Tricky remainder problem- Experts input required

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by tutorphd » Thu Jun 28, 2012 10:32 pm
The remainder of a product is a product of the remainders of each factor.

Break down into large factors:
2^39 = 2^9 * 2^9 * 2^9 * 2*9 * 8

Calculate remainder of each factor:
2^9 = 512, divided by 39 gives a remainder of 5
8 divided by 39 gives a remainder of 8

Multiply the remainders of the 5 factors and try to reduce to a remainder less than 39 by subtracting multiples of 39:
5*5*5*5*8 = 125*40 = 125*(39+1) = 125*39 + 125 = 125*39 + 3*39 + 8

So the reduced remainder is 8.

In general, such a laborous problem won't appear on GMAT, unless there is a faster shortcut.
Last edited by tutorphd on Thu Jun 28, 2012 10:44 pm, edited 1 time in total.
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by Anurag@Gurome » Thu Jun 28, 2012 10:44 pm
khandelwal.ab wrote:What is the remainder when 2^39 is divided by 39?
This can be easily solved using modular arithmetic. According to modular arithmetic, (a mod b) = c means when a is divided by b, the remainder is c. A fundamental formula in modular arithmetic is
  • (ab) mod c = [a mod c * b mod c] mod c

    Hence, (a^b) mod c = [(a mod c)^b] mod c
Now, 2^39 = (2^3)*(2^36) = 8*[(2^9)^4] = 8*[(512)^4]

Now, (8 mod 39) = 8
And, (512 mod 39) = 5
And, (512^4 mod 39) = ((512 mod 39)^4 mod 39) = (5^4 mod 39) = (625 mod 39) = 1

Hence, (2^39) mod 39 = 8*[(512)^4] mod 39 = [(8 mod 39)*(512^4 mod 39)] mod 39 = [8*1] mod 39 = 8

The correct answer is D.
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by \'manpreet singh » Thu Jun 28, 2012 11:58 pm
Anurag i feel that the modular method stated here consumes the same time as the previous method.
I doubt such a hardworking problem is asked in the GMAT as it will surely eat away a lot of time!!!
Kindly conform if there is possibility of such question. :?: :?:

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by tutorphd » Fri Jun 29, 2012 4:50 am
They are the same method only mine is explained without formulas.
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