Question

2^301 divide by 5, what is the least positive reminder?

Answer 1

Try pattern recognition.

\(2^1\) divided by 5: 2
\(2^2\): 4
\(2^3\): 3
\(2^4\): 1
\(2^5\): 2
.
.
.

Answer 2

So basically, the remainder is 1 for all powers divisible by 4.

Answer 3

300 is divisible by 4, so it would leave a remainder of 1.

The series of remainders goes like this: 1, 2, 4, 3, 1, 2, 4, 3. So 2 always comes after 1.

Answer 4

Do you understand?

Answer 5

nope :(

Answer 6

OK, let me explain clearly this time. I was a little messy the last time.

Answer 7

\(2^0\) divided by \(5\) leaves remainder: 1.
\(2^1\) divided by \(5\) leaves remainder: 2.
\(2^2\) divided by \(5\) leaves remainder: 4.
\(2^3\) divided by \(5\) leaves remainder: 3.
\(2^4\) divided by \(5\) leaves remainder: 1.
\(2^5\) divided by \(5\) leaves remainder: 2.
\(2^6\) divided by \(5\) leaves remainder: 4.
\(2^7\) divided by \(5\) leaves remainder: 3.
\(2^8\) divided by \(5\) leaves remainder: 1.

Answer 8

Do you see how the cycle is repeating?

The remainders are coming out to be 1, 2, 4, 3, 1, 2, 4, 3, 1...

Also, the remainder comes out to be 1 when the power of 2 is divisible by 4.

Answer 9

So we have come up with a pattern and we can safely say that when the power of 2 is divisible by 4, the remainder is 1!

Now 300 is divisible by 4, right? So what would you get as the remainder when you divide \(\large 2^{300}\) by 5?

Answer 10

1

Answer 11

Very good. Now, do you see how the power that comes after the powers divisible by 4 left a remainder of 2?

Answer 12

ya

Answer 13

Look at the pattern.

\(2^1\) came right after \(2^0\) and left the remainder of 2.
\(2^5\) came right after \(2^4\) and left the remainder of 2.

Answer 14

So what can you say about \(2^{301}\) which comes right after \(2^{300}\)?

Answer 15

reminder is 2

Answer 16

Exactly. Also, if I may ask, what class are you in?

Answer 17

11

Answer 18

can u answer this on the basis of binomial theorem?

Answer 19

can u answer this on the basis of binomial theorem? @ganeshie8

Answer 20

try to write it as (x-1)^n or (x+1)^n and expand

Answer 21

2(65-1)^50

Answer 22

expand

Answer 23

2M(5)-2

Answer 24

\[
2^{301} = (5-3)^{301} = \sum_{k=0}^{301}5^k(-3)^{n-k}
\]The only term which will not be divisible by \(5\) is the one where \(k=0\).

Answer 25

\(\large (x+y)^n = \sum \limits_{k=0}^n \binom{n}{k} x^{n-k}y^k\)

Answer 26

wio has it !

Answer 27

so what is the reminder?

Answer 28

Well, when \(k=0\) you get \[
(-3)^{301}
\]

Answer 29

so?

Answer 30

2(65-1)^50


after expanding, you should get :
2[M(5) + 1]

right ?