Remainders | Number Properties | Quantitative Aptitude

We have seen a number of posts on divisibility, odd-even concepts and perfect squares. Individually, each topic has very simple concepts but when they all come together in one question, it can be difficult to wrap one’s head around so many ideas. The tests excel at giving questions where multiple concepts are tested. Let’s take a look at one such Data Sufficiency question today:

Question 1: If ${p}$ , ${x}$ , and ${y}$ are positive integers, ${y}$ is odd, and ${p = x^2 + y^2}$ , is ${x}$ divisible by 4?

1) When ${p}$ is divided by 8, the remainder is 5.

2) ${x - y = 3}$

This Data Sufficiency question has a lot of information in the question stem. First, we need to sort through this information before we move on to the statements.

We know that $p$ , $x$ and $y$ are positive integers. $y$ is an unknown odd number, so it can be written in the form $2n + 1$ . We also know that $p = x^2 + y^2$ .

Because $y$ is written in the form $2n + 1$ , $y^2$ can be written as:

$y^2 =(2n + 1)^2$

$y^2 = 4n^2 + 4n + 1$

$y^2 = 4n(n + 1) + 1$

An interesting thing to note here is that one case of $n$ and $(n+1)$ will be odd and the other will be even. In every case, $n(n + 1)$ is even. Therefore, $y^2$ is 1 more than a multiple of 8. In other words, we can write it as $y^2 = 8m + 1$ .

Now we can say $p = x^2 + 8m + 1$ .

With this in mind, is $x$ divisible by 4? Let’s examine the statements to find out:

Statement 1: When $p$ is divided by 8, the remainder is 5.

Because $y^2 = 8m + 1$ , we can see that when $y^2$ is divided by 8, the remainder will be 1. Therefore, to get a remainder of 5 when $p$ is divided by 8, when $x^2$ is divided by 8, we should get a remainder of 4.

Now we know that $x^2$ can be written in the form $8a + 4$ (i.e. we can make “ $a$ ’” groups of 8 each and have 4 leftover).

$x^2 = 4\times(2a + 1)$

So $x = 2 \times \sqrt{an\ odd\ number}$

Note that square root of an odd number will be an odd number only. If there is no 2 in the perfect square, obviously there was no 2 in the number, too.

So, $x = 2 \times {some\ other\ odd\ number}$ , which means $x$ will be a multiple of 2, but not of 4 definitely. This statement alone is sufficient.

Now let’s look at the next statement:

Statement 2: $x - y = 3$

Since $y$ is odd, we can say that $x$ will be even (an even – an odd = an odd). But whether $x$ is divisible by 2 only or by 4 as well, we cannot say since we have no constraints on $p$ .

This statement alone is not sufficient to answer the question.

Therefore, our answer is (A).

Test takers might feel that not every step in this solution is instinctive. For example, how do we know that we should put $y^2$ in the form $4n(n+1) + 1$ ? Keep the target in mind – we know that we need to find whether $x$ is divisible by 4. Hence, try to get everything in terms of multiples of 4 + a remainder.

Keep practicing!

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