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

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== Problem ==
 
== Problem ==
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<!-- 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 ==
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{{solution}}
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== Solution 1 ==
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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,
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<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
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<cmath>f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}</cmath>
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satisfies the conditions and has no other roots.
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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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== Solution 2 (non-rigorous) ==
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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>
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Solution by [[User:a1b2|a1b2]]
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== Solution 3 ==
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Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that <math>x = 0, \pm 5, \pm 10, \pm 15... \pm 1000</math> so the answer is 400 + 1 = <math>\boxed{401}</math>
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== Solution 4==
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Let <math>z</math> be an arbitrary zero. If <math>z=2-x</math>, then <math>x=2-z</math> and <math>2+x=4-z</math>. Repeat with other equation to find if <math>z</math> is a zero then so are <math>4-z</math> and <math>14-z</math>. From <math>0</math>, we get <math>4</math> and <math>14</math>. Now note that applying either of these twice will return <math>z</math>, so we must apply them in an alternating fashion for distinct roots. Doing so to <math>4</math> and <math>14</math> returns <math>10</math> and <math>-10</math>, respectively. A pattern will emerge of each path hitting a multiple of <math>10</math> after <math>2</math> moves. Hence, we will reach <math>\pm 1000</math> after <math>200</math> jumps in either direction. Including zero, there are <math>2\cdot200+1=\boxed{401}</math>
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~[[User:N828335|N828335]]
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== See also ==
 
== See also ==
* [[1984 AIME Problems/Problem 11 | Previous problem]]
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{{AIME box|year=1984|num-b=11|num-a=13}}
* [[1984 AIME Problems/Problem 13 | Next problem]]
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* [[1984 AIME Problems]]
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 14:37, 22 July 2020

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 1

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 $\boxed{401}$.

Solution 2 (non-rigorous)

We notice that the function has reflectional symmetry across both $x=2$ and $x=7$. We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$. There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$. $QED \blacksquare$

Solution by a1b2

Solution 3

Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$

Solution 4

Let $z$ be an arbitrary zero. If $z=2-x$, then $x=2-z$ and $2+x=4-z$. Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$. From $0$, we get $4$ and $14$. Now note that applying either of these twice will return $z$, so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$, respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$

~N828335

See also

1984 AIME (ProblemsAnswer KeyResources)
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

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