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2013 Stanford Mathematics Tournament
1
SMT 2013 Team #1
SMT 2013 Team #1
Source:
February 4, 2013
Problem Statement
Let
f
1
(
n
)
f_1(n)
f
1
(
n
)
be the number of divisors that
n
n
n
has, and define
f
k
(
n
)
=
f
1
(
f
k
−
1
(
n
)
)
f_k(n)=f_1(f_{k-1}(n))
f
k
(
n
)
=
f
1
(
f
k
−
1
(
n
))
. Compute the smallest integer
k
k
k
such that
f
k
(
201
3
2013
)
=
2
f_k(2013^{2013})=2
f
k
(
201
3
2013
)
=
2
.
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