2017 AIME I Problems/Problem 9
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
See also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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