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9
2023 Algebra NT #9 s_{20}(n) - s_{23}(n)
2023 Algebra NT #9 s_{20}(n) - s_{23}(n)
Source:
February 28, 2024
number theory
Problem Statement
For any positive integers
a
a
a
and
b
b
b
with
b
>
1
b > 1
b
>
1
, let
s
b
(
a
)
s_b(a)
s
b
(
a
)
be the sum of the digits of
a
a
a
when it is written in base
b
b
b
. Suppose
n
n
n
is a positive integer such that
∑
i
=
1
⌊
log
23
n
⌋
s
20
(
⌊
n
2
3
i
⌋
)
=
103
and
∑
i
=
1
⌊
log
20
n
⌋
s
23
(
⌊
n
2
0
i
⌋
)
=
115
\sum^{\lfloor \log_{23} n\rfloor}_{i=1} s_{20} \left( \left\lfloor \frac{n}{23^i} \right\rfloor \right)= 103 \,\,\, \text{and} \,\,\, \sum^{\lfloor \log_{20} n\rfloor}_{i=1} s_{23} \left( \left\lfloor \frac{n}{20^i} \right\rfloor \right)= 115
i
=
1
∑
⌊
l
o
g
23
n
⌋
s
20
(
⌊
2
3
i
n
⌋
)
=
103
and
i
=
1
∑
⌊
l
o
g
20
n
⌋
s
23
(
⌊
2
0
i
n
⌋
)
=
115
Compute
s
20
(
n
)
−
s
23
(
n
)
s_{20}(n) - s_{23}(n)
s
20
(
n
)
−
s
23
(
n
)
.
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