Difference between revisions of "1989 AIME Problems/Problem 8"
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<cmath>3 \cdot 123 - 3 \cdot 12 + 1 = \boxed{334}.</cmath> | <cmath>3 \cdot 123 - 3 \cdot 12 + 1 = \boxed{334}.</cmath> | ||
− | I first saw this in a Mathologer video | + | I first saw this in a Mathologer video: https://www.youtube.com/watch?v=4AuV93LOPcE |
~Ritwin | ~Ritwin |
Latest revision as of 04:15, 20 November 2023
Contents
Problem
Assume that are real numbers such that
Find the value of
.
Solution 1 (Quadratic Function)
Note that each given equation is of the form for some
When we expand and combine like terms, we obtain a quadratic function of
where
and
are linear combinations of
and
We are given that
and we wish to find
We eliminate by subtracting the first equation from the second, then subtracting the second equation from the third:
By either substitution or elimination, we get
and
Substituting these back produces
Finally, the answer is
~Azjps ~MRENTHUSIASM
Solution 2 (Linear Combination)
For simplicity purposes, we number the given equations and
in that order. Let
Subtracting
from
subtracting
from
and subtracting
from
we obtain the following equations, respectively:
Subtracting
from
and subtracting
from
we obtain the following equations, respectively:
Finally, applying the Transitive Property to
and
gives
from which
~Duohead ~MRENTHUSIASM
Solution 3 (Finite Differences by Arithmetic)
Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:
Label equations and
as Solution 2 does. Since the coefficients of
or
respectively, all form quadratic sequences with second differences
we conclude that the second differences of equations
must be constant.
It follows that the second differences of must be constant, as shown below:
Finally, we have from which
~MRENTHUSIASM
Solution 4 (Finite Differences by Algebra)
Notice that we may rewrite the equations in the more compact form as:
where
and
is what we are trying to find.
Now consider the polynomial given by (we are only treating the
as coefficients).
Notice that is in fact a quadratic. We are given
and are asked to find
. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find
.
Alternatively, applying finite differences, one obtains
Solution 5 (Assumption)
The idea is to multiply the first, second and third equations by and
respectively.
We can only consider the coefficients of and
Subtracting
from
we get
Subtracting
from
we get
Subtracting
from
we get
From
and
we have
Substituting this into
gives
Therefore, the answer is
Solution 6 (Assumption)
We let . Thus, we have
Grinding this out, we have
which gives
as our final answer.
~Pleaseletmewin
Solution 7 (Similar to Solutions 3 and 4)
Let be the sequence of perfect squares.
By either expanding or via finite differences, one can prove the miraculous recursion
Hence, the answer is simply
I first saw this in a Mathologer video: https://www.youtube.com/watch?v=4AuV93LOPcE
~Ritwin
Video Solution
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.