Difference between revisions of "1984 AIME Problems/Problem 12"
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<cmath>f(t)=f(14-t)=f(14-(4-t))=f(t+10)</cmath> | <cmath>f(t)=f(14-t)=f(14-(4-t))=f(t+10)</cmath> | ||
− | Since <math>0</math> is a root, all multiples of <math>10</math> are roots, and anything congruent to <math>4\pmod{10}</math> | + | 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> | <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 <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>. | ||
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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}} |
Revision as of 18:30, 30 June 2013
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 |