Difference between revisions of "2017 AIME I Problems/Problem 9"
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Without checking whether there are other sums congruent to <math>45 \pmod{99}</math>, we can just write the answer to be <math>\boxed{045}</math>. | Without checking whether there are other sums congruent to <math>45 \pmod{99}</math>, we can just write the answer to be <math>\boxed{045}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>b_n = 2a_{n+10}</math>. We can find a formula for <math>b_n</math>: | ||
+ | |||
+ | <math>b_n = (20+n)(n+1)</math>. | ||
+ | |||
+ | Notice that both can't have a factor of 3. Thus we can limit our search range of n to <math>n \equiv 7,8 \pmod{9}</math>. Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, <math>b_n = 2a_{n+10}</math>, so, the 35th term in b corresponds to the 45th term in a. Thus our answer is <math>\boxed{045}</math> | ||
+ | |||
+ | -AlexLikeMath | ||
==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 00:28, 21 August 2019
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 .
Note that we can also construct the solution using CRT by assuming either divides and divides , or divides and divides , and taking the smaller solution.
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
Solution 4 (not good, risky)
We just notice that , so we are just trying to find modulo , or modulo . Also, the sum to is divisible by , and is the first one that is. Thus, if we sum to the is cut off and thus is just a sum to .
Without checking whether there are other sums congruent to , we can just write the answer to be .
Solution 5
Let . We can find a formula for :
.
Notice that both can't have a factor of 3. Thus we can limit our search range of n to . Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, , so, the 35th term in b corresponds to the 45th term in a. Thus our answer is
-AlexLikeMath
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.