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human
|
Compute the sum of the squares of the roots of the equation \[x^{2018} +44x^{2015} + 3x^3 + 404 = 0.\]
|
gpt
|
Let $r_1, r_2, \dots, r_{2018}$ be the roots. By Vieta's formulas, $r_1+r_2+\dots+r_{2018}=0.$ To get the squared terms we want, we square both sides, giving \[(r_1^2+r_2^2+\dots+r_{2018}^2) + 2(r_1r_2+r_1r_3+\dotsb) = 0,\]where the second term on the left-hand side is the sum of all terms $r_ir_j,$ where $i < j.$ By Vieta's formulas, this also equals $0,$ so \[r_1^2+r_2^2+\dots+r_{2018}^2=\boxed{0}\,.\]
|
human
|
Find the number of functions $f(n),$ taking the integers to the integers, such that
\[f(a + b) + f(ab) = f(a) f(b) + 1\]for all integers $a$ and $b.$
|
gpt
|
Setting $a = b = 0,$ we get
\[2f(0) = f(0)^2 + 1.\]Then $f(0)^2 - 2f(0) + 1 = (f(0) - 1)^ 2 = 0,$ so $f(0) = 1.$
Setting $a = 1$ and $b = -1,$ we get
\[f(0) + f(-1) = f(1) f(-1) + 1,\]so $f(-1) (f(1) - 1) = 0.$ This means either $f(-1) = 0$ or $f(1) = 1.$
First, we look at the case where $f(1) = 1.$ Setting $b = 1,$ we get
\[f(a + 1) + f(a) = f(a) + 1,\]so $f(a + 1) = 1.$ This means $f(n) = 1$ for all integers $n.$
Next, we look at the case where $f(-1) = 0.$ Setting $a = b = -1,$ we get
\[f(-2) + f(1) = f(-1)^2 + 1 = 1.\]Setting $a = 1$ and $b = -2,$ we get
\[f(-1) + f(-2) = f(1) f(-2) + 1,\]which simplifies to $f(-2) = f(1) f(-2) + 1.$ Substituting $f(-2) = 1 - f(1),$ we get
\[1 - f(1) = f(1) (1 - f(1)) + 1,\]which simplifies to $f(1)^2 - 2f(1) = f(1) (f(1) - 2) = 0.$ Hence, either $f(1) = 0$ or $f(1) = 2.$
First, we look at the case where $f(1) = 0.$ Setting $b = 1,$ we get
\[f(a + 1) + f(a) = 1,\]so $f(a + 1) = 1 - f(a).$ This means $f(n)$ is 1 if $n$ is even, and 0 if $n$ is odd.
Next, we look at the case where $f(1) = 2.$ Setting $b = 1,$ we get
\[f(a + 1) + f(a) = 2f(a) + 1,\]so $f(a + 1) = f(a) + 1.$ Combined with $f(1) = 2,$ this means $f(n) = n + 1$ for all $n.$
Thus, there a total of $\boxed{3}$ functions: $f(n) = 1$ for all $n,$ $f(n) = n + 1$ for all $n,$ and
\[f(n) = \left\{
\begin{array}{cl}
1 & \text{if $n$ is even}, \\
0 & \text{if $n$ is odd}.
\end{array}
\right.\]We check that all three functions work.
|
human
|
Let $x$ be a positive real number. Find the minimum value of $4x^5 + 5x^{-4}.$
|
gpt
|
By AM-GM,
\begin{align*}
4x^5 + 5x^{-4} &= x^5 + x^5 + x^5 + x^5 + x^{-4} + x^{-4} + x^{-4} + x^{-4} + x^{-4} \\
&\ge 9 \sqrt[9]{(x^5)^4 \cdot (x^{-4})^5} \\
&= 9.
\end{align*}Equality occurs when $x = 1,$ so the minimum value is $\boxed{9}.$
|
human
|
Let $a,$ $b,$ $c$ be the roots of the cubic $x^3 + 3x^2 + 5x + 7 = 0.$ Given that $P(x)$ is a cubic polynomial such that $P(a) = b + c,$ $P(b) = a + c,$ $P(c) = a + b,$ and $P(a + b + c) = -16,$ find $P(x).$
|
gpt
|
By Vieta's formulas, $a + b + c = -3,$ so $P(-3) = -16.$
Let $Q(x) = P(x) + x + 3.$ Then
\begin{align*}
Q(a) &= b + c + a + 3 = 0, \\
Q(b) &= a + c + b + 3 = 0, \\
Q(c) &= a + b + c + 3 = 0, \\
Q(-3) &= P(-3) - 3 + 3 = -16.
\end{align*}Hence, $Q(x) = k(x - a)(x - b)(x - c) = k(x^3 + 3x^2 + 5x + 7)$ for some constant $k.$ Setting $x = -3,$ we get
\[-16 = -8k,\]so $k = 2.$ Then $Q(x) = 2(x^3 + 3x^2 + 5x + 7),$ so
\[P(x) = Q(x) - x - 3 = 2(x^3 + 3x^2 + 5x + 7) - x - 3 = \boxed{2x^3 + 6x^2 + 9x + 11}.\]
|
human
|
The sequences of positive integers $1,a_2, a_3, \dots$ and $1,b_2, b_3, \dots$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$. There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$. Find $c_k$.
|
gpt
|
Let $d$ be the common difference, and let $r$ be the common ratio, so $d$ and $r$ are positive integers. Then $a_n = 1 + (n - 1) d$ and $b_n = r^{n - 1},$ so
\begin{align*}
1 + (k - 2) d + r^{k - 2} &= 100, \\
1 + kd + r^k &= 1000.
\end{align*}Then
\begin{align*}
(k - 2) d + r^{k - 2} &= 99, \\
kd + r^k &= 999.
\end{align*}From the second equation, $r^k < 999.$ If $k \ge 4,$ then $r < 999^{1/4},$ so $r \le 5.$
Since the geometric sequence is increasing, $r \neq 1,$ so the possible values of $r$ are 2, 3, 4, and 5. We can write the equations above as
\begin{align*}
(k - 2) d &= 99 - r^{k - 2}, \\
kd &= 999 - r^k.
\end{align*}Thus, $99 - r^{k - 2}$ is divisible by $k - 2,$ and $999 - r^k$ is divisible by $k.$
If $r = 2,$ then the only possible values of $k$ are 4, 5, 6, 7, and 8. We find that none of these values work.
If $r = 3,$ then the only possible values of $k$ are 4, 5, and 6. We find that none of these values work.
If $r = 4,$ then the only possible values of $k$ is 4. We find that this value does not work.
If $r = 4,$ then the only possible values of $k$ is 4. We find that this value does not work.
Therefore, we must have $k = 3,$ so
\begin{align*}
d + r &= 99, \\
3d + r^3 &= 999.
