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

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== Problem ==
 
== Problem ==
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Find <math>a_{}^{}x^5 + b_{}y^5</math> if the [[real number]]s <math>a_{}^{}</math>, <math>b_{}^{}</math>, <math>x_{}^{}</math>, and <math>y_{}^{}</math> satisfy the [[equation]]s
+
Find <math>ax^5 + by^5</math> if the real numbers <math>a,b,x,</math> and <math>y</math> satisfy the equations
−
<cmath>ax + by = 3^{}_{},</cmath>
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<cmath>\begin{align*}
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<cmath>ax^2 + by^2 = 7^{}_{},</cmath>
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ax + by &= 3, \\
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<cmath>ax^3 + by^3 = 16^{}_{},</cmath>
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ax^2 + by^2 &= 7, \\
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<cmath>ax^4 + by^4 = 42^{}_{}.</cmath>
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ax^3 + by^3 &= 16, \\
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ax^4 + by^4 &= 42.
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\end{align*}</cmath>
  
 
== Solution 1 ==
 
== Solution 1 ==
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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>.  
 
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{20}</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 ==
 
== Solution 3 ==
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<math>a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16</math>
 
<math>a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16</math>
−
<math>ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 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:
 
Similarly take the first two terms, yielding:
−
 
+
−
<math>ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10</math>.
+
<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,
 
Lastly take an alternating three-term sum,
  
 
<math>a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12</math>
 
<math>a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12</math>
−
<math>ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 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  
 
Now to get the solution, let the answer be <math>k</math>, so  
  
−
<math>ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68</math>.
+
<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
 
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.
 
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
  
 
== See also ==
 
== See also ==

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

See also

1990 AIME (Problems • Answer Key • Resources)
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