Speed and Accuracy Trade-Off | Number Properties

We know that speed is important in GMAT. We have about 2 mins per question and we always have questions in which we get stuck, waste 3-4 mins and probably still answer incorrectly. So we are always trying to go faster, rush, complete the easy ones in less time! In our bid to save time, sometimes we sacrifice accuracy. We should know that accuracy is most important. No point running through questions and completing all of them before time if at the end of it all, most of our answers are incorrect – there are no bonus points for completing the test before time, after all!

In your haste to complete the test on time, don’t overlook the important details. Getting too many easy questions wrong is certainly disastrous. Take a step back and ensure that what they asked is what you have found and that your logic is solid. To illustrate the problem, let’s give you a question – people gloss over it, consider it an easy remainders problem, answer it incorrectly and move on.

But guess what, it isn’t as easy as it looks!

Question: If ${m}$ and ${n}$ are positive integers such that ${m > n}$ , what is the remainder when ${m^2 - n^2}$ is divided by 21?

Statement 1: The remainder when ${(m + n)}$ is divided by 7 is 1.

Statement 2: The remainder when ${(m - n)}$ is divided by 3 is 1.

First let’s give you the incorrect solution provided by many.

Question: What is the remainder when $(m^2 - n^2)$ is divided by 21?

Statement 1: The remainder when $(m + n)$ is divided by 7 is 1.

$(m + n) = 7a + 1$

Statement 2: The remainder when $(m - n)$ is divided by 3 is 1.

$(m - n) = 3b + 1$

Therefore, remainder of product $(m^2 - n^2) = (m + n) \times (m - n) = (7a + 1)(3b + 1)$ when it is divided by 21 is 1.

Answer (C).

This would have been correct had the statements been:

Statement 1: The remainder when $(m + n)$ is divided by 21 is 1.

Statement 2: The remainder when $(m - n)$ is divided by 21 is 1.

Statement 1: $(m + n) = 21a + 1$

Statement 2: $(m - n) = 21b + 1$

( $m^2 - n^2$ ) = $(m + n)\times(m - n) = (21a + 1)\times(21b + 1) = 21\times21ab + 21a + 21b + 1$

Here, every term is divisible by 21 except the last term 1. So when we divide $(m^2 - n^2)$ by 21, the remainder will be 1.

But let’s go back to our original question. If you solved it the way given above and got the answer as (C), you are not the only one who jumped the gun. Many people end up doing just that. But here is the correct solution:

The statements given are:

Statement 1: The remainder when $(m + n)$ is divided by 7 is 1.

$(m + n) = 7a + 1$

Statement 2: The remainder when $(m - n)$ is divided by 3 is 1.

$(m - n) = 3b + 1$

This gives us $(m^2 - n^2) = (m + n) \times (m - n) = (7a + 1)(3b + 1) = 21ab + 7a + 3b + 1$

Here only the first term is divisible by 21. We have no clue about the other terms. We cannot say that 7a is divisible by 21. It may or may not be depending on the value of $a$ . Similarly, $3b$ may or may not be divisible by 21 depending on the value of $b$ . So how can we say here that the remainder must be 1? We cannot. We do not know what the remainder will be in this case even with both statements together.