\end{align*}From the first equation, $d = 99 - r.$ Substituting, we get
\[3(99 - r) + r^3 = 999,\]so $r^3 - 3r - 702 = 0.$ This factors as $(r - 9)(r^2 + 9r + 78) = 0,$ so $r = 9,$ so $d = 90.$ Then $a_3 = 1 + 2 \cdot 90 = 181$ and $c_3 = 9^2 = 81,$ and $c_3 = 181 + 81 = \boxed{262}.$
|
human
|
Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$.
|
gpt
|
If $x^2-x-1$ is a factor of $ax^{17}+bx^{16}+1,$ then both the roots of $x^2-x-1$ must also be roots of $ax^{17}+bx^{16}+1.$ Let $s$ and $t$ be the roots of $x^2-x-1.$ Then we must have \[as^{17} + bs^{16} + 1 = at^{17} + bt^{16} + 1 = 0.\]Since $s$ is a root of $s^2-s-1=0,$ we have $s^2=s+1.$ This equation lets us express higher powers of $s$ in the form $Ms+N,$ for constants $M$ and $N.$ We have \[\begin{aligned} s^3 &= s^2 \cdot s = (s+1)s = s^2+s=(s+1)+s=2s+1, \\ s^4 &= s^3 \cdot s = (2s+1)s = 2s^2 + s = 2(s+1) + s = 3s+2, \\ s^5 &= s^4 \cdot s =(3s+2)s = 3s^2+2s=3(s+1)+2s=5s+3, \end{aligned}\]and so on. Seeing a pattern, we guess that \[s^n = F_ns + F_{n-1},\]where $\{F_n\}$ are the Fibonacci numbers (with $F_1 = F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3$). We can prove this formula with induction (see below). This means that \[s^{16} = F_{16}s + F_{15} = 987s + 610 \; \text{ and } \; s^{17} = F_{17}s + F_{16} = 1597s + 987.\]Thus, \[as^{17} + bs^{16} + 1 = (1597a+987b)s + (987a+610b) + 1,\]so it must be the case that $1597a + 987b = 0$ and $987a + 610b =- 1.$ This system has solutions $a = \boxed{987}$ and $b = -1597.$
Proof of formula: We already did the base cases of the induction. If $s^n = F_ns + F_{n-1}$ for some value of $n,$ then \[\begin{aligned} s^{n+1} = s^n \cdot s &= (F_ns + F_{n-1}) \cdot s \\ &= F_ns^2 + F_{n-1}s\\ & = F_n(s+1) + F_{n-1}s\\ & = (F_n+F_{n-1})s + F_n = F_{n+1}s + F_n. \end{aligned}\]This completes the inductive step. $\square$
|
human
|
The equation \[\frac{x}{x+1} + \frac{x}{x+2} = kx\]has exactly two complex roots. Find all possible complex values for $k.$
Enter all the possible values, separated by commas.
|
gpt
|
Multiplying both sides by $(x+1)(x+2),$ we get \[x(x+2) + x(x+1) = kx(x+1)(x+2),\]or \[2x^2 + 3x = kx^3 + 3kx^2 + 2kx.\]This rearranges to the equation \[0 = kx^3 + (3k-2)x^2 + (2k-3)x,\]or \[0 = x(kx^2 + (3k-2)x + (2k-3)).\]Clearly $x = 0$ is a root of this equation. All the other roots must satisfy the equation \[0 = kx^2 + (3k-2)x + (2k-3).\]If $k = 0,$ then the equation becomes $-2x - 3 = 0,$ so $x = -\frac{3}{2}.$ Thus, $k = 0$ works.
Otherwise, the $x^2$ coefficient of the right-hand side is nonzero, so the equation is a proper quadratic equation. For the given equation to have exactly two roots, one of the following must be true:
The quadratic has $0$ as a root, and the other root is nonzero. Setting $x = 0,$ we get $0 = 2k-3,$ so $k = \tfrac32.$ This is a valid solution, because then the equation becomes $0 = \tfrac32 x^2 + \tfrac52 x,$ which has roots $x = 0$ and $x = -\tfrac53.$
The quadratic has two equal, nonzero roots. In this case, the discriminant must be zero: \[(3k-2)^2 - 4k(2k-3) = 0,\]which simplifies to just $k^2 + 4 = 0.$ Thus, $k = \pm 2i.$ These are both valid solutions, because we learned in the first case that $k = \tfrac32$ is the only value of $k$ which makes $0$ a root of the quadratic; thus, the quadratic has two equal, nonzero roots for $k = \pm 2i.$
The possible values for $k$ are $k = \boxed{0,\tfrac32, 2i, -2i}.$
|
human
|
Let $a$ and $b$ be positive real numbers such that $a^3 + b^3 = a + b.$ Simplify
\[\frac{a}{b} + \frac{b}{a} - \frac{1}{ab}.\]
|
gpt
|
From the equation $a^3 + b^3 = a + b,$
\[(a + b)(a^2 - ab + b^2) = a + b.\]Since $a$ and $b$ are positive, $a + b$ is positive, so we can cancel the factors of $a + b$ to get
\[a^2 - ab + b^2 = 1.\]Then
\[\frac{a^2 + b^2 - 1}{ab} = \frac{ab}{ab} = \boxed{1}.\]
|
human
|
Find constants $A,$ $B,$ and $C$ so that
\[\frac{x^2 - 7}{(x - 2)(x - 3)(x - 5)} = \frac{A}{x - 2} + \frac{B}{x - 3} + \frac{C}{x - 5}.\]Enter the ordered triple $(A,B,C).$
|
gpt
|
Multiplying both sides by $(x - 2)(x - 3)(x - 5),$ we get
\[x^2 - 7 = A(x - 3)(x - 5) + B(x - 2)(x - 5) + C(x - 2)(x - 3).\]Setting $x = 2,$ we get $3A = -3,$ so $A = -1.$
Setting $x = 3,$ we get $-2B = 2,$ so $B = -1.$
Setting $x = 5,$ we get $6C = 18,$ so $C = 3.$ Thus, $(A,B,C) = \boxed{(-1,-1,3)}.$
|
human
|
The sequence $a_1,$ $a_2,$ $a_3,$ $\dots$ satisfies $a_1 = 19,$ $a_9 = 99,$ and for all $n \ge 3,$ $a_n$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2.$
|
gpt
|
For $n \ge 3,$ we have that
\[a_n = \frac{a_1 + a_2 + \dots + a_{n - 1}}{n - 1},\]or
\[(n - 1) a_n = a_1 + a_2 + \dots + a_{n - 1}.\]Likewise,
\[n a_{n + 1} = a_1 + a_2 + \dots + a_{n - 1} + a_n.\]Subtracting these equations, we get
\[n a_{n + 1} - (n - 1) a_n = a_n,\]so $n a_{n + 1} = n a_n.$ Then $a_{n + 1} = a_n.$
This means that the terms $a_3,$ $a_4,$ $a_5,$ $\dots$ are all equal. In particular, $a_3 = 99,$ so
\[\frac{19 + a_2}{2} = 99.\]We find $a_2 = \boxed{179}.$
|
human
|
A certain ellipse is tangent to both the $x$-axis and the $y$-axis, and its foci are at $(2, -3 + \sqrt{5})$ and $(2, -3 - \sqrt{5}).$ Find the length of the major axis.
