Difference between revisions of "1984 AIME Problems/Problem 7"
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We start by finding values of the function right under 1000 since they require little repetition. | We start by finding values of the function right under 1000 since they require little repetition. | ||
− | < | + | <math>f(999)=f(f(1004))=f(1001)=998</math> |
− | + | <math>f(998)=f(f(1003))=f(1000)=997</math> | |
− | f(997)=f(f(1002))=f(999)=998 | + | <math>f(997)=f(f(1002))=f(999)=998</math> |
− | f(996)=f(f(1001))=f(998)=997 | + | <math>f(996)=f(f(1001))=f(998)=997</math> |
Soon we realize the <math>f(k)</math> for integers <math>k<1000</math> either equal <math>998</math> or <math>997</math> based on it parity. (If short on time, a guess of 998 or 997 can be taken now.) | Soon we realize the <math>f(k)</math> for integers <math>k<1000</math> either equal <math>998</math> or <math>997</math> based on it parity. (If short on time, a guess of 998 or 997 can be taken now.) |
Revision as of 18:31, 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 1000 since they require little repetition.
Soon we realize the for integers
either equal
or
based on it parity. (If short on time, a guess of 998 or 997 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 |