1990 AIME Problems/Problem 15

Problem

Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations $\begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*}$

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, $ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}$ .

Solution 3

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

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

Similarly take the first two terms, yielding:

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

Lastly take an alternating three-term sum,

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

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

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

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.

~lpieleanu (reformatting and minor edits)

Solution 4

We first let the answer to this problem be $k.$ Multiplying the first equation by $x$ gives $ax^2 + bxy=3x$ .

Subtracting this equation from the second equation gives $by^2-bxy=7-3x$ . Similarly, doing the same for the other equations, we obtain: $by^2-bxy=7-3x$ , $by^3-bxy^2=16-7x$ , $by^4-bxy^3=42-16x$ , and $by^5-bxy^4=k-42x$

Now lets take the first equation. Multiplying this by $y$ and subtracting this from the second gives us $by^3-bxy^2=(7-3x)y$ . We can also obtain $by^4-bxy^3=(16-7x)y$ .

Now we can solve for $x$ and $y$ ! $(7-3x)y = 16-7x$ and $(16-7x)y=42-16x$ . Solving for $x$ and $y$ gives us $(-7+\sqrt{87},-7-\sqrt{87})$ (It can be switched, but since the given equations are symmetric, it doesn't matter). $k-42x= (42-16x)y$ , and solving for $k$ gives us $k= \boxed{020}$ .

~pi_is_3.141

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last question
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