MathDB
Problems
Contests
National and Regional Contests
Russia Contests
All-Russian Olympiad Regional Round
1995 All-Russian Olympiad Regional Round
10.2
gcd(m,n)+lcm(m,n) = m+n All-Russian MO 1995 Regional (R4) 10.2
gcd(m,n)+lcm(m,n) = m+n All-Russian MO 1995 Regional (R4) 10.2
Source:
August 26, 2024
LCM
GCD
number theory
Problem Statement
Natural numbers
m
m
m
and
n
n
n
satisfy
g
c
d
(
m
,
n
)
+
l
c
m
(
m
,
n
)
=
m
+
n
.
gcd(m,n)+lcm(m,n) = m+n.
g
c
d
(
m
,
n
)
+
l
c
m
(
m
,
n
)
=
m
+
n
.
Prove that one of numbers
m
,
n
m,n
m
,
n
divides the other.
Back to Problems
View on AoPS