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September 1, 2019

Last three digit of the last year: TIFR GS 2018 Part B Problem 9

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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
(d) 912.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part B Problem 9[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Number theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Burton[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/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?)
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]
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.

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Watch the video

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Similar Problems

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