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

Question

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

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

souvonik2k · Accepted Answer

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

vitaliyGMAT · Answer

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.

vitaliyGMAT · Answer

\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.

vitaliyGMAT · Answer

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

chetan2u · Answer

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

Fdambro294 · Answer

Rule: when a Number is Divided by 100, the Last 2 Digits of the Numerator/Dividend will be the Remainder

To find the last 2 Digits of (2)^100:

(2)^10 = 1024

Now if we just look at the last 2 Digits when we take (2)^20th Power:

(2)^20 =   ^2 =   ^2 = .......576 ----- Last 2 Digits will be 76

Again, looking at the last 2 Digits when we take (2)^30th Power

(2)^30 = (2)^10 * (2)^20 =   *   = .......24 ----- Last 2 Digits will be 24

Again, looking at the last 2 Digits when we take (2)^40th Power

(2)^40 = (2)^20 * (2)^20 =   *   = .....76 ----- LAST 2 Digits will be 76

this Pattern will continue such that:

I.   ^ODD = Last 2 Digits will be = 24

II.   ^EVEN = Last 2 Digits will be = 76

(2)^100 =   ^10 =   ^EVEN ----- LAST 2 Digits will be 76

-D-

sagar23691 · Answer

Can anyone please explain me, why only last two digits were find, when the question was asked to find the remainder ?

chetan2u · Answer

Because each number can be written as a multiple of 100+last 2 digits.
 Example - 765 =700+65, when you divide it by 100, 700 is divisible by 100 and 65 is the remainder.

ScottTargetTestPrep · Answer

Solution:
 
The remainder when a number is divided by 100 is the last two digits of the number. Also, recall that if a number raised to a power has a remainder of r when it is divided by 100, the last two digits of the number raised to the same power also have r as the remainder when divided by 100. For example, 123^2 = 15129 has a remainder of 29 when divided by 100, 23^2 = 529 also has a remainder of 29 when divided by 100. Now, let’s use the fact that 2^10 = 1024. So we have:

2^20 = (2^10)^2 will have the same remainder as 24^2 = 576 when they are divided by 100 and notice that the remainder is 76.

2^40 = (2^20)^2 will have the same remainder as 76^2 = 5776 when they are divided by 100 and notice that the remainder is 76.

2^80 = (2^40)^2 will have the same remainder as 76^2 = 5776 when they are divided by 100 and notice that the remainder is 76.

As you can see, we are trying to get to 2^100 by squaring repeatedly, starting from 2^10. However, if we square again, we will have 2^160, which exceeds 2^100, the number we need to get to. However, notice that 2^100 = 2^80 x 2^20; therefore, we can just multiply the remainders of 2^80 and 2^20 to get the remainder of 2^100 when it’s divided by 100. Since the remainders of 2^80 and 2^20 are both 76, and 76 x 76 = 5776, the remainder is 76 when 2^80 x 2^20 = 2^100 is divided by 100.

Answer: D

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=HgBq7rluoLw

We need to find the Remainder when  2^{100}  by 100 and we know that:

Remainder of a number by 100 = Last two digits of the numbers
 
  2^{100}  =  2^{10*10}  =  (2^{10})^{10} 
 =  1024^{10}

Last two digits of  1024^{10}  = Last two digits of  24^{10}

Now, last two digits of 24 * 24 = 76
And, last two digits of 76 * 24 = 24

=> Last two digits of  24^{Even Power}  = 76
 => Last two digits of  24^{Odd Power}  = 24

=> Last two digits of  24^{10}  = 76

So,  Answer will be D 
Hope it Helps!

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

https://www.youtube.com/watch?v=h1HTEhmUb1w

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

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