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Poland - Second Round
1983 Poland - Second Round
4
4
Part of
1983 Poland - Second Round
Problems
(1)
largest odd number by which k is divisible, sum
Source: Polish MO Recond Round 1983 p4
9/9/2024
Let
a
(
k
)
a(k)
a
(
k
)
be the largest odd number by which
k
k
k
is divisible. Prove that
∑
k
=
1
2
n
a
(
k
)
=
1
3
(
4
n
+
2
)
.
\sum_{k=1}^{2^n} a(k) = \frac{1}{3}(4^n+2).
k
=
1
∑
2
n
a
(
k
)
=
3
1
(
4
n
+
2
)
.
number theory
divisor
Sum