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1990 AIME Problems/Problem 15

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Problem

Find a_{}^{}x^5 + b_{}y^5 if the real numbers a_{}^{}, b_{}^{}, x_{}^{}, and y_{}^{} satisfy the equations ax + by = 3^{}_{}, ax^2 + by^2 = 7^{}_{}, ax^3 + by^3 = 16^{}_{}, ax^4 + by^4 = 42^{}_{}.

Solution

Set S = (x + y) and P = xy. Then the relationship

(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})

can be exploited:

\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\(ax^3 + by^3)(x + y) & = & (ax^4 ...

Therefore:

\begin{eqnarray*}7S & = & 16 + 3P \\16S & = & 42 + 7P\end{eqnarray*}

Consequently, S = - 14 and P = - 38. Finally:

\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\(42)(S) & = & (ax^5 + by^5) +...

See also

1990 AIME (ProblemsResources)
Preceded by
Problem 14
Followed by
Last question
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