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Remainder when 2^100 is divided by 9

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Remainder when 2^100 is divided by 9  [#permalink]

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New post Updated on: 10 Jan 2019, 08:42
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What is the remainder when 2^100 is divided by 9?

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

Originally posted by kanakdaga on 10 Jan 2019, 08:29.
Last edited by kanakdaga on 10 Jan 2019, 08:42, edited 1 time in total.
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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 10 Jan 2019, 08:35
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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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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 10 Jan 2019, 08:41
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I think the OA is wrong

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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 10 Jan 2019, 08:42
deepverma wrote:
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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couldn't really understand! can you elaborate ?
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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 10 Jan 2019, 08:43
rajatvermaenator wrote:
I think the OA is wrong

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apologies!
Corrected now!
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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 10 Jan 2019, 08:58
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Please check if my approach is correct!

9=8+1
9= 2^3 + 1
or 2^3 = 9 - 1

When there are 100 2s and we need groups of 3 2s , it will be possible by the following way:

2^(99) * 2^(1) = 2^(100)
2^(3*33) * 2^(1) = 2^(100)
8^(33) * 2^(1) = 2^(100)
(9-1)^(33) * 2^(1) = 2^(100)

9 will leave remainder 0
-1^33 will leave remainder -1
2^1 will leave remainder 2

Thus , -1 * 2 = -2

Now we add divisor to negative remainder : -2 + 9 = 7

Hence, 7 is the actual remainder.

KUDOS IF THIS HELPS! :D
Please correct me if wrong!
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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 11 Jan 2019, 00:11
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2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
thus the cyclicity is 4m+1, 4m+2, 4m+3, 4m
2^100=2^4*25=2^4m, which is the same as 2^4=16=2^4*4=2^4m.

thus I simply use 16 divided by 9 and its remainder is 7, so D
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Remainder when 2^100 is divided by 9  [#permalink]

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New post 11 Jan 2019, 20:43
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!
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Remainder when 2^100 is divided by 9  [#permalink]

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New post 11 Jan 2019, 22:15
jfranciscocuencag wrote:
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. - This was not required as such

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.
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Re: Remainder when 2^100 is divided by 9  [#permalink]

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New post 12 Jan 2019, 15:29
1
KanishkM wrote:
jfranciscocuencag wrote:
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. - This was not required as such

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.


Now its clear for me, thank you!!!
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Re: Remainder when 2^100 is divided by 9   [#permalink] 12 Jan 2019, 15:29
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