# Understand the problem

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

# 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?

# 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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