utils_math.py

def get_examples(mode) -> str:
        """Get examples based on the selected mode."""
        if mode == "simple":
            return """Problem: There are 15 trees in the grove. Grove workers will plant trees in the grove today. After they are done, there will be 21 trees. How many trees did the grove workers plant today?
Answer: There are 15 trees originally. Then there were 21 trees after the Grove workers planted some more. So there must have been 21 - 15 = 6 trees that were planted. The answer is 6.
###
Problem: If there are 3 cars in the parking lot and 2 more cars arrive, how many cars are in the parking lot?
Answer: There are originally 3 cars. Then 2 more cars arrive. Now 3 + 2 = 5 cars are in the parking lot. The answer is 5.
###
Problem: Leah had 32 chocolates and her sister had 42. If they ate 35, how many pieces do they have left in total?
Answer: Originally, