Difference between revisions of "1984 AIME Problems/Problem 12"
(→Problem) |
|||
Line 2: | Line 2: | ||
<!-- 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> | <!-- 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 1 == |
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, | 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> | <cmath>f(t)=f(14-t)=f(14-(4-t))=f(t+10)</cmath> | ||
Line 12: | Line 12: | ||
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>. | ||
+ | == Solution 2 (non-rigorous) == | ||
+ | We notice that the function has reflectional symmetry across both <math>x=2</math> and <math>x=7</math>. We also use the fact that <math>x=0</math> is a root. This shows that <math>x=4</math> and <math>x=14</math> are also roots. We then apply the reflection across the other axis to form <math>x=\pm 10</math> as roots. Continuing this shows that the roots are <math>0 \mod 10</math> or <math>4 \mod 10</math>. There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of <math>\boxed{401}</math>. <math>QED \blacksquare</math> | ||
+ | |||
+ | Solution by [[User:a1b2|a1b2]] | ||
== 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 23:21, 13 April 2018
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 1
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 .
Solution 2 (non-rigorous)
We notice that the function has reflectional symmetry across both and . We also use the fact that is a root. This shows that and are also roots. We then apply the reflection across the other axis to form as roots. Continuing this shows that the roots are or . There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of .
Solution by a1b2
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 |