|
gpt
|
We see that the center of the ellipse is $(2,-3),$ and that the major axis lies along the line $x = 2.$ Since the ellipse is tangent to the $x$-axis, one end-point of the major axis must be $(2,0),$ and the other end-point must be $(2,-6).$ Thus, the length of the major axis is $\boxed{6}.$
[asy]
unitsize(1 cm);
draw(shift((2,-3))*xscale(2)*yscale(3)*Circle((0,0),1));
draw((-1,0)--(4,0));
draw((0,1)--(0,-6));
draw((2,0)--(2,-6));
draw((0,-3)--(4,-3));
dot("$(2,0)$", (2,0), N);
dot("$(2,-6)$", (2,-6), S);
dot("$(2,-3)$", (2,-3), SE);
dot((2,-3 + sqrt(5)));
dot((2,-3 - sqrt(5)));
label("$(2, -3 + \sqrt{5})$", (2, -3 + sqrt(5)), E, UnFill);
label("$(2, -3 - \sqrt{5})$", (2, -3 - sqrt(5)), E, UnFill);
[/asy]
|
human
|
Find the remainder when $x^4 + 2$ is divided by $(x - 2)^2.$
|
gpt
|
Dividing $x^4 + 2$ by $x - 2,$ we get
\[x^4 + 2 = (x - 2)(x^3 + 2x^2 + 4x + 8) + 18.\]Dividing $x^3 + 2x^2 + 4x + 8$ by $x - 2,$ we get
\[x^3 + 2x^2 + 4x + 8 = (x - 2)(x^2 + 4x + 12) + 32.\]Thus,
\begin{align*}
x^4 + 2 &= (x - 2)(x^3 + 2x^2 + 4x + 8) + 18 \\
&= (x - 2)((x - 2)(x^2 + 4x + 12) + 32) + 18 \\
&= (x - 2)^2 (x^2 + 4x + 12) + 32(x - 2) + 18 \\
&= (x -2)^2 (x^2 + 4x + 12) + 32x - 46,
\end{align*}so the remainder is $\boxed{32x - 46}.$
|
human
|
Solve the inequality
\[\frac{1}{x - 1} - \frac{4}{x - 2} + \frac{4}{x - 3} - \frac{1}{x - 4} < \frac{1}{30}.\]
|
gpt
|
Subtracting $\frac{1}{30}$ from both sides, we get
\[\frac{1}{x - 1} - \frac{4}{x - 2} + \frac{4}{x - 3} - \frac{1}{x - 4} - \frac{1}{30} < 0.\]Putting everything over a common denominator, we get
\[\frac{-x^4 + 10x^3 - 5x^2 - 100x - 84}{30(x - 1)(x - 2)(x - 3)(x - 4)} < 0,\]which factors as
\[-\frac{(x + 2)(x + 1)(x - 6)(x - 7)}{(x - 1)(x - 2)(x - 3)(x - 4)} < 0.\]We can build a sign chart, but since all of the factors are linear, we can track what happens to the expression as $x$ increases. At $x = -3,$ the expression is negative. As $x$ increases past $-2,$ the expression becomes positive. As $x$ increases past $-1,$ the expression becomes negative, and so on. Thus, the solution is
\[x \in \boxed{(-\infty,-2) \cup (-1,1) \cup (2,3) \cup (4,6) \cup (7,\infty)}.\]
|
human
|
The function $f(x)$ satisfies
\[f(x + y) = f(x) + f(y)\]for all real numbers $x$ and $y,$ and $f(4) = 5.$ Find $f(5).$
|
gpt
|
We can write
\begin{align*}
f(4) &= f(3) + f(1) \\
&= f(2) + f(1) + f(1) \\
&= f(1) + f(1) + f(1) + f(1),
\end{align*}so $4f(1) = 5,$ which means $f(1) =\frac{5}{4}.$ Therefore,
\[f(5) = f(1) + f(4) = 5 + \frac{5}{4} = \boxed{\frac{25}{4}}.\]
|
human
|
Let $\omega$ be a nonreal root of $x^3 = 1.$ Compute
\[(1 - \omega + \omega^2)^4 + (1 + \omega - \omega^2)^4.\]
|
gpt
|
We know that $\omega^3 - 1 = 0,$ which factors as $(\omega - 1)(\omega^2 + \omega + 1) = 0.$ Since $\omega$ is not real, $\omega^2 + \omega + 1 = 0.$
Then
\[(1 - \omega + \omega^2)^4 + (1 + \omega - \omega^2)^4 = (-2 \omega)^4 + (-2 \omega^2)^4 = 16 \omega^4 + 16 \omega^8.\]Since $\omega^3 = 1,$ this reduces to $16 \omega + 16 \omega^2 = 16(\omega^2 + \omega) = \boxed{-16}.$
|
human
|
The polynomial $f(x)$ satisfies $f(x + 1) - f(x) = 6x + 4.$ Find the leading coefficient of $f(x).$
|
gpt
|
Let $n$ be a positive integer. Setting $x = 0,$ $1,$ $2,$ $\dots,$ $n - 1,$ we get
\begin{align*}
f(1) - f(0) &= 6 \cdot 0 + 4, \\
f(2) - f(1) &= 6 \cdot 1 + 4, \\
f(3) - f(2) &= 6 \cdot 2 + 4, \\
&\dots, \\
f(n) - f(n - 1) &= 6 \cdot (n - 1) + 4.
\end{align*}Adding all the equations, we get
\[f(n) - f(0) = 6 (0 + 1 + 2 + \dots + (n - 1)) + 4n = 6 \cdot \frac{n(n - 1)}{2} + 4n = 3n^2 + n.\]Since this holds for all positive integers $n,$
\[f(x) = 3x^2 + x + c\]for some constant $c.$ Hence, the leading coefficient of $f(x)$ is $\boxed{3}.$
|
human
|
Let $r$ and $s$ denote the two real roots of $x^2 - x \sqrt{5} + 1 = 0.$ Then determine $r^8 + s^8.$
|
gpt
|
By Vieta's formulas, $r + s = \sqrt{5}$ and $rs = 1.$ Squaring the equation $r + s = \sqrt{5},$ we get
\[r^2 + 2rs + s^2 = 5,\]so $r^2 + s^2 = 5 - 2rs = 3.$ Squaring this equation, we get
\[r^4 + 2r^2 s^2 + s^4 = 9,\]so $r^4 + s^4 = 9 - 2r^2 s^2 = 9 - 2 = 7.$ Squaring once more, we get
\[r^8 + 2r^4 s^4 + s^8 = 49,\]so $r^8 + s^8 = 49 - 2r^4 s^4 = \boxed{47}.$
|
human
|
Compute the sum of the roots of the equation \[x\sqrt{x} - 6x + 7\sqrt{x} - 1 = 0,\]given that all of the roots are real and nonnegative.
|
gpt
|
The given equation is not a polynomial equation, so we can't use Vieta's formulas directly. To create a related polynomial equation, we substitute $y = \sqrt{x},$ or $x = y^2,$ giving \[y^3 - 6y^2 + 7y - 1 = 0.\]For each value of $y$ which satisfies this equation, the corresponding value of $x$ which satisfies the original equation is $x = y^2.$ Therefore, we want to find the sum of the squares of the roots of this equation.
