Our Divisibility and Remainders module 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*2 = 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*4 = 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)*(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.

Answer (D)

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