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Simple one would be :

REM(2^100/12)

=> REM [ 16^25 /12]

Since REM(16/12) = 4

=> REM [ 4^25/12 ]
=> REM [ (4^5)^5/12 ]
=> REM [ (16*16*4)^5 /12]
=> REM[ 64^5/12]

Since REM[64/12] =4

=> REM (4^5/12) =4
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I have problems with this one.

2^100 / 12 = 2^100 / (2^2*3) = 2^98 / 3

I tried to find the remainder when 2^98 is divided by 3. I got the solution R = 1
Why can't I split the 12 up to 2^2 * 3?

Thanks
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hsinam3
Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none


\(2^{100} = 4^{50}\) and all +ive power of 4 always give remainder 4 when divided by 12.
so answer is A .
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lucky1829
I have problems with this one.

2^100 / 12 = 2^100 / (2^2*3) = 2^98 / 3

I tried to find the remainder when 2^98 is divided by 3. I got the solution R = 1
Why can't I split the 12 up to 2^2 * 3?

Thanks

Following post by Karishma should clarify this issue: when-2-is-divided-by-194911.html#p1504691
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2^100 can be rewritten as (12-10)^100. Expanding this expression using the binomial theorem, all terms will contain 12^100 except the last term, this means that all terms will be divisible by 12 except the last term. finding the remainder when (-10)^100 divided by 12.

10/12 remainder is 4.
100/12 remainder is again 4.

Therefore similarly (-10)^100 remainder will be 4.
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hsinam3
Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none

2^100/12 = (2*2*2^98)/12 = 2^98/3

or, 2^98/3 = (3-1)^98 / 3

All the terms in the expression are divisible by 3 but the last which is 1^98 ( Check this piece on use of binomial theorem by Veritas prep https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... ek-in-you/ super easy and very handy for remainder Qs)

Hence it boils down to 1/3 and the remainder is 1.

But we are not done yet. Remember how we simplified the expression in the beginning by dividing the divisor by 4? In order to get the correct remainder we have to multiply what we got as answer by 4. (Check how 1/3 and 2/6 despite being the same fraction do not have the same remainder!)

Answer A
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Bunuel
Following post by Karishma should clarify this issue: when-2-is-divided-by-194911.html#p1504691
Oh that's where I got it wrong. My approach was:

2^100/12 = 2^98/3

Remainder when:
2^1/3: 2
2^2/3: 1
2^3/3: 2
2^4/3: 1
2^5/3: 2
2^6/3: 1
2^7/3: 2

So, it is clear that for ever even power of 2, the remainder is 1 when that even power of 2 is divided by 3.

Since 98 is also even power, remainder will be 1.

Now that I read Karishma's post, I understand that I now need to multiply this 1 by 4 (because I had initially divided by 4)! Tricky.
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Alternatively from a binomial point of you=>
2^100 => (12+4)^25 => 12P+ 4^25
No consider 4^25 => 4^1 divided by 12 => remainder => 4
4^2 divided by 12 => remainder => 4
4^3 divided by 12 => remainder => 4
hence 4^25 => 12Q+4
hence the expression becomes => 12T+4 => 4 is the remainder
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Bunuel - I approached this question in the same manner you did. However, I don't understand why 2^1 is not included on your list.

Can you please elaborate as to why you went straight for 2^2 and so on...?
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hsinam3
Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none

\(\frac{2^{100}}{12} = \frac{2^{100}}{2^2*3} = \frac{2^{98}}{3}\)

\(2= -1 (mod 3)\)

\(\frac{(-1)^{98}}{3} = 1 (mod 3)\)

Multiplying this by \(2^2\)

\(1*4=(mod 3*4) = 4 (mod 12)\)

Remainder \(4\).
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hsinam3
Find the remainder of 2^100/12

A. 4
B. 2
C. 8
D. 1
E. none
value of remainder of 2^ even power (>=4) when divided by 12 ; 4
and remainder when divided by 2^ odd power (>=5) is 8

2^100 is even power so remainder would be 4
option A
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