Difference between revisions of "2017 AIME I Problems/Problem 9"
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<math>a_n=a_{n-1} + n \pmod{99}</math>. Using the steps of the previous solution we get up to <math>n^2+n+ \equiv 90 \pmod{99}</math>. This gives away the fact that <math>(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}</math> so either <math>n</math> or <math>n+1</math> must be a multiple of 9. | <math>a_n=a_{n-1} + n \pmod{99}</math>. Using the steps of the previous solution we get up to <math>n^2+n+ \equiv 90 \pmod{99}</math>. This gives away the fact that <math>(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}</math> so either <math>n</math> or <math>n+1</math> must be a multiple of 9. | ||
− | Case 1 (<math>n|9</math>): Say <math>n=9x</math> and after simplification <math>x(9x+1) = | + | Case 1 (<math>n|9</math>): Say <math>n=9x</math> and after simplification <math>x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}</math>. |
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+ | Case 2: (<math>n+1|9</math>): Say <math>n=9a</math> and after simplification <math>(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}</math>. | ||
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+ | As a result <math>a</math> must be a divisor of <math>10</math> and after doing some testing in both cases the smallest value that works is <math>x=5 \implies \boxed{45}</math>. | ||
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+ | ~First | ||
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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 17:40, 9 March 2017
Contents
[hide]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 .
Solution 3
. Using the steps of the previous solution we get up to . This gives away the fact that so either or must be a multiple of 9.
Case 1 (): Say and after simplification .
Case 2: (): Say and after simplification .
As a result must be a divisor of and after doing some testing in both cases the smallest value that works is .
~First
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 |
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