To do this, let $r,$ $s,$ and $t$ denote the roots of this equation. Then by Vieta's formulas, $r+s+t=6$ and $rs+st+tr=7,$ so \[r^2+s^2+t^2=(r+s+t)^2-2(rs+st+tr) = 6^2 - 2 \cdot 7 = \boxed{22}.\]
|
human
|
Find the sum $$\frac{2^1}{4^1 - 1} + \frac{2^2}{4^2 - 1} + \frac{2^4}{4^4 - 1} + \frac{2^8}{4^8 - 1} + \cdots.$$
|
gpt
|
Notice that $$\frac{2^{2^k}}{4^{2^k} - 1} = \frac{2^{2^k} + 1}{4^{2^k} - 1} - \frac{1}{4^{2^k} - 1} = \frac{1}{2^{2^k}-1} - \frac{1}{4^{2^k}-1} = \frac{1}{4^{2^{k-1}}-1} - \frac{1}{4^{2^k}-1}.$$Therefore, the sum telescopes as $$\left(\frac{1}{4^{2^{-1}}-1} - \frac{1}{4^{2^0}-1}\right) + \left(\frac{1}{4^{2^0}-1} - \frac{1}{4^{2^1}-1}\right) + \left(\frac{1}{4^{2^1}-1} - \frac{1}{4^{2^2}-1}\right) + \cdots$$and evaluates to $1/(4^{2^{-1}}-1) = \boxed{1}$.
|
human
|
For a positive real number $x > 1,$ the Riemann zeta function $\zeta(x)$ is defined by
\[\zeta(x) = \sum_{n = 1}^\infty \frac{1}{n^x}.\]Compute
\[\sum_{k = 2}^\infty \{\zeta(2k - 1)\}.\]Note: For a real number $x,$ $\{x\}$ denotes the fractional part of $x.$
|
gpt
|
For $x \ge 2,$
\begin{align*}
\zeta(x) &= 1 + \frac{1}{2^x} + \frac{1}{3^x} + \dotsb \\
&\le 1 + \frac{1}{2^2} + \frac{1}{3^2} + \dotsb \\
&< 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dotsb \\
&= 1 + \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dotsb \\
&= 2,
\end{align*}so $\lfloor \zeta(x) \rfloor = 1.$ Then
\[\{\zeta(x)\} = \zeta(x) - 1.\]Thus, we want to sum
\[\sum_{k = 2}^\infty (\zeta(2k - 1) - 1) = \sum_{k = 2}^\infty \sum_{n = 2}^\infty \frac{1}{n^{2k - 1}}.\]We switch the order of summation, to get
\begin{align*}
\sum_{n = 2}^\infty \sum_{k = 2}^\infty \frac{1}{n^{2k - 1}} &= \sum_{n = 2}^\infty \left( \frac{1}{n^3} + \frac{1}{n^5} + \frac{1}{n^7} + \dotsb \right) \\
&= \sum_{n = 2}^\infty \frac{1/n^3}{1 - 1/n^2} \\
&= \sum_{n = 2}^\infty \frac{1}{n^3 - n}.
\end{align*}By partial fractions,
\[\frac{1}{n^3 - n} = \frac{1/2}{n - 1} - \frac{1}{n} + \frac{1/2}{n + 1}.\]Therefore,
\begin{align*}
\sum_{n = 2}^\infty \frac{1}{n^3 - n} &= \sum_{n = 2}^\infty \left( \frac{1/2}{n - 1} - \frac{1}{n} + \frac{1/2}{n + 1} \right) \\
&= \left( \frac{1/2}{1} - \frac{1}{2} + \frac{1/2}{3} \right) + \left( \frac{1/2}{2} - \frac{1}{3} + \frac{1/2}{4} \right) + \left( \frac{1/2}{3} - \frac{1}{4} + \frac{1/2}{5} \right) + \dotsb \\
&= \frac{1/2}{1} - \frac{1}{2} + \frac{1/2}{2} = \boxed{\frac{1}{4}}.
\end{align*}
|
human
|
Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(x) f(y) - f(xy) = x + y\]for all real numbers $x$ and $y.$
Let $n$ be the number of possible values of $f(2),$ and let $s$ be the sum of all possible values of $f(2).$ Find $n \times s.$
|
gpt
|
Setting $x = y = 1,$ we get
\[f(1)^2 - f(1) = 2,\]so $f(1)^2 - f(1) - 2 = 0.$ This factors as $(f(1) + 1)(f(1) - 2) = 0,$ so $f(1) = -1$ or $f(1) = 2.$
Setting $y = 1,$ we get
\[f(x) f(1) - f(x) = x + 1\]for all $x.$ Then $f(x) (f(1) - 1) = x + 1.$ Since $f(1) \neq 1,$ we can write
\[f(x) = \frac{x + 1}{f(1) - 1}.\]If $f(1) = -1,$ then
\[f(x) = \frac{x + 1}{-2},\]and we can check that this function does not work.
If $f(1) = 2,$ then
\[f(x) = x + 1\]and we can check that this function works.
Therefore, $n = 1$ and $s = 3,$ so $n \times s = \boxed{3}.$
|
human
|
Let $P(x) = 0$ be the polynomial equation of least possible degree, with rational coefficients, having $\sqrt[3]{7} + \sqrt[3]{49}$ as a root. Compute the product of all of the roots of $P(x) = 0.$
|
gpt
|
We recognize the number $\sqrt[3]{7} + \sqrt[3]{49}$ from the difference-of-cubes factorization \[7 - 1 = \left(\sqrt[3]{7} - 1\right)\left(1 + \sqrt[3]{7} + \sqrt[3]{49}\right).\]Solving for $\sqrt[3]{7} + \sqrt[3]{49},$ we get \[\sqrt[3]{7} + \sqrt[3]{49} = \frac{7-1}{\sqrt[3]{7}-1} - 1 = \frac{6}{\sqrt[3]{7}-1} - 1.\]We can use this expression to build a polynomial which has $\sqrt[3]{7} + \sqrt[3]{49}$ as a root. First, note that $\sqrt[3]{7}$ is a root of $x^3 - 7 = 0.$ Then, $\sqrt[3]{7}-1$ is a root of $(x+1)^3 - 7 = 0,$ because $(\sqrt[3]{7}-1+1)^3 - 7 = (\sqrt[3]{7})^3 - 7 = 0.$ (You could also note that the graph of $y=(x+1)^3-7$ is a one-unit leftward shift of the graph of $y=x^3-7,$ so the roots of $(x+1)^3-7=0$ are one less than the roots of $x^3-7=0.$)
It follows that $\frac{6}{\sqrt[3]{7}-1}$ is a root of the equation \[\left(\frac{6}{x}+1\right)^3 - 7= 0,\]because when $x = \frac{6}{\sqrt[3]{7}-1},$ we have $\frac{6}{x} = \sqrt[3]{7}-1.$ We multiply both sides by $x^3$ to create the polynomial equation \[(6+x)^3 - 7x^3 = 0.\]Finally, replacing $x$ with $x+1$ like before, we see that $\frac{6}{\sqrt[3]{7}-1} - 1$ is a root of the equation \[(7+x)^3 - 7(x+1)^3 = 0.\]This equation is equivalent to \[x^3 - 21x - 56 = 0,\]so by Vieta's formulas, the product of the roots is $\boxed{56}.$
|
human
|
In a certain sequence the first term is $a_1=2007$ and the second term is $a_2=2008$. Furthermore, the values of the remaining terms are chosen so that $a_n+a_{n+1}+a_{n+2}=n$ for all $n\ge 1$. Determine $a_{1000}$.
