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 17:30, 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,
![\[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\]](http://latex.artofproblemsolving.com/3/d/1/3d1d66df8c4d5231f7ea007a2a059bf59798b2a1.png)
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
![\[f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}\]](http://latex.artofproblemsolving.com/d/1/f/d1f6e47e1c7f4c6baa45e814fa7223f5adc22adb.png)
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 | ||
