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1977 IMO Shortlist
11
Prove that min(x1,x2,...,xn) = sqrt n
Prove that min(x1,x2,...,xn) = sqrt n
Source:
September 20, 2010
algebra
recurrence relation
Sequence
equation
floor function
IMO Shortlist
Problem Statement
Let
n
n
n
be an integer greater than
1
1
1
. Define
x
1
=
n
,
y
1
=
1
,
x
i
+
1
=
[
x
i
+
y
i
2
]
,
y
i
+
1
=
[
n
x
i
+
1
]
,
for
i
=
1
,
2
,
…
,
x_1 = n, y_1 = 1, x_{i+1} =\left[ \frac{x_i+y_i}{2}\right] , y_{i+1} = \left[ \frac{n}{x_{i+1}}\right], \qquad \text{for }i = 1, 2, \ldots\ ,
x
1
=
n
,
y
1
=
1
,
x
i
+
1
=
[
2
x
i
+
y
i
]
,
y
i
+
1
=
[
x
i
+
1
n
]
,
for
i
=
1
,
2
,
…
,
where
[
z
]
[z]
[
z
]
denotes the largest integer less than or equal to
z
z
z
. Prove that
min
{
x
1
,
x
2
,
…
,
x
n
}
=
[
n
]
\min \{x_1, x_2, \ldots, x_n \} =[ \sqrt n ]
min
{
x
1
,
x
2
,
…
,
x
n
}
=
[
n
]
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