|
gpt
|
To get started, we compute the first ten terms as: \[ 2007, 2008, -4014, 2008, 2009, -4013, 2009, 2010, -4012, 2010, \ldots \]It appears that each term is 1 greater than the number three terms previous. We can demonstrate that this will always occur using the given recurrence relation. We know that $a_n+a_{n+1}+a_{n+2}=n$ and that $a_{n+1}+a_{n+2}+a_{n+3}=n+1$. Subtracting the former from the latter yields $a_{n+3}-a_n=1$, which is the pattern that we observed. Therefore we find that \[ a_1 = 2007, \ a_4=2008, \ a_7=2009, \ldots, a_{1000}=2007+333=\boxed{\mathbf{2340}}. \]
|
human
|
Let $P(x) = (x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P\left(Q(x)\right) = P(x)\cdot R(x)$?
|
gpt
|
The polynomial $P(x)\cdot R(x)$ has degree 6, so $Q(x)$ must have degree 2. Therefore $Q$ is uniquely determined by the ordered triple $(Q(1), Q(2),Q(3))$. When $x = 1$, 2, or 3, we have
\[0 = P(x)\cdot R(x) = P\left(Q(x)\right).\]It follows that $(Q(1), Q(2), Q(3))$ is one of the 27 ordered triples $(i, j, k)$, where $i$, $j$, and $k$ can be chosen from the set $\{1, 2, 3\}$.
However, the choices $(1, 1, 1)$, $(2, 2, 2)$, $(3, 3, 3)$, $(1, 2, 3)$, and $(3, 2, 1)$ lead to polynomials $Q(x)$ defined by $Q(x) = 1$, $2,$ $3,$ $x,$ and $4-x$, respectively, all of which have degree less than 2. The other $\boxed{22}$ choices for $(Q(1),Q(2),Q(3))$ yield non-collinear points, so in each case $Q(x)$ is a quadratic polynomial.
|
human
|
What is the value of $\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)$? Express your answer as a common fraction.
|
gpt
|
We have that
\[\left(\frac{2}{\cancel{3}}\right)\left(\frac{\cancel{3}}{\cancel{4}}\right)\left(\frac{\cancel{4}}{\cancel{5}}\right)\left(\frac{\cancel{5}}{6}\right)=\frac{2}{6}=\boxed{\frac{1}{3}}. \]
|
human
|
A hyperbola is centered at the origin and opens either horizontally or vertically. It passes through the points $(-3, 4),$ $(-2, 0),$ and $(t, 2).$ Find $t^2.$
|
gpt
|
Because the hyperbola has center at $(0,0)$ and intersects the $x-$axis at $(-2,0)$, it must open horizontally, and $(-2,0)$ must be one of its vertices. Therefore, it has an equation of the form \[\frac{x^2}{2^2} - \frac{y^2}{b^2} = 1\]for some $b>0.$ Setting $x=-3$ and $y=4,$ we get the equation \[\frac{9}{4} - \frac{16}{b^2} = 1,\]which gives $b^2 = \frac{64}{5}.$ Therefore, the equation of the hyperbola is \[\frac{x^2}{4} - \frac{5y^2}{64} = 1.\]Setting $x=t$ and $y=2,$ we get \[\frac{t^2}{4} - \frac{5}{16} = 1,\]which gives $t^2= \boxed{\frac{21}{4}}.$[asy]
void axes(real x0, real x1, real y0, real y1)
{
draw((x0,0)--(x1,0),EndArrow);
draw((0,y0)--(0,y1),EndArrow);
label("$x$",(x1,0),E);
label("$y$",(0,y1),N);
for (int i=floor(x0)+1; i<x1; ++i)
draw((i,.1)--(i,-.1));
for (int i=floor(y0)+1; i<y1; ++i)
draw((.1,i)--(-.1,i));
}
path[] yh(real a, real b, real h, real k, real x0, real x1, bool upper=true, bool lower=true, pen color=black)
{
real f(real x) { return k + a / b * sqrt(b^2 + (x-h)^2); }
real g(real x) { return k - a / b * sqrt(b^2 + (x-h)^2); }
if (upper) { draw(graph(f, x0, x1),color, Arrows); }
if (lower) { draw(graph(g, x0, x1),color, Arrows); }
path [] arr = {graph(f, x0, x1), graph(g, x0, x1)};
return arr;
}
void xh(real a, real b, real h, real k, real y0, real y1, bool right=true, bool left=true, pen color=black)
{
path [] arr = yh(a, b, k, h, y0, y1, false, false);
if (right) draw(reflect((0,0),(1,1))*arr[0],color, Arrows);
if (left) draw(reflect((0,0),(1,1))*arr[1],color, Arrows);
}
void e(real a, real b, real h, real k)
{
draw(shift((h,k))*scale(a,b)*unitcircle);
}
size(7cm);
axes(-4, 4, -5, 5);
xh(2, 8/sqrt(5), 0, 0, -5, 5);
dot((-3,4)^^(-2,0)^^(sqrt(21/4),2));
label("$(-3,4)$",(-3,4),ENE);
label("$(-2,0)$",(-2,0),NW);
label("$(t,2)$",(sqrt(21/4),2),NW);
[/asy]
|
human
|
Let $r$, $s$, and $t$ be the three roots of the equation $$
8x^3 + 1001x + 2008 = 0.
$$Find $(r + s)^3 + (s + t)^3 + (t + r)^3.$
|
gpt
|
By Vieta's formulas, the sum of the three roots is $r+s+t=0$. Thus, we can write \[(r+s)^3 + (s+t)^3 + (t+r)^3 = (-t)^3 + (-r)^3 + (-s)^3 = -(r^3+s^3+t^3).\]Since each root satisfies the given equation, we have \[8r^3 + 1001r + 2008 = 0,\]so $r^3 = -\frac{1001}{8}r - 251$. Similar equations hold for $s$ and $t$. Thus, \[-(r^3+s^3+t^3) = \frac{1001}{8}(r+s+t) + 3 \cdot 251.\]Since $r+s+t=0,$ the answer is $3 \cdot 251 = \boxed{753}$.
|
human
|
The Fibonacci sequence is defined $F_1 = F_2 = 1$ and $F_n = F_{n - 1} + F_{n - 2}$ for all $n \ge 3.$
The Fibonacci numbers $F_a,$ $F_b,$ $F_c$ form an increasing arithmetic sequence. If $a + b + c = 2000,$ compute $a.$
|
gpt
|
We claim that if $F_a,$ $F_b,$ $F_c$ form an increasing arithmetic sequence, then $(a,b,c)$ must be of the form $(n,n + 2,n + 3)$ for some positive integer $n.$ (The only exception is $(2,3,4).$)
From $F_c - F_b = F_b - F_a,$ we get
\[F_c = F_b + (F_b - F_a) < F_b + F_{b + 1} = F_{b + 2}.\]Also, $F_c > F_b.$ Therefore, $F_c = F_{b + 1}.$
Then
\begin{align*}
F_a &= 2F_b - F_c \\
&= 2F_b - F_{b + 1} \\
&= F_b - (F_{b + 1} - F_b) \\
&= F_b - F_{b - 1} \\
&= F_{b - 2}.
\end{align*}Then $a$ must be equal to $b - 2$ (unless $b = 3,$ which leads to the exceptional case of $(2,3,4)$). Taking $n = b - 2,$ we get $(a,b,c) = (n,n + 2,n + 3).$
Then $a + (a + 2) + (a + 3) = 2000,$ so $a = \boxed{665}.$
|
human
|
Let $x$ and $y$ be real numbers such that
\[xy - \frac{x}{y^2} - \frac{y}{x^2} = 3.\]Find the sum of all possible values of $(x - 1)(y - 1).$
|
gpt
|
From the given equation, $x^3 y^3 - x^3 - y^3 = 3x^2 y^2,$ or
\[x^3 y^3 - x^3 - y^3 - 3x^2 y^2 = 0.\]We have the factorization
\[a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).\]Taking $a = xy,$ $b = -x,$ and $c = -y,$ we get
\[x^3 y^3 - x^3 - y^3 - 3x^2 y^2 = (xy - x - y)(a^2 + b^2 + c^2 - ab - ac - bc) = 0.\]If $xy - x - y = 0,$ then
\[(x - 1)(y - 1) = xy - x - y + 1 = 1.\]If $a^2 + b^2 + c^2 - ab - ac - bc = 0,$ then $2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 0,$ which we can write as
\[(a - b)^2 + (a - c)^2 + (b - c)^2 = 0.\]This forces $a = b = c,$ so $xy = -x = -y.$ We get that $x = y,$ so $x^2 + x = x(x + 1) = 0.$ Hence, $x = 0$ or $x = -1.$ From the given condition, we cannot have $x = 0,$ so $x = -1,$ and $y = -1,$ so $(x - 1)(y - 1) = 4.$
Thus, the possible values of $(x - 1)(y - 1)$ are 1 and 4, and their sum is $\boxed{5}.$
|
human
|
Let $r$ be the positive real solution to $x^3 + \frac{2}{5} x - 1 = 0.$ Find the exact numerical value of
\[r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb.\]
|
gpt
|
Let $S = r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb.$ Then
\[r^3 S = r^5 + 2r^8 + 3r^{11} + 4r^{13} + \dotsb.\]Subtracting this equation from $S = r^2 + 2r^5 + 3r^8 + 4r^{11} + \dotsb,$ we get
\[S (1 - r^3) = r^2 + r^5 + r^8 + r^{11} + \dotsb = \frac{r^2}{1 - r^3}.\]Hence,
\[S = \frac{r^2}{(1 - r^3)^2}.\]Since $r^3 + \frac{2}{5} r - 1 = 0,$ $1 - r^3 = \frac{2}{5} r.$ Therefore,
\[S = \frac{r^2}{\frac{4}{25} r^2} = \boxed{\frac{25}{4}}.\]
|
human
|
Find the real root of the equation \[\sqrt{x} + \sqrt{x+2} = 10.\]
|
gpt
|
Subtracting $\sqrt{x}$ from both sides and then squaring, we get \[x+2 = (10-\sqrt x)^2 = x - 20\sqrt x + 100.\]Therefore, $20\sqrt x = 98,$ so $\sqrt x = \frac{98}{20} = \frac{49}{10}.$ Therefore, $x = \left(\frac{49}{10}\right)^2 = \boxed{\frac{2401}{100}},$ or $x = 24.01.$
|
human
|
What is the remainder when $x^2+7x-5$ divides $2x^4+11x^3-42x^2-60x+47$?
|
gpt
|
\[
\begin{array}{c|cc ccc}
\multicolumn{2}{r}{2x^2} & -3x & -11 \\
\cline{2-6}
x^2+7x-5 & 2x^4 & +11x^3 & -42x^2 & -60x & +47 \\
\multicolumn{2}{r}{-2x^4} & -14x^3 & +10x^2 \\
\cline{2-4}
\multicolumn{2}{r}{0} & -3x^3 & -32x^2 & -60x \\
\multicolumn{2}{r}{} & +3x^3 & +21x^2 & -15x \\
\cline{3-5}
\multicolumn{2}{r}{} & 0 & -11x^2 & -75x & +47 \\
\multicolumn{2}{r}{} & & +11x^2 & +77x & -55 \\
\cline{4-6}
\multicolumn{2}{r}{} & & 0 & 2x & -8 \\
\end{array}
\]Since the degree of $2x-8$ is lower than that of $x^2+7x-5$, we cannot divide any further. So our remainder is $\boxed{2x-8}$.
|
human
|
Find the distance between the vertices of the hyperbola
\[\frac{x^2}{99} - \frac{y^2}{36} = 1.\]
|
gpt
|
We read that $a^2 = 99,$ so $a = \sqrt{99} = 3 \sqrt{11}.$ Therefore, the distance between the vertices is $2a = \boxed{6 \sqrt{11}}.$
|
human
|
If $x = 101$ and $x^3y - 2x^2y + xy = 101000$, what is the value of $y$?
|
gpt
|
Note that the left hand side of the second equation can be factored: $(x^2 - 2x + 1)xy = (x - 1)^2xy = 101000$. We are given that $x = 101$, so we have $(101- 1)^2(101)y = 1010000y = 101000$. It follows that $y = \boxed{\frac{1}{10}}$.
