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# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]What are the last 3 digits of $$2^{2017}$$?
(a) 072
(b) 472
(c) 512

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]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?)
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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.
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]
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 theory

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

# Watch the video

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# Similar Problems

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