Divisibility + Odd/Even + Perfect Squares | Number Properties

We know how to deal with 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. GMAT excels at giving questions where multiple concepts are tested. Let’s take a look at one such question today.

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

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

Statement 2: ${x - y = 3}$

Solution:

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

$p$ , $x$ and $y$ are positive integers.

$y$ is odd so it is of the form $2n + 1$

$p = x^2 + y^2$

$y$ is of the form $(2n + 1)$ so $y^2$ is of the form

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

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

$p = x^2 + 8m + 1$

Question – Is $x$ divisible by 4?

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

When $y^2$ is divided by 8, remainder is 1. To get a remainder of 5, when $x^2$ is divided by 8, we should get a remainder of 4.

So $x^2$ must be of 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 $x = 2 \times \sqrt{Odd\ number}$

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

So $x$ will be a multiple of 2 but not of 4 definitely.

This statement alone is sufficient.

Statement 2: $x - y = 3$

Since $y$ is odd, we can say that $x$ will be even (Even – Odd = 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.

Answer (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 + Remainder.