MathDB
Problems
Contests
National and Regional Contests
India Contests
Postal Coaching
2005 Postal Coaching
4
4
Part of
2005 Postal Coaching
Problems
(1)
One on gcds
Source: Indian Postal Coaching 2005
9/22/2005
Let
m
,
n
m,n
m
,
n
be natural numbers and let
d
=
g
c
d
(
m
,
n
)
d = gcd(m,n)
d
=
g
c
d
(
m
,
n
)
. Let
x
=
2
m
−
1
x = 2^{m} -1
x
=
2
m
−
1
and
y
=
2
n
+
1
y= 2^n +1
y
=
2
n
+
1
(a) If
m
d
\frac{m}{d}
d
m
is odd, prove that
g
c
d
(
x
,
y
)
=
1
gcd(x,y) = 1
g
c
d
(
x
,
y
)
=
1
(b) If
m
d
\frac{m}{d}
d
m
is even, Find
g
c
d
(
x
,
y
)
gcd(x,y)
g
c
d
(
x
,
y
)
number theory
greatest common divisor
algorithm
Euclidean algorithm
number theory solved