*Friday, February 15th, 2013*

**Basic Applications of Remainder Theorem**

In my previous post, we discussed the cyclical nature of the remainders when a^{n} is divided by d. In this post, we will see how problems on finding out the remainder can be broken down into smaller parts.

**Funda 1: Remainder of a sum when it is being divided is going to be the same as the sum of the individual remainders.**

Let us look at an example for this case:

Eg: Find out the remainder when (79+80+81) is divided by 7.

If we add it up first, we get the sum as 240 and the remainder as 2 as shown below:

However, it would be easier to find out the individual remainders of 79, 80 & 81; which come out to be 2, 3 & 4 respectively and adding them up later to get the answer. This process is shown below:

I hope you would agree that the second method is easier. May be the difference in difficulty level is not highlighted here. Let us look at another idea on the same lines.

** ****Funda 2: Remainder of a product when it is being divided is going to be the same as the product of the individual remainders.**

Let us look at an example for this case:

Eg: Find out the remainder when (79 x 80 x 81) is divided by 7.

If we multiply it first, we get the product as 511920 and the remainder as 3 as shown below:

However, it would be easier to find out the individual remainders of 79, 80 & 81; which come out to be 2, 3 & 4 respectively and multiply them to get 24, which will eventually lead to the remainder of 3. This process is shown below:

I guess there is no doubt in this question that the second method is easier. To be honest, I would take more time to just find out the product of (79 x 80 x 81) than to solve the entire question.

That is the reason I recommend breaking down the problem into smaller parts.

**Funda 3: Negative Remainders – When the absolute value of the -ive remainder is lesser than the absolute value of the positive remainder, it is recommended that you consider a multiple greater than the divisor.**

When 7 is divided by 4, the remainder can be considered as 3 or -1.

When 18 is divided by 7, the remainder can be considered as 4 or -3.

When 689 is divided b 23, the remainder can be considered as 22 or -1.

As you can see from above, the calculations would reduce drastically in the third case if you consider a negative remainder. As a tip, in remainder questions, *you should always think of multiples or powers which can lead to a remainder of 1 or -1*.

Till now, the examples I have taken are too simple to be asked in CAT or for that matter any other MBA entrance exam. Let us look at an example that uses all the above-mentioned ideas and is of a slightly higher difficulty level.

Eg: Find out the remainder when 83^{261} is divided by 17.

First of all we need to break down 83^{261} into smaller parts.

It would be easier if consider the remainder as -2 because our calculations would be lesser. So essentially, our question reduces to:

Now, referring to the tip I gave above, think of a power of 2 that would give a remainder of 1 or -1 from 17.

2^{4} is 16 and would give a remainder of -1 from 17.

We have a 2^{261} here. We will have to break it down to (2^{260} x 2) so that we can convert it to a power of 16. This step is shown below:

I highly recommend that in this question and other questions of this type, you should verify the answer from Wolfram Alpha.

Hope you found this article useful. I look forward for your suggestions to my future posts.

[…] Preparation Factor Theory Dealing With Factorials Baisc Idea of Remainders Cyclicity Of Remainders Basic Application of Remainder Theorem Remainders Advanced Determining the second last digit and the last two digits Application of LCM […]

[…] Preparation Factor Theory Dealing With Factorials Baisc Idea of Remainders Cyclicity Of Remainders Basic Application of Remainder Theorem Remainders Advanced Determining the second last digit and the last two digits Application of LCM […]