Remainder when 2^100 is divided by 9

2^6=64 ND 9*7=63 so remainder is 1 for 2^96 is 1*2^4 = 16 divided by 9...remainder 7

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Hello!

Can someone please provide a short cut for this problem?

I took a lot of time to understand this one:

2/9 = 2
4/9 = 4
8/9 = 8
16/9 = 7
32/9 = 5
64/9 = 1
128/9 = 2
256/9 = 4
512/9 = 8
1024/9 = 7
../9 = 5
.../9 = 1

Hence the cyclicity of remainders is 2,4,8,7,5,1.

Since 100 is even then the cyclicity for evens is 4,7,1.

From here I've got a doubt, ¿what should I choose?

I took 7 cuz 2 power 100 is similar to 2 power 10, which holds a remainder of 7.

Kind regards!

Hi, Actually you were on the right track, and got the correct sequence of cyclicity

Just that you didnt have to check for even numbers as such

cyclicity of remainders is 2,4,8,7,5,1

So if you notice that its a series of 6 numbers, which will be repeated all the time

16*6 =96 this means for 2^96 the remainder will be 1, after this move 4 times, in the series of cyclcity, To give you the position for 2^100.

Kindly let me know if this helps you.

When a power of 2 is divided by 9, the remainder start repeating itsel after 2^6:

2^1 => remainder = 2
2^2 => remainder = 4
2^3 => remainder = 8
2^4 => remainder = 7
2^5 => remainder = 5
2^6 => remainder = 1

2^7 => remainder = 2

Therefore,
100/6 gives 4 as reminder => 2^100 and 2^4 have the same remainder when divided by 9, ie 7.

Cyclicity is the best way to go for this

2^1 2
2^2 4
2^3 8
2^4 16
2^5 2

Every n×4 th power will have 6 as last digit.

100 25x4

=> when we divide 16 with 9 remainder will be 7.

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Asked: What is the remainder when 2^100 is divided by 9?

2^3mod9 = 8mod9 = - 1
100 = 3*33 + 1

2^100mod9 = 2^99*2mod9 = (-1)^33*2mod9 = -2mod9 = 7mod9

IMO D

2 nice concepts of remainders which are often used.

1) The concept of negative remainders. When the remainder is 1 less than the divisor, the remainder is -1.

In short |Negative Remainder| + Positive Remainder = Divisor


2) We should attempt to bring most of the terms in the form \((xa \pm 1)^n\), where x is a multiple of \(a\)

(i) \(\frac{(xa + 1)^n}{a}\) gives us a remainder of 1, irrespective whether n is even or odd.

(ii) \(\frac{(xa + 1)^n}{a}\) gives us a remainder of 1 when n is even and a remainder of -1 when n is odd.


The above question is to find the remainder when \(\frac{2^{100}}{9}\)

Now \(2^3 = 8 = 9 - 1\)

So, \(R[\frac{2^{100}}{9}] = R[\frac{(2^3)^{33} * 2}{9}] = R[\frac{(9-1)^{33}}{9}] * R[\frac{2}{9}]\)


Since 33 is an odd value the remainder of \(\frac{(9-1)^{33}}{9} = -1 \space or \space 8\)

The question now becomes the remainder of 8 * 2 divided by 9 = \(R[\frac{19}{9}] = 7\)


Option D

Arun Kumar

CrackVerbalGMAT, can we use the rule of cyclicity to solve this question ?

Hi kntombat. We can definitely use the rule of cyclicity here.

Look out for the cycle in which we get a remainder of 1 (or -1, whichever comes first).

(1) \(R[\frac{2}{9}] = 2\)

(2) \(R[\frac{2 * 2}{9}] = R[\frac{4}{9}] = 4\)

(3) \(R[\frac{4 * 2}{9}] = R[\frac{8}{9}] = 8 \space or \space -1\). This is sufficient, because we will get a remainder of 1 on the double of this cycle, i.e on the 6th cycle.

(4) \(R[\frac{8 * 2}{9}] = R[\frac{16}{9}] = 7\)

(5) \(R[\frac{7 * 2}{9}] = R[\frac{14}{9}] = 5\)

(6) \(R[\frac{5 * 2}{9}] = R[\frac{10}{9}] = 1\)


So every 6 powers of 2 will give us 1 as the remainder and hence \(R[\frac{(2)^{100}}{9}] = R[\frac{(2^6)^{16} * 2^4}{9}]\)


= \(R[\frac{(2^6)^{16}}{9}] * R[\frac{2^4}{9}] = 1 * R[\frac{16}{9}] = 7\)


Arun Kumar

The units digit of the 2 exponent is in a cycle of: (2, 4, 8, 6)

2^100 = units digit of 6

16 / 9 = remainder 7.

Answer is D.

What is the remainder when 2^100 is divided by 9?

a. 1
b. 3
c. 4
d. 7
e. 8

Concept: Binomial Theorem
Break the exponent 2^100 = (2^3)^33 * 2
=(9 - 1)^33 * 2/9
First term in the bracket 9 is fully divisible by the denominator 9, second term in the bracket (-1)^33 divided by 9 leaves a negative remainder 1 which is a positive remainder 8 (Negative remainders concept)
Therefore 8*2/9 leaves a remainder 7
D

↧↧↧ Detailed Video Solution to the Problem ↧↧↧



We need to find the remainder when \(2^{100}\) is divided by 9?

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 9

Remainder of \(2^1\) (=2) by 9 = 2
Remainder of \(2^2\) (=4) by 9 = 4
Remainder of \(2^3\) (=8) by 9 = 8
Remainder of \(2^4\) (=16) by 9 = 7
Remainder of \(2^5\) (=32) by 9 = 5
Remainder of \(2^6\) (=64) by 9 = 1
Remainder of \(2^7\) (=64) by 9 = 2

=> Cycle is 6

=> We need to find Remainder of 100 by 6 = 4
=> Remainder of \(2^{100}\) is divided by 9 = Remainder of 2^4 by 9 = 7

So, Answer will be D
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

Hey but how would you know that it is 16 .. could be 36 too leading to a completely different result..

I'll try to simplify what they did,

\(2^{100} = 2 * 2^{99}\)

\(= 2 * 8^{33}\) (we did \(2^{3*33}\))

Now, remainder of \(8^{33} = -1\), we'll add divisor 9 to it as remainder can't be negative, thus remainder of \(8^{33} = -1 + 9 = 8\)

= \(2 * 8 = 16\)

16 / 9 will now give remainder 7