|
human
|
The hyperbola given by the equation \[\frac{x^2}{100} - \frac{y^2}{64} = 1\]has asymptotes $y = \pm mx,$ where $m$ is positive. Find $m.$
|
gpt
|
To get the equation of the asymptotes, we replace the $1$ on the right-hand side with $0,$ giving the equation\[\frac{x^2}{100} - \frac{y^2}{64} = 0.\](Notice that there are no points $(x, y)$ which satisfy both this equation and the given equation, so as expected, the hyperbola never intersects its asymptotes.) This is equivalent to $\frac{x^2}{100} = \frac{y^2}{64},$ or $\frac{y}{8} = \pm \frac{x}{10}.$ Thus, $y = \pm \frac{4}{5} x,$ so $m = \boxed{\frac45}.$[asy]
void axes(real x0, real x1, real y0, real y1)
{
draw((x0,0)--(x1,0),EndArrow);
draw((0,y0)--(0,y1),EndArrow);
label("$x$",(x1,0),E);
label("$y$",(0,y1),N);
for (int i=floor(x0)+1; i<x1; ++i)
draw((i,.1)--(i,-.1));
for (int i=floor(y0)+1; i<y1; ++i)
draw((.1,i)--(-.1,i));
}
path[] yh(real a, real b, real h, real k, real x0, real x1, bool upper=true, bool lower=true, pen color=black)
{
real f(real x) { return k + a / b * sqrt(b^2 + (x-h)^2); }
real g(real x) { return k - a / b * sqrt(b^2 + (x-h)^2); }
if (upper) { draw(graph(f, x0, x1),color, Arrows); }
if (lower) { draw(graph(g, x0, x1),color, Arrows); }
path [] arr = {graph(f, x0, x1), graph(g, x0, x1)};
return arr;
}
void xh(real a, real b, real h, real k, real y0, real y1, bool right=true, bool left=true, pen color=black)
{
path [] arr = yh(a, b, k, h, y0, y1, false, false);
if (right) draw(reflect((0,0),(1,1))*arr[0],color, Arrows);
if (left) draw(reflect((0,0),(1,1))*arr[1],color, Arrows);
}
void e(real a, real b, real h, real k)
{
draw(shift((h,k))*scale(a,b)*unitcircle);
}
size(10cm);
axes(-15,15,-10,10);
xh(10,8,0,0,-8,8);
draw((-12,-48/5)--(12,48/5),dotted);
draw((12,-48/5)--(-12,48/5),dotted);
[/asy]
|
human
|
Find $s$ given that \[3 \log_2 s = \log_2 (3s).\]
|
gpt
|
By logarithm identities, we have \[\log_2 (s^3) = \log_2 (3s).\]Thus, $s^3 = 3s$, and since $s$ must be positive, we can divide by $s$ to get $s^2 = 3.$ Thus, $s = \boxed{\sqrt{3}}.$
|
human
|
If $z^2 - 8z + 37 = 0$, how many possible values are there for $|z|$?
|
gpt
|
We could use the quadratic formula, but there is a shortcut: note that if the quadratic is not a perfect square, the solutions will be of the form $p \pm \sqrt{q}$ or $p \pm i \sqrt{q}$. In the first case, if both solutions are real, there are 2 different values of $|z|$, whereas in the second case, there is only one value, since $|p + i\sqrt{q}| = |p - i\sqrt{q}| = \sqrt{p^2 + q}$. So all we have to do is check the sign of the discriminant: $b^2 - 4ac = 64 - 4(37) < 0$. Since the discriminant is negative, there are two nonreal solutions, and thus only $\boxed{1}$ possible value for the magnitude.
|
human
|
Let $x$ and $y$ be positive real numbers. Find the minimum value of
\[\frac{\sqrt{(x^2 + y^2)(3x^2 + y^2)}}{xy}.\]
|
gpt
|
By Cauchy-Schwarz,
\[(y^2 + x^2)(3x^2 + y^2) \ge (xy \sqrt{3} + xy)^2,\]so
\[\frac{\sqrt{(x^2 + y^2)(3x^2 + y^2)}}{xy} \ge 1 + \sqrt{3}.\]Equality occurs when $\frac{y^2}{3x^2} = \frac{x^2}{y^2},$ or $y = x \sqrt[4]{3},$ so the minimum value is $\boxed{1 + \sqrt{3}}.$
|
human
|
If $a,$ $b,$ $c,$ $x,$ $y,$ and $z$ are positive and $a^2 + b^2 + c^2 = 25,$ $x^2 + y^2 + z^2 = 36,$ and $ax + by + cz = 30,$ compute
\[\frac{a + b + c}{x + y + z}.\]
|
gpt
|
By the Cauchy-Schwarz inequality,
\[(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) \ge (ax + by + cz)^2.\]This becomes $25 \cdot 36 \ge 30^2.$ Since $25 \cdot 36 = 900 = 30^2,$ we get equality.
From the equality condition in Cauchy-Schwarz,
\[\frac{a}{x} = \frac{b}{y} = \frac{c}{z}.\]Let
\[k = \frac{a}{x} = \frac{b}{y} = \frac{c}{z}.\]Then $a = kx,$ $b = ky$ and $c = kz,$ so
\[a^2 + b^2 + c^2 = k^2 x^2 + k^2 y^2 + k^2 z^2 = k^2 (x^2 + y^2 + z^2).\]Then $36k^2 = 25,$ so $k^2 = \frac{25}{36},$ which means $k = \frac{5}{6}.$ Therefore,
\[\frac{a + b + c}{x + y + z} = \boxed{\frac{5}{6}}.\]
|
human
|
If $x^2 + 2x + 5$ is a factor of $x^4 + Px^2 + Q,$ find $P + Q.$
|
gpt
|
We have that
\[(x^2 + 2x + 5)(x^2 + bx + c) = x^4 + Px^2 + Q.\]for some coefficients $b$ and $c.$ Expanding, we get
\[x^4 + (b + 2) x^3 + (2b + c + 5) x^2 + (5b + 2c) x + 5c = x^4 + Px^2 + Q.\]Matching coefficients, we get
\begin{align*}
b + 2 &= 0, \\
2b + c + 5 &= P, \\
5b + 2c &= 0, \\
5c &= Q.
\end{align*}Solving $b + 2 = 0$ and $5b + 2c = 0,$ we get $b = -2$ and $c = 5.$ Then $P = 2b + c + 5 = 6$ and $Q = 5c = 25,$ so $P + Q = \boxed{31}.$
|
human
|
Compute the product of the roots of the equation \[x^3 - 12x^2 + 48x + 28 = 0.\]
|
gpt
|
By Vieta's formulas, the product of the roots is the negation of the constant term divided by the leading ($x^3$) coefficient. Therefore, the answer is \[\frac{-28}{1} = \boxed{-28}.\]
|
human
|
Which of the following quantities is the largest? (Write $A$, $B$, or $C$.) \[ A.\ \ \frac{2006}{2005}+\frac{2006}{2007} \qquad B.\ \ \frac{2006}{2007}+\frac{2008}{2007} \qquad C.\ \ \frac{2007}{2006}+\frac{2007}{2008} \]
|
gpt
|
This problem illustrates how algebra can clarify arithmetic. Comparing these quantities directly is a chore. Instead, we note that the first and third choices are both of the form $\frac{n}{n-1}+\frac{n}{n+1}$ for $n=2006$ and $n=2007$. Rewriting this expression algebraically leads to \[ \frac{n(n+1)}{n^2-1}+\frac{n(n-1)}{n^2-1} = \frac{2n^2}{n^2-1} = 2 + \frac{2}{n^2-1}. \]In particular both $A$ and $C$ are larger than 2. It is easy to verify that choice $B$ is equal to 2, so it cannot be the answer. Finally, note that larger values of $n$ produce smaller results, implying that $\boxed{\text{A}}$ is the largest. (It is also possible to guess the answer by trying examples involving much smaller numbers.)
|
human
|
Find the remainder when $x^{44} + x^{33} + x^{22} + x^{11} + 1$ is divided by $x^4 + x^3 + x^2 + x + 1.$
|
gpt
|
Let $\alpha$ be a root of $x^4 + x^3 + x^2 + x + 1 = 0,$ so
\[\alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1 = 0.\]Then $(\alpha - 1)(\alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1) = 0,$ which simplifies to $\alpha^5 = 1.$ Then
\begin{align*}
\alpha^{44} + \alpha^{33} + \alpha^{22} + \alpha^{11} + 1 &= (\alpha^5)^8 \cdot \alpha^4 + (\alpha^5)^6 \cdot \alpha^3 + (\alpha^5)^4 \cdot \alpha^2 + (\alpha^5)^2 \cdot \alpha + 1 \\
&= \alpha^4 + \alpha^3 + \alpha^2 + \alpha + 1 \\
&= 0.
