Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1984 AIME Problems/Problem 12"

(→‎See also)
m
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
A function <math>\displaystyle f</math> is defined for all real numbers and satisfies <math>\displaystyle f(2+x)=f(2-x)</math> and <math>\displaystyle f(7+x)=f(7-x)</math> for all <math>\displaystyle x</math>. If <math>\displaystyle x=0</math> is a root for <math>\displaystyle f(x)=0</math>, what is the least number of roots <math>\displaystyle f(x)=0</math> must have in the interval <math>\displaystyle -1000\leq x \leq 1000</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>\displaystyle f(2+x)=f(2-x)</math>, then substituting <math>\displaystyle t=2+x</math> gives <math>\displaystyle f(t)=f(4-t)</math>. Similarly, <math>\displaystyle 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>
−
<math>\displaystyle f(t)=f(14-t)=f(14-(4-t))=f(t+10)</math>
 
−
 
 
−
 
 
−
Since 0 is a root, all multiples of 10 are roots, and anything of the form "4 minus a multiple of 10" (that is, anything congruent to 4 modulo 10) are also roots. To see that these may be the only integer roots, observe that the function
 
−
 
 
−
<math>\sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{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>
 
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 modulo 10, therefore the minimum number of roots is
 
−
 
 
−
<math>401</math>
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=11|num-a=13}}
 
{{AIME box|year=1984|num-b=11|num-a=13}}
−
* [[AIME Problems and Solutions]]
+
 
−
* [[American Invitational Mathematics Examination]]
+
[[Category:Intermediate Algebra Problems]]
−
* [[Mathematics competition resources]]
+
[[Category:Intermediate Number Theory Problems]]

Revision as of 19:06, 9 April 2008

Problem

A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$. If $x=0$ is a root for $f(x)=0$, what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$?

Solution

If $f(2+x)=f(2-x)$, then substituting $t=2+x$ gives $f(t)=f(4-t)$. Similarly, $f(t)=f(14-t)$. In particular, \[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\]

Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$) are also roots. To see that these may be the only integer roots, observe that the function \[f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}\] satisfies the conditions and has no other roots.

In the interval $-1000\leq x\leq 1000$, there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$, therefore the minimum number of roots is $\boxed{401}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_12&oldid=24637"