Understand the problem

What are the last 3 digits of \(2^{2017}\)?
(a) 072
(b) 472
(c) 512
(d) 912.
Source of the problem
TIFR GS 2018 Part B Problem 9
Number theory
Difficulty Level
Suggested Book

Start with hints

Do you really need a hint? Try it first!

Pretty convenient problem for number theory lovers. I’m going to give some insight from Group Theory and solve with basic congruence techniques.
  1. Now, an obvious fact is \(2^{2017}\) ≅ \(0(mod8)\)
  2. so my idea is to find modul0 1000. so where does group theory lend you a hand?
  3. see that 2 is a generator of (Z/125Z)* (why?)
So we get \(2^{2017}\) ≅ \(2^{2017}(mod125)\) ≅ \(2^{17}\) ≅ \(72 (mod 125)\)
4. Now, what is the most demanding step after this?
combining 1 and 3 we get \(2^{2017}\) ≅ \(072(mod 1000)\) [as \(125|2^{2017}-72, 8|2^{2017}-72 \)=> third digit is 0]
Hence the answer is 072.
Bonus Problems:
Q. find last two digits of 2^2016 like this process.

Q. Prove Wilson's theorem using basic group theory

Q. Prove Wilson's theorem by using Sylows theorems.

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