Difference between revisions of "1984 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | 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>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> | ||
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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>. | 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]] | ||
+ | |||
+ | == 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 <math>x = 0, \pm 5, \pm 10, \pm 15... \pm 1000</math> so the answer is 400 + 1 = <math>\boxed{401}</math> | ||
+ | == Solution 4== | ||
+ | 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> | ||
+ | |||
+ | ~[[User:N828335|N828335]] | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:37, 22 July 2020
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
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 so the answer is 400 + 1 =
Solution 4
Let be an arbitrary zero. If
, then
and
. Repeat with other equation to find if
is a zero then so are
and
. From
, we get
and
. Now note that applying either of these twice will return
, so we must apply them in an alternating fashion for distinct roots. Doing so to
and
returns
and
, respectively. A pattern will emerge of each path hitting a multiple of
after
moves. Hence, we will reach
after
jumps in either direction. Including zero, there are
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 |
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