Difference between revisions of "2016 AIME II Problems/Problem 9"
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Case <math>3.2.\hspace{10mm} b = 3.</math> | Case <math>3.2.\hspace{10mm} b = 3.</math> | ||
<cmath>c_{k-1} = 3^{k-2} + (k-2) a + 1 = 100, c_{k+1} = 3^k + ka + 1 = 1000\implies a = 450-4\cdot 3^{k-2} \implies 3^{k-4}(9-4k) +50k = 3\cdot 37 \implies \O.</cmath> | <cmath>c_{k-1} = 3^{k-2} + (k-2) a + 1 = 100, c_{k+1} = 3^k + ka + 1 = 1000\implies a = 450-4\cdot 3^{k-2} \implies 3^{k-4}(9-4k) +50k = 3\cdot 37 \implies \O.</cmath> | ||
− | ''' | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=8|num-a=10}} | {{AIME box|year=2016|n=II|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:38, 1 September 2022
Contents
Problem
The sequences of positive integers and
are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let
. There is an integer
such that
and
. Find
.
Solution 1
Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for . When we get to
and
, we have
and
, which works, therefore, the answer is
.
Solution 2 (No trial and error)
We have and
. First,
implies
, so
.
It follows that , i.e.,
Moreover, since
is atleast
we get
, i.e.
. For every value of
in this range, define
, and define
. We are looking for values of
such that
. Let's make a table:
The only admissible values for
are
. However, since
, we must have
. This does not hold for
because
does not divide
. This leaves
as the only option.
For and
, we check:
implies
, i.e.
. Then
and
and
; so it works! Then
.
Solution 3
Using the same reasoning ( isn't very big), we can guess which terms will work. The first case is
, so we assume the second and fourth terms of
are
and
. We let
be the common ratio of the geometric sequence and write the arithmetic relationships in terms of
.
The common difference is , and so we can equate:
. Moving all the terms to one side and the constants to the other yields
, or
. Simply listing out the factors of
shows that the only factor
less than a square that works is
. Thus
and we solve from there to get
.
Solution by rocketscience
Solution 4 (More Robust Bash)
The reason for bashing in this context can also be justified by the fact 100 isn't very big.
Let the common difference for the arithmetic sequence be , and the common ratio for the geometric sequence be
. The sequences are now
, and
. We can now write the given two equations as the following:
Take the difference between the two equations to get . Since 900 is divisible by 4, we can tell
is even and
is odd. Let
,
, where
and
are positive integers. Substitute variables and divide by 4 to get:
Because very small integers for yield very big results, we can bash through all cases of
. Here, we set an upper bound for
by setting
as 3. After trying values, we find that
, so
. Testing out
yields the correct answer of
. Note that even if this answer were associated with another b value like
, the value of
can still only be 3 for all of the cases.
Solution 5 (Casework)
Let and
be in the form of
Case
Case
Case
Case
Case
vladimir.shelomovskii@gmail.com, vvsss
See also
2016 AIME II (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.