Difference between revisions of "1984 AIME Problems/Problem 7"
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== Solution 2== | == Solution 2== | ||
− | We start by finding values of the function right under 1000 since they require | + | We start by finding values of the function right under <math>1000</math> since they require iteration of the function. |
<cmath>f(999)=f(f(1004))=f(1001)=998</cmath> | <cmath>f(999)=f(f(1004))=f(1001)=998</cmath> |
Revision as of 18:33, 25 July 2011
Contents
Problem
The function f is defined on the set of integers and satisfies
Find .
Solution 1
Define , where the function
is performed
times. We find that
.
. So we now need to reduce
.
Let’s write out a couple more iterations of this function:
So this function reiterates with a period of 2 for
. It might be tempting at first to assume that
is the answer; however, that is not true since the solution occurs slightly before that. Start at
:
Solution 2
We start by finding values of the function right under since they require iteration of the function.
Soon we realize the for integers
either equal
or
based on it parity. (If short on time, a guess of
or
can be taken now.)
If
is even
if
is odd
.
has even parity, so
.
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |