Difference between revisions of "1984 AIME Problems/Problem 12"
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== Solution 3 == | == 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> | 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> | ||
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+ | ~[[User:N828335|N828335]] | ||
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== See also == | == See also == | ||
{{AIME box|year=1984|num-b=11|num-a=13}} | {{AIME box|year=1984|num-b=11|num-a=13}} | ||
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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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