Difference between revisions of "2017 AIME I Problems/Problem 9"
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<cmath>0 \equiv (2n+1)^2+35 \pmod {99} </cmath> | <cmath>0 \equiv (2n+1)^2+35 \pmod {99} </cmath> | ||
<cmath>64 \equiv (2n+1)^2 \pmod {99} </cmath> | <cmath>64 \equiv (2n+1)^2 \pmod {99} </cmath> | ||
− | The | + | The only squares that are congruent to <math>64 \pmod {99}</math> are <math>(\pm 8)^2</math> and <math>(\pm 19)^2</math>, so |
<cmath>2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}</cmath> | <cmath>2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}</cmath> | ||
<math>2n+1 \equiv -8 \pmod {99}</math> yields <math>n=45</math> as the smallest integer solution. | <math>2n+1 \equiv -8 \pmod {99}</math> yields <math>n=45</math> as the smallest integer solution. | ||
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The smallest positive integer solution greater than <math>10</math> is <math>n=\boxed{045}</math>. | The smallest positive integer solution greater than <math>10</math> is <math>n=\boxed{045}</math>. | ||
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==See also== | ==See also== | ||
{{AIME box|year=2017|n=I|num-b=8|num-a=10}} | {{AIME box|year=2017|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:21, 8 March 2017
Contents
Problem 9
Let , and for each integer let . Find the least such that is a multiple of .
Solution 1
Writing out the recursive statement for and summing them gives Which simplifies to Therefore, is divisible by 99 if and only if is divisible by 99, so needs to be divisible by 9 and 11. Assume that is a multiple of 11. Writing out a few terms, , we see that is the smallest that works in this case. Next, assume that is a multiple of 11. Writing out a few terms, , we see that is the smallest that works in this case. The smallest is .
Solution 2
By looking at the first few terms, we can see that This implies Since , we can rewrite the equivalence, and simplify The only squares that are congruent to are and , so yields as the smallest integer solution.
yields as the smallest integer solution.
yields as the smallest integer solution.
yields as the smallest integer solution. However, must be greater than .
The smallest positive integer solution greater than is .
See also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.