# Understand the problem

What are the last 3 digits of \(2^{2017}\)?

(a) 072

(b) 472

(c) 512

(d) 912.

(a) 072

(b) 472

(c) 512

(d) 912.

##### Source of the problem

TIFR GS 2018 Part B Problem 9

##### Topic

Number theory

##### Difficulty Level

Easy

##### Suggested Book

Burton

# 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.

- Now, an obvious fact is \(2^{2017}\) ≅ \(0(mod8)\)
- so my idea is to find modul0 1000. so where does group theory lend you a hand?
- 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. SOME NUMBER THEORIC PROBLEMS: Q. Prove Wilson's theorem using basic group theoryQ. Prove Wilson's theorem by using Sylows theorems.

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