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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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Updated on: 23 Nov 2016, 16:45
2
18
00:00

Difficulty:

95% (hard)

Question Stats:

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

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What is the remainder when divide $$2^{100}$$ by 100?

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

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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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23 Nov 2016, 14:42
4
4
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

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

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23 Nov 2016, 10:51
5
1
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.

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

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23 Nov 2016, 12:50
1
$$\frac{2^{100}}{100} = \frac{2^{100}}{4*25}$$

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 $$N = 25*3+1 =76$$

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

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23 Nov 2016, 13:24
2
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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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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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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