Difference between revisions of "1984 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | A function <math> | + | 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>? |
== Solution == | == Solution == | ||
− | If <math> | + | If <math>f(2+x)=f(2-x)</math>, then substituting <math>t=2+x</math> gives <math>f(t)=f(4-t)</math>. Similarly, <math>f(t)=f(14-t)</math>. In particular, |
− | + | <cmath>f(t)=f(14-t)=f(14-(4-t))=f(t+10)</cmath> | |
− | < | ||
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+ | Since <math>0</math> is a root, all multiples of <math>10</math> are roots, and anything congruent to <math>4\pmod{10}</math>) are also roots. To see that these may be the only integer roots, observe that the function | ||
+ | <cmath>f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}</cmath> | ||
satisfies the conditions and has no other roots. | satisfies the conditions and has no other roots. | ||
− | + | In the interval <math>-1000\leq x\leq 1000</math>, there are <math>201</math> multiples of <math>10</math> and <math>200</math> numbers that are congruent to <math>4 \pmod{10}</math>, therefore the minimum number of roots is <math>\boxed{401}</math>. | |
− | In the interval <math>-1000\leq x\leq 1000</math>, there are 201 multiples of 10 and 200 numbers that are congruent to 4 | ||
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== See also == | == See also == | ||
{{AIME box|year=1984|num-b=11|num-a=13}} | {{AIME box|year=1984|num-b=11|num-a=13}} | ||
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− | + | [[Category:Intermediate Algebra Problems]] | |
− | + | [[Category:Intermediate Number Theory Problems]] |
Revision as of 19:06, 9 April 2008
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 |