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

Latest revision as of 15:24, 28 May 2023

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Find <math>a_{}^{}x^5 + b_{}y^5</math> if the real numbers <math>a_{}^{}</math>, <math>b_{}^{}</math>, <math>x_{}^{}</math>, and <math>y_{}^{}</math> satisfy the equations
+Find <math>ax^5 + by^5</math> if the real numbers <math>a,b,x,</math> and <math>y</math> satisfy the equations
-<center><math>ax + by = 3^{}_{},</math></center>
+<cmath>\begin{align*}
-<center><math>ax^2 + by^2 = 7^{}_{},</math></center>
+ax + by &= 3, \\
-<center><math>ax^3 + by^3 = 16^{}_{},</math></center>
+ax^2 + by^2 &= 7, \\
-<center><math>ax^4 + by^4 = 42^{}_{}.</math></center>
+ax^3 + by^3 &= 16, \\
+ax^4 + by^4 &= 42.
+\end{align*}</cmath>
+== Solution 1 ==
+Set <math>S = (x + y)</math> and <math>P = xy</math>. Then the relationship
+<cmath>(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})</cmath>
+can be exploited:
+<cmath>\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*}</cmath>
+Therefore:
+<cmath>\begin{eqnarray*}7S & = & 16 + 3P \\
+S & = & 42 + 7P\end{eqnarray*}</cmath>
+Consequently, <math>S = - 14</math> and <math>P = - 38</math>. Finally:
+<cmath>\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*}</cmath>
+== Solution 2 ==
+A [[linear recurrence | recurrence]] of the form <math>T_n=AT_{n-1}+BT_{n-2}</math> will have the closed form <math>T_n=ax^n+by^n</math>, where <math>x,y</math> are the values of the starting term that make the sequence geometric, and <math>a,b</math> 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 <math>T_1=3</math> and <math>T_2=7</math>. Then <math>T_3=ax^3+by^3=16=7A+3B</math>, and <math>T_4=ax^4+by^4=42=16A+7B</math>.
+Solving these simultaneous equations for <math>A</math> and <math>B</math>, we see that <math>A=-14</math> and <math>B=38</math>. So, <math>ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}</math>.
+== Solution 3 ==
+Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:
+<math>a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16</math>
+<math>ax\left( \frac{x^3-1}{x-1} \right) + by\left( \frac{y^3-1}{y-1} \right) = 16</math>.
+Similarly take the first two terms, yielding:
+<math>ax \left( \frac{x^2-1}{x-1} \right) + by \left( \frac{y^2-1}{y-1} \right) = 10</math>.
+Lastly take an alternating three-term sum,
+<math>a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12</math>
+<math>ax \left( \frac{x^3+1}{x+1} \right) + by \left( \frac{y^3+1}{y+1} \right) = 12</math>.
+Now to get the solution, let the answer be <math>k</math>, so
+<math>ax \left( \frac{x^4-1}{x-1} \right) + by \left(\frac{y^4-1}{y-1} \right) = 68</math>.
+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 <math>k.</math> Multiplying the first equation by <math>x</math> gives <math>ax^2 + bxy=3x</math>.
+Subtracting this equation from the second equation gives <math>by^2-bxy=7-3x</math>. Similarly, doing the same for the other equations, we obtain:
+<math>by^2-bxy=7-3x</math>, <math>by^3-bxy^2=16-7x</math>, <math>by^4-bxy^3=42-16x</math>, and <math>by^5-bxy^4=k-42x</math>
+Now lets take the first equation. Multiplying this by <math>y</math> and subtracting this from the second gives us  <math>by^3-bxy^2=(7-3x)y</math>. We can also obtain  <math>by^4-bxy^3=(16-7x)y</math>.
+Now we can solve for <math>x</math> and <math>y</math>! <math>(7-3x)y = 16-7x</math> and <math>(16-7x)y=42-16x</math>. Solving for <math>x</math> and <math>y</math> gives us <math>(-7+\sqrt{87},-7-\sqrt{87})</math> (It can be switched, but since the given equations are symmetric, it doesn't matter). <math>k-42x= (42-16x)y</math>, and solving for <math>k</math> gives us <math>k= \boxed{020}</math>.
+~pi_is_3.141
-== Solution ==
-{{solution}}
 == See also ==
 {{AIME box|year=1990|num-b=14|after=Last question}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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