Difference between revisions of "1984 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | A [[function]] <math>f</math> is defined for all real numbers and satisfies <math>f(2+x)=f(2-x)</math> and <math>f(7+x)=f(7-x)</math> for all <math>x</math>. If <math>x=0</math> is a root for <math>f(x)=0</math>, what is the least number of roots <math>f(x)=0</math> must have in the interval <math>-1000\leq x \leq 1000</math>? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>A [[function]] <math>f</math> is defined for all real numbers and satisfies <math>f(2+x)=f(2-x)</math> and <math>f(7+x)=f(7-x)</math> for all <math>x</math>. If <math>x=0</math> is a root for <math>f(x)=0</math>, what is the least number of roots <math>f(x)=0</math> must have in the interval <math>-1000\leq x \leq 1000</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
== Solution == | == Solution == |
Revision as of 17:53, 27 March 2015
Problem
A function is defined for all real numbers and satisfies
and
for all
. If
is a root for
, what is the least number of roots
must have in the interval
?
Solution
If , then substituting
gives
. Similarly,
. In particular,
Since is a root, all multiples of
are roots, and anything congruent to
are also roots. To see that these may be the only integer roots, observe that the function
satisfies the conditions and has no other roots.
In the interval , there are
multiples of
and
numbers that are congruent to
, therefore the minimum number of roots is
.
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |