Difference between revisions of "1990 AIME Problems/Problem 15"

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{20}\end{eqnarray*}</cmath>
+
ax^5 + by^5 & = & \boxed{020}\end{eqnarray*}</cmath>
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 12:00, 17 December 2018

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 1

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 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}

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) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{020}\end{eqnarray*}

Solution 2

A recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$, where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ 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 $T_1=3$ and $T_2=7$. Then $T_3=ax^3+by^3=16=7A+3B$, and $T_4=ax^4+by^4=42=16A+7B$.

Solving these simultaneous equations for $A$ and $B$, we see that $A=-14$ and $B=38$. So, $ax^5+by^5=T_5=-14(42)+38(16)= \boxed{20}$.

Solution 3

Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:

$a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16$ $ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16$.

Similarly take the first two terms, yielding:

$ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10$.

Lastly take an alternating three-term sum,

$a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12$ $ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12$.

Now to get the solution, let the answer be $k$, so

$ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68$.

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 (ProblemsAnswer KeyResources)
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