A Take on Takeaways | Number Properties

Once you have covered your fundamentals, we suggest you to practice advanced questions and jot down your takeaways from them. Sometimes students wonder how to find that all important “takeaway”. Today, let’s discuss how to elicit a takeaway from a question which seems to have none.

What is a takeaway? It is a small note to yourself which you would do well to remember while going for the exam. Even if you don’t remember the exact property you jotted down, knowing that such a property exists is enough. You can always try it on a couple of numbers in the test to recall the exact content.

The question we will discuss today serves another purpose – it discusses properties of squares of odd and even integers so in a sense is a continuation of our advanced number properties discussion.

Question: Given ${x}$ and ${y}$ are positive integers such that ${y}$ is odd, is ${x}$ divisible by 4?

Statement 1: When ${(x^2 + y^2)}$ is divided by 8, the remainder is 5.

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

Solution: As of now, we don’t know any specific properties of squares of odd and even integers. However, we do have a good (presumably!) understanding of divisibility. To recap quickly, divisibility is nothing but grouping. To take an example, if we divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is the quotient and 1 is the remainder. For more on these concepts, check out our divisibility module in the Products section.

Coming back to our question,

First thing that comes to mind is that if $y$ is odd, $y = (2k + 1)$ .

We have no information on $x$ so let’s proceed to the two statements.

Statement 1: When $(x^2 + y^2)$ is divided by 8, the remainder is 5.

The statement tells us something about $y^2$ so let’s get that.

If $y = (2k + 1)$

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

Since one of $k$ and $(k+1)$ will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), $k(k+1)$ will be even. So $4k(k+1)$ will be divisible by $4\times2$ i.e. by 8. So when $y^2$ is divided by 8, it will leave a remainder 1.

When $y^2$ is divided by 8, remainder is 1. To get a remainder of 5 when $x^2 + y^2$ is divided by 8, we should get a remainder of 4 when $x^2$ is divided by 8. So $x$ must be even. If $x$ were odd, the remainder when $x^2$ were divided by 8 would have been 1. So we know that $x$ is divisible by 2 but we don’t know whether it is divisible by 4 yet.

$x^2 = 8a + 4$ (when $x^2$ is divided by 8, it leaves remainder 4)

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

So $x = 2 \times \sqrt{Odd Number}$

Square root of an odd number will be an odd number so we can see that $x$ is even but not divisible by 4. This statement alone is sufficient to say that $x$ is NOT divisible by 4.

Statement 2: $x - y = 3$

Since $y$ is odd, we can say that $x$ will be even (Since Even – Odd = Odd). But whether $x$ is divisible by 2 only or by 4 as well, we cannot say. This statement alone is not sufficient.

Answer (A).

So could you point out the takeaway from this question?

Note that when we were analyzing y, we used no information other than that it is odd. We found out that the square of any odd number when divided by 8 will always yield a remainder of 1.

Now what can you say about the square of an even number? Say you have an even number $x$ .

$x = 2a$

$x^2 = 4a^2$

This tells us that $x^2$ will be divisible by 4 i.e. we can make groups of 4 with nothing leftover. What happens when we try to make groups of 8? We join two groups of 4 each to make groups of 8. If the number of groups of 4 is even, we will have no remainder leftover. If the number of groups of 4 is odd, we will have 1 group leftover i.e. 4 leftover. So when the square of an even number is divided by 8, the remainder is either 0 or 4.

Looking at it in another way, we can say that if $a$ is odd, $x^2$ will be divisible by 4 and will leave a remainder of 4 when divided by 8. If $a$ is even, $x^2$ will be divisible by 16 and will leave a remainder of 0 when divided by 8.

Takeaways

– The square of any odd number when divided by 8 will always yield a remainder of 1.

– The square of any even number will be either divisible by 4 but not by 8 or it will be divisible by 16 (obvious from the fact that squares have even powers of prime factors so 2 will have a power of 2 or 4 or 6 etc). In the first case, the remainder when it is divided by 8 will be 4; in the second case the remainder will be 0.

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