Difference between revisions of "1990 AIME Problems/Problem 15"
Hashtagmath (talk | contribs) m (→Solution 1) |
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(42)(S) & = & (ax^5 + by^5) + (P)(16) \\ | (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ | ||
(42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ | (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ | ||
− | ax^5 + by^5 & = & \boxed{ | + | ax^5 + by^5 & = & \boxed{020}\end{eqnarray*}</cmath> |
== Solution 2 == | == Solution 2 == |
Revision as of 12:00, 17 December 2018
Problem
Find if the real numbers , , , and satisfy the equations
Solution 1
Set and . Then the relationship
can be exploited:
Therefore:
Consequently, and . Finally:
Solution 2
A recurrence of the form will have the closed form , where are the values of the starting term that make the sequence geometric, and are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppose we have such a recurrence with and . Then , and .
Solving these simultaneous equations for and , we see that and . So, .
Solution 3
Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:
.
Similarly take the first two terms, yielding:
.
Lastly take an alternating three-term sum,
.
Now to get the solution, let the answer be , so
.
Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated as done in the first solution.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
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.