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November 11, 2019

RMO 2019 (Maharashtra Goa) Adding GCDs

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Understand the problem

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For each \( n  \in \mathbb{N} \) let \( d_n \) denote the G.C.D. of n and (2019 - n). Find the value of \(  d_1 +  d_2 + ... + d_{2019} \).

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Regional Math Olympiad, 2019, Maharashtra, Goa Region Problem 1[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]5/10

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But first, can you try these problems? 

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  1. Show that G.C.D. of k and 0 is k for any positive integer k.
  2. Show rigorously that G.C.D. (a, b)  = G.C.D. (a, a+b) for any non-negative integers a and b
  3. Can you find and prove a similar result with a negative sign? 

Send the solutions to support@cheenta.com

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Similar Problems

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One comment on “RMO 2019 (Maharashtra Goa) Adding GCDs”

  1. #1)Show that G.C.D. of k and 0 is k for any positive integer k.
    We know GCD is highest common factor of both k &0;
    k=kX1;
    0=kX0;
    so,GCD=k proved
    #2)Show rigorously that G.C.D. (a, b) = G.C.D. (a, a+b) for any non-negative integers a and b;
    let GCD of a &b be h;
    so a=h*q1; b=h*q2; q1 &q2 do not have any common factor other than 1;or GCD of h1&h2 is 1;
    so, (a+b)=h*(q1+q2);
    so GCD of a =q1*h& (a+b)=h*(q1+q2);
    GCD of a&b=GCD of a&(a+b)=h proved
    #3Can you find and prove a similar result with a negative sign?
    yes let a=-m=-12(say)& b=-n=-15(say); m &n natural numbers if m&n has GCDas h=3

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