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Topic: f(84) (Read 1042 times) |
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tony123
Junior Member
Posts: 61
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let f function from z to z(integers)
and
f(n)=n - 3 at n > 999
f(n)= f(f(n+5) at n < 1000
find
f(84)
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Sameer
Uberpuzzler
Pie = pi * e
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Posts: 1261
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Re: f(84)
« Reply #1 on: Jun 27th, 2007, 11:20am » |
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hmmm flip flop .. flip flop .... flip flop ... thinking... choosing f(84) = f(0)
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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SMQ
wu::riddles Moderator
Uberpuzzler
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Posts: 2084
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Re: f(84)
« Reply #3 on: Jun 27th, 2007, 11:45am » |
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Just to be thorough:
working forward using rule 1:
f(1000) = 997
f(1001) = 998
f(1002) = 999
f(1003) = 1000
f(1004) = 1001
working backward using rule 2:
f(999) = f(f(1004)) = f(1001) = 998
f(998) = f(f(1003)) = f(1000) = 997
f(997) = f(f(1002)) = f(999) = 998
f(996) = f(f(1001)) = f(998) = 997
And in general, for integer x: 0 <= x <= 497,
f(2x+1) = f(f(2x+6)) = f(997) = 998
f(2x) = f(f(2x+5)) = f(998) = 997
Therefore f(84) = 997
--SMQ
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