\end{align*}Since the first polynomial is 0 for every root $\alpha$ of the second polynomial, the first polynomial is divisible by the second polynomial, which means the remainder is $\boxed{0}.$
|
human
|
A circle rests in the interior of the parabola with equation $y = x^2,$ so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency?
|
gpt
|
Let one of the points of tangency be $(a,a^2).$ By symmetry, other point of tangency is $(-a,a^2).$ Also by symmetry, the center of the circle lies on the $y$-axis. Let the center be $(0,b),$ and let the radius be $r.$
[asy]
unitsize(1.5 cm);
real func (real x) {
return(x^2);
}
pair A = (1,1), O = (0,3/2);
draw(Circle(O,sqrt(5)/2));
draw(graph(func,-1.5,1.5));
draw((-1.5,0)--(1.5,0));
draw((0,-0.5)--(0,3));
dot("$(a,a^2)$", A, SE);
dot("$(-a,a^2)$", (-1,1), SW);
dot("$(0,b)$", O, E);
[/asy]
The equation of the parabola is $y = x^2.$ The equation of the circle is $x^2 + (y - b)^2 = r^2.$ Substituting $y = x^2,$ we get
\[x^2 + (x^2 - b)^2 = r^2.\]This expands as
\[x^4 + (1 - 2b)x^2 + b^2 - r^2 = 0.\]Since $(a,a^2)$ and $(-a,a^2)$ are points of tangency, $x = a$ and $x = -a$ are double roots of this quartic. In other words, it is the same as
\[(x - a)^2 (x + a)^2 = (x^2 - a^2)^2 = x^4 - 2a^2 x^2 + a^4 = 0.\]Equating the coefficients, we get
\begin{align*}
1 - 2b &= -2a^2, \\
b^2 - r^2 &= a^4.
\end{align*}Then $2b - 2a^2 = 1.$ Therefore, the difference between the $y$-coordinates of the center of the circle $(0,b)$ and the point of tangency $(a,a^2)$ is
\[b - a^2 = \boxed{\frac{1}{2}}.\]
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human
|
Is $f(x) = \frac{3}{2x^{6}-5}$ an even function, odd function, or neither?
Enter "odd", "even", or "neither".
|
gpt
|
$$f(-x) = \frac{3}{2(-x)^{6}-5} = \frac{3}{2x^{6}-5} = f(x)$$Hence $f$ is $\boxed{\text{even}}.$
|
human
|
What is $(-1)^1+(-1)^2+\cdots+(-1)^{2006}$ ?
|
gpt
|
Because $(-1)^k$ equals 1 if $k$ is even and $-1$ if $k$ is odd, the sum can be written as \[
(-1+1)+(-1+1)+\cdots+(-1+1) =0+0+\cdots+0=\boxed{0}.
\]
|
human
|
Find the minimum value of
\[\frac{x^2}{x - 8}\]for $x > 8.$
|
gpt
|
We can write
\[\frac{x^2}{x - 8} = \frac{x^2 - 64 + 64}{x - 8} = \frac{(x - 8)(x + 8) + 64}{x - 8} = x + 8 + \frac{64}{x - 8} = x - 8 + \frac{64}{x - 8} + 16.\]By AM-GM,
\[x - 8 + \frac{64}{x - 8} \ge 2 \sqrt{(x - 8) \cdot \frac{64}{x - 8}} = 16,\]so
\[\frac{x^2}{x - 8} \ge 32.\]Equality occurs when $x = 16,$ so the minimum value is $\boxed{32}.$
|
human
|
Factor completely over the set of polynomials with integer coefficients:
\[4(x + 5)(x + 6)(x + 10)(x + 12) - 3x^2.\]
|
gpt
|
First, we can multiply the factors $x + 5$ and $x + 12$ to get
\[(x + 5)(x + 12) = x^2 + 17x + 60.\]We can then multiply the factors $x + 6$ and $x + 10$ to get
\[(x + 6)(x + 10) = x^2 + 16x + 60.\]So, let $u = x^2 + 16x + 60.$ Then
\begin{align*}
4(x + 5)(x + 6)(x + 10)(x + 12) - 3x^2 &= 4(u + x)(u) - 3x^2 \\
&= 4u^2 + 4ux - 3x^2 \\
&= (2u + 3x)(2u - x) \\
&= (2(x^2 + 16x + 60) + 3x)(2(x^2 + 16x + 60) - x) \\
&= (2x^2 + 35x + 120)(2x^2 + 31x + 120) \\
&= \boxed{(2x^2 + 35x + 120)(x + 8)(2x + 15)}.
\end{align*}
|
human
|
Compute the smallest positive integer $n$ such that $n + i,$ $(n + i)^2,$ and $(n + i)^3$ are the vertices of a triangle in the complex plane whose area is greater than 2015.
|
gpt
|
We have that
\[(n + i)^2 = n^2 + 2ni + i^2 = (n^2 - 1) + (2n)i,\]and
\[(n + i)^3 = n^3 + 3n^2 i + 3ni^2 + i^3 = (n^3 - 3n) + (3n^2 - 1)i.\]By the Shoelace Theorem, area of the triangle with vertices $(n,1),$ $(n^2 - 1,2n),$ and $(n^3 - 3n,3n^2 - 1)$ is
\begin{align*}
&\frac{1}{2} \left|(n)(2n) + (n^2 - 1)(3n^2 - 1) + (n^3 - 3n)(1) - (1)(n^2 - 1) - (2n)(n^3 - 3n) - (3n^2 - 1)(n)\right| \\
&= \frac{1}{2} (n^4 - 2n^3 + 3n^2 - 2n + 2) = \frac{1}{2} [(n^2 - n + 1)^2 + 1].
\end{align*}Thus, we want $n$ to satisfy
\[\frac{1}{2} [(n^2 - n + 1)^2 + 1] > 2015,\]or $(n^2 - n + 1)^2 > 4029.$ Checking small values, we find the smallest positive integer $n$ that works is $\boxed{9}.$
|
human
|
Evaluate $\left|{-4+\frac{7}{6}i}\right|$.
|
gpt
|
We have $\left|{-4+\frac{7}{6}i}\right|=\frac{1}{6}|{-24+7i}|=\frac{1}{6}\sqrt{(-24)^2+7^2}=\boxed{\frac{25}{6}}$
|
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