that when $z^2 - 6z + 9$ is divided by 8, the remainder is 4. So no matter what $z$ is, $z^2 - 6z + 9 + 8z$ , when divided by 8, will only give us a remainder of 4 ( $8z$ is a multiple of 8, so will give remainder 0).

$z^2 - 6z + 9 + 8z = z^2 + 2z + 9$

$z^2 + 2z + 9$ when divided by 8, gives remainder 4. This means $z^2 + 2z + 5$ is divisible by 8 and would give remainder 0, further implying that $z^2 + 2z + 4$ would be 1 less than a multiple of 8, and hence, would give us a remainder of 7 when divided by 8. This statement alone is sufficient.

Let’s look at the second statement:

Statement 2: When $2z$ is divided by 8, the remainder is 2

$2z = 8a + 2$

$z = 4a + 1$

$z^2 = (4a + 1)^2 = 16a^2 + 8a + 1$

When $z^2$ is divided by 8, the remainder is 1. When $2z$ is divided by 8, the remainder is 2. So when $z^2 + 2z$ is divided by 8 the remainder will be 1+2 = 3.

When $z^2 + 2z + 4$ is divided by 8, remainder will be 3 + 4 = 7. This statement alone is also sufficient. Because both statements alone are sufficient, our answer is (D).

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