Difference between revisions of "1984 AIME Problems/Problem 12"
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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> | <!-- 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]] | ||
== 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 22:21, 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 |