Friends in this post we will learn about How to find the remainder of a number with a power b i.e a^{b}

Let us take a problem for better understanding.

>>Question

Q. What is the remainder when you divide 2^{200} by 7?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Answer

We cannot compute 2^{200} in anywhere near the time allotted, so we should look for a pattern in much simpler problems that we can scale up to 2^{200}.

The simpler problems we should solve are these:

What is the remainder when you divide 2 by 7?

What is the remainder when you divide 2^{2} by 7?

What is the remainder when you divide 2^{3} by 7?

What is the remainder when you divide 2^{4} by 7?

... and so on with the powers of 2.

The answers follow this pattern:

2 divided by 7 leaves remainder 2.

2^{2} (which equals 4) divided by 7 leaves remainder 4.

2^{3} (which equals 8) divided by 7 leaves remainder 1.

2^{4} (which equals 16) divided by 7 leaves remainder 2.

2^{5} (which equals 32) divided by 7 leaves remainder 4.

2^{6} (which equals 64) divided by 7 leaves remainder 1.

We should stop as soon as we notice that the cycle will repeat itself forever in this pattern: [2, 4, 1]. Every third remainder is the same. (From here on out, "remainder" always means "remainder after we divide by 7.") Since every third remainder is the same, we should look at the remainder when the power is a multiple of 3. The remainders of 2^{3} and 2^{6} are 1. Thus, the remainder of 2 raised to a power that is any multiple of 3 is 1.

Now, 200 is not a multiple of 3, but we can look for a multiple of 3 near 200. 201 is a multiple of 3 (its digits add to 3), so 2^{201} has a remainder of 1. Finally, we notice that the remainder of 2^{200} must be one position earlier in the cycle than the remainder of 2^{201}. Since the cycle is [2, 4, 1], the remainder of 2^{200} is 4.

The correct answer is D.

In the above problem we take a point which was 2^{3} when divided by 7 gives 1.This is the main point which i mean to explain you.Just adhere to this point in any question of this type we only need to find the power at which the number becomes near to the divisor.

Take an example.

Before taking any problem points to remember.

1. when (x+1)^{a} is divided by x the remainder is always 1.

2. when (x-1)^{a} is divided by x there are 2 cases.

i) when the power "a" is odd the remainder will be x-1 always.

ii) when the power "a" is even the remainder will be 1 always.

easy way of remembering the 2. point is when power is even just think how we pronounce even i.e " e one" the remainder is one.

Q.Suppose we need to find the remainder of 32^{31} when divided by 5?

In this as i told you earlier we try to find the power at which the number is near to divisor but in this case as the number is greater than the divisor so divide the number with the divisor.

32/5 remainder is 2.

Now we will take 2 as consideration i.e 2^{31} not 32^{31}

Now 2 is less than 5 we take a power of 2 which is near to 5 i.e 2^{2}=4.

when we divide 4 with 5 we get remainder as 4.

So expression becomes as 2x(2^{2})^{15}. When we divide 4^{15} by 5 we get 4 (from point 2.)

Now when 4x2 is divided by 5 we get 3 as remainder.

Hence our answer is 3.

Try some own your own.

Learn about cyclicity Find unit digit in A^{b}.

Crack SSC CGL 2016

## 9 Jan 2016

# Finding the Remainder of 2^200 when divided by 7

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