Difference between revisions of "2016 AIME II Problems/Problem 9"
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Solution by rocketscience | Solution by rocketscience | ||
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| + | == Solution 3 (More Robust Bash) == | ||
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| + | The reason for bashing in this context can also be justified by the fact 100 isn't very big. | ||
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| + | Let the common difference for the arithmetic sequence be <math>a</math>, and the common ratio for the geometric sequence be <math>b</math>. The sequences are now <math>1, a+1, 2a+1, \ldots</math>, and <math>1, b, b^2, \ldots</math>. We can now write the given two equations as the following: | ||
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| + | <math>1+(k-2)a+b^{k-2} = 100</math> | ||
| + | <math>1+ka+b^k = 1000</math> | ||
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| + | Take the difference between the two equations to get <math>2a+(b^2-1)b^{k-2} = 900</math>. Since 900 is divisible by 4, we can tell <math>a</math> is even and <math>b</math> is odd. Let <math>a=2m</math>, <math>b=2n+1</math>, where <math>m</math> and <math>n</math> are positive integers. Substitute variables and divide by 4 to get: | ||
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| + | <math>m+(n+1)(n)(2n+1)^{k-2} = 225</math> | ||
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| + | Because very small integers for <math>n</math> yield very big results, we can bash through all cases of <math>n</math>. Here, we set an upper bound for <math>n</math> by setting <math>k</math> as 3. After trying values, we find that <math>n\leq 4</math>, so <math>b=9, 7, 5, 3</math>. Testing out <math>b=9</math> yields the correct answer of <math>\boxed{262}</math>. Note that even if this answer were associated with another b value like <math>b=3</math>, the value of <math>k</math> is still 3 for all of the cases. | ||
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| + | -Dankster42 | ||
== 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}} | ||
Revision as of 19:30, 17 March 2020
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
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 3 (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 is still 3 for all of the cases.
-Dankster42
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 | ||
