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2003 Federal Math Competition of S&M
Problem 1
10^n|floor((5+√35)^(2n-1))
10^n|floor((5+√35)^(2n-1))
Source: Serbia 2003 3&4th Grade P1
May 13, 2021
number theory
floor function
Divisibility
recursion
Problem Statement
Prove that the number
⌊
(
5
+
35
)
2
n
−
1
⌋
\left\lfloor\left(5+\sqrt{35}\right)^{2n-1}\right\rfloor
⌊
(
5
+
35
)
2
n
−
1
⌋
is divisible by
1
0
n
10^n
1
0
n
for each
n
∈
N
n\in\mathbb N
n
∈
N
.
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