Finding the Remainder of 2^200 when divided by 7

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 2²⁰⁰ by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Answer
We cannot compute 2²⁰⁰ in anywhere near the time allotted, so we should look for a pattern in much simpler problems that we can scale up to 2²⁰⁰.

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 2² by 7?
What is the remainder when you divide 2³ by 7?
What is the remainder when you divide 2⁴ by 7?
... and so on with the powers of 2.

The answers follow this pattern:
2 divided by 7 leaves remainder 2.
2² (which equals 4) divided by 7 leaves remainder 4.
2³ (which equals 8) divided by 7 leaves remainder 1.
2⁴ (which equals 16) divided by 7 leaves remainder 2.
2⁵ (which equals 32) divided by 7 leaves remainder 4.
2⁶ (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 2³ and 2⁶ 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 2²⁰¹ has a remainder of 1. Finally, we notice that the remainder of 2²⁰⁰ must be one position earlier in the cycle than the remainder of 2²⁰¹. Since the cycle is [2, 4, 1], the remainder of 2²⁰⁰ is 4.

The correct answer is D.

In the above problem we take a point which was 2³ 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 32³¹ 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 2³¹ not 32³¹
Now 2 is less than 5 we take a power of 2 which is near to 5 i.e 2²=4.
when we divide 4 with 5 we get remainder as 4.
So expression becomes as 2x(2²)¹⁵. When we divide 4¹⁵ 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

Share on Google Plus

About PrepHelp Educators

This is a short description in the author block about the author. You edit it by entering text in the "Biographical Info" field in the user admin panel.