Continuing our quest to master ‘pattern recognition’, let’s discuss a tricky little question today. It is best done using divisibility and remainders logic we discussed in some previous posts. We suggest you check out these divisibility posts if you haven’t yet.

We are first going to see how to solve the question conceptually. The interesting thing is – what do you do if you are under immense pressure during the test and are unable to remember anything you ever read on divisibility? It is a fairly common phenomenon – students have reported that they had blanked out during the test and couldn’t think of an appropriate approach. Our suggestion is that in that case, you should lean on trying to figure out the pattern. Try out a couple of values and see what you get. You may not understand why you are getting what you are getting but that doesn’t stop you from getting the correct answer. Let’s jump on to the question – we will first discuss the ideal approach and then go on to what happens if you don’t have a clue of what to do in the question.

**Question**: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

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

Statement 2: m – n = 3

**Solution**: Given that p, m and n are positive integers. If n is odd, n^2 must also be odd. How do you represent an odd integer? As (2k + 1)

n^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 – discussed in detail here), 4k(k+1) will be divisible by 8. Therefore, when n^2 is divided by 8, it will leave a remainder of 1.

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

When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. When n^2 is divided by 8, the remainder will be 1. To get a remainder of 5, when m^2 is divided by 8, we should get a remainder of 4.

m^2 = 8a + 4 (i.e. we can make ‘a’ groups of 8 and 4 will be leftover)

m^2 = 4(2a + 1)

This implies m = 2*√(Odd Number) because (2a+1) is an odd number. Square root of an odd number will also be odd.

Therefore, we can say that m is not divisible by 4.

This statement alone is sufficient.

**Statement 2**: m – n = 3

The difference between m and n is 3 i.e. an odd number. Since n is odd, we can say that m will be even (Even – Odd = Odd). But whether m is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p.

This statement alone is not sufficient.

**Answer (A)**

In this question, analyzing the question stem and statement 1 is a little complicated. Say we don’t analyze the question stem and jump to statement 1. Let’s see how we can use pattern recognition to make it easier.

**Question**: If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

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

Statement 2: m – n = 3

**Solution**:

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

When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. We need a remainder of 5 when m^2 + n^2 is divided by 8. Let’s try to find the remainders when m^2 and n^2 are divided by 8.

We are given that n is odd. Let’s try to figure out what this implies.

n = 1; When n^2 ( = 1) is divided by 8, the remainder is 1.

n = 3; When n^2 ( = 9) is divided by 8, the remainder is 1.

n = 5; When n^2 ( = 25) is divided by 8, the remainder is 1.

n = 7; When n^2 (= 49) is divided by 8, the remainder is 1.

There is a pattern here! Whenever you divide the square of an odd number by 8, you get the remainder 1. (We have discussed ‘why’ above in the logical explanation of statement 1)

This implies that when we divide m^2 by 8, we get 4 as remainder. If m^2 gives 4 as remainder, it means it is of the form m^2 = 8a + 4 = 4(2a + 1). So m must be of the form 2√(Odd Number). Hence m is not divisible by 4.

This statement is sufficient alone.

Through this example, you can see that pattern recognition is a very important tool (in easy as well as difficult questios)

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!*