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Putnam
1973 Putnam
A3
A3
Part of
1973 Putnam
Problems
(1)
Putnam 1973 A3
Source: Putnam 1973
5/25/2022
Let
n
n
n
be a fixed positive integer and let
b
(
n
)
b(n)
b
(
n
)
be the minimum value of
k
+
n
k
,
k+\frac{n}{k},
k
+
k
n
,
where
k
k
k
is allowed to range through all positive integers. Prove that
⌊
b
(
n
)
⌋
=
⌊
4
n
+
1
⌋
.
\lfloor b(n) \rfloor= \lfloor \sqrt{4n+1} \rfloor.
⌊
b
(
n
)⌋
=
⌊
4
n
+
1
⌋
.
Putnam
floor function
inequalities
minimum