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Evaluate this remainder (Posted on 2008-08-13) Difficulty: 2 of 5
What is the remainder when you divide 299 by 99?

See The Solution Submitted by pcbouhid    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 3 of 6 |

At the outset, we note that 33^2 is divisible by 99, so that 33^x is divisible by 99 for all positive integers x >= 2.

Now,
2^15 =  32^3
=(33-1)^3
= (33^3 33*3(33 -1) 1)
= M(99) 1

Thus: 2^15 = -1 (mod 99)
or, 2^90  = 1 (mod 99)

Again:
2^9
= 16*32
= (15+1)(33-1)(mod 99)
= 15*33 +(33-15-1)
= M(99) + 17, and so: 2^9 = 17 (mod 99);  since 15 is divisible by 3, so that 33*15 is divisible by 99

Thus, 2^99
= (2^90)*(2^9) (mod 99)
= 1*17 (mod 99)
= 17

Consequently, the required remainder is 17.

Edited on August 14, 2008, 2:04 am
  Posted by K Sengupta on 2008-08-13 11:36:25

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