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What is the remainder when divide 2^100 by 100?

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What is the remainder when divide 2^100 by 100?  [#permalink]

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New post Updated on: 23 Nov 2016, 16:45
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A
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D
E

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  95% (hard)

Question Stats:

32% (02:31) correct 68% (02:13) wrong based on 117 sessions

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Originally posted by broall on 23 Nov 2016, 09:01.
Last edited by broall on 23 Nov 2016, 16:45, edited 1 time in total.
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Re: What is the remainder when divide 2^100 by 100?  [#permalink]

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New post 23 Nov 2016, 14:42
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Approach #3 very fast

\(\frac{2^{100}}{100}\) means we need to find last two digits of \(2^{100}\)

We know that \(2^{10}=1024\)

\(24^{odd}\) last 2 digits = \(24\)

\(24^{even}\) last 2 digits = \(76\)

\(2^{100}\) = \((2^{10})^{10}\) = \((1024)^{10}\) = \(24^{even}\)

Last two digits are 76

Answer D.
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Re: What is the remainder when divide 2^100 by 100?  [#permalink]

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New post 23 Nov 2016, 10:51
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We need the last two digits of 2^100.
Since 100 is divisible by 4, by the cyclicity of 2, last digit must be 6.
2^4=16
2^8=256 (difference of 56-16=40)
2^12=4096 (difference of 96-56=40)
2^16=___36 (last 2 digits)
2^20=___76 (last 2 digits)
2^24=___16
Observe that last 2 digits start repeating.
Hence 2^100 must have last 2 digits as 76 since 100 is divisible by 5 and the series repeats after 5 terms.
Answer D

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Re: What is the remainder when divide 2^100 by 100?  [#permalink]

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New post 23 Nov 2016, 12:50
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\(\frac{2^{100}}{100} = \frac{2^{100}}{4*25}\)

We’ll find a remainder for 4 and 25 separately and use a Chinese remainder theorem to find a number.

\(2^{100} = 0 (mod 4)\) so our \(N = 0 (mod 4)\)

For 25 we’ll use slightly different approach.

GCF of \(2^{100}\) and \(25\) is \(1\), they are co-prime and we can apply Euiler's theorem.

\(2^{20} = 1 (mod 25)\) ---> \(2^{100} = (2^{20})^5 = 1^5 = 1\)

\(N = 1 (mod 25)\)

Combining two equations we’ll get

\(25x + 1 = 4y\)

or \(25x + 1 = 0 (mod 4)\) Solving this linear congruence we’ll get \(x=3 (mod 4)\)

and our number \(N = 25*3+1 =76\)

Answer D.
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Re: What is the remainder when divide 2^100 by 100?  [#permalink]

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New post 23 Nov 2016, 13:24
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nguyendinhtuong wrote:
What is the remainder when divide \(2^{100}\) by 100?

A. 16
B. 36
C. 56
D. 76
E. 96


Another approach

\(\frac{2^{100}}{100} = \frac{2^{100}}{2^2*25} = \frac{2^{98}}{25}\)

We know that \(2^7 = 128 = 3 (mod 25)\)

\(\frac{2^{98}}{25} = \frac{(2^7)^{14}}{25} = \frac{3^{14}}{25}\)

This will not help us much, but \(3^3 = 27 = 2 (mod 25)\)

And we have:

\(\frac{(3^3)^4*3^2}{25} = \frac{2^4*3^2}{25} = \frac{144}{25} = -6 (mod 25)\)

Now we need to multiply by cancelled factor \(2^2\)

\(-6*4 (mod 25*4) = -24 (mod 100) = 76 (mod 100)\)

Our remainder is 76
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Re: What is the remainder when divide 2^100 by 100?  [#permalink]

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New post 23 Nov 2016, 19:10
nguyendinhtuong wrote:
What is the remainder when divide \(2^{100}\) by 100?

A. 16
B. 36
C. 56
D. 76
E. 96



Hi,
One approach that can be used in all such Qs is to find closest multiples taking power..
Example here itself
\(2^100= (2^9)^{11}*2=512^{11}*2\)...
Now 512 to some power will leave a remainder same as 12 to that power..
So the remainder will be same as \(12^{11}*2\)...
\(12^{11}*2=(10+2)^{11}*2\)
Now when you expand this all terms except two terms willbe div by 100..
\(2^{11}*10^0+2^{10}*10^1=2048+1024*10\)
Last two digits are 48+40=88..
Remainder=88*2=176
That is 76
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Re: What is the remainder when divide 2^100 by 100?  [#permalink]

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New post 30 Jul 2019, 02:03
Bunuel can you please take this one?

I tried to use this method but got the wrong answer
https://gmatclub.com/forum/remainder-wh ... l#p2327183
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Re: What is the remainder when divide 2^100 by 100?   [#permalink] 30 Jul 2019, 02:03
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