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another remainder exercise

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by tpr-becky » Sun Apr 25, 2010 7:35 am
You can takle this problem with simple long division - divide 291 by 7 - sorry I can't figure out how to format it here but the result is 41 remainer 4 - your answer doesn't seem correct. ( you leave the decimal place off when dividing for remainders)
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by GmatTakerNo.1 » Sun Apr 25, 2010 7:41 am
Sorry, just forget the ^ between 2 and 91, but I edit it.
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by this_time_i_will » Sun Apr 25, 2010 9:54 am
GmatTakerNo.1 wrote:Can you help me with this:

What is the remainder when 2^91 is divided by 7?

The answer is 2, but I don´t know how to tackle this problem.
I am not sure whether fermat's little theorem is covered in GMAT or not. But this theorem is really handly when we the divisor is a prime number.
It states:
for any integer a and prime number divisor p:
a^(p-1)#1 mod p. (# : is congruent to).
This means whenever a^(p-1) is divided by p, the remainder is 1.
Using this theorem in our question:
2^6#1mod7
=2^90#1mod7.....(a)
Also,
=2#2mod7....(b).
From (a) and (b)(infact multiplying (a) and (b) )
=2^91#2 mod 7. so remainder = 2.
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by eaakbari » Wed Apr 28, 2010 6:33 am
GmatTakerNo.1 wrote:Can you help me with this:

What is the remainder when 2^91 is divided by 7?

The answer is 2, but I don´t know how to tackle this problem.
In such questions. try to identify a pattern.
Consider 2 ^ n
Remainder when n = 1 is 2
Remainder when n = 2 is 4
Remainder when n = 3 is 1
Remainder when n = 4 is 2
Remainder when n = 5 is 4

Hence we have a repeating pattern of 2,4,1. Hence the 91st will have remainder as 2.

HTH
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