Difference between revisions of "1984 AIME Problems/Problem 12"
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== See also == | == See also == | ||
− | * [[ | + | {{AIME box|year=1984|num-b=11|num-a=13}} |
− | * [[ | + | * [[AIME Problems and Solutions]] |
− | * [[ | + | * [[American Invitational Mathematics Examination]] |
+ | * [[Mathematics competition resources]] |
Revision as of 14:26, 6 May 2007
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
If , then substituting gives . Similarly, . In particular,
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
satisfies the conditions and has no other roots.
In the interval , there are 201 multiples of 10 and 200 numbers that are congruent to 4 modulo 10, therefore the minimum number of roots is
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 |