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National and Regional Contests
Netherlands Contests
Dutch Mathematical Olympiad
1976 Dutch Mathematical Olympiad
5
f(k_0) < f(k), f(k) = k + [ n/k ] , k_0 = [ \sqrt{n} ] + 1
f(k_0) < f(k), f(k) = k + [ n/k ] , k_0 = [ \sqrt{n} ] + 1
Source: Netherlands - Dutch NMO 1976 p5
January 28, 2023
inequalities
algebra
floor function
Problem Statement
f
(
k
)
=
k
+
[
n
k
]
f(k) = k + \left[ \frac{n}{k}\right ]
f
(
k
)
=
k
+
[
k
n
]
,
k
∈
{
1
,
2
,
.
.
.
,
n
}
k \in \{1,2,..., n\}
k
∈
{
1
,
2
,
...
,
n
}
,
k
0
=
[
n
]
+
1
k_0 =\left[ \sqrt{n} \right] + 1
k
0
=
[
n
]
+
1
.Prove that
f
(
k
0
)
<
f
(
k
)
f(k_0) < f(k)
f
(
k
0
)
<
f
(
k
)
if
k
∈
{
1
,
2
,
.
.
.
,
n
}
k \in \{1,2,..., n\}
k
∈
{
1
,
2
,
...
,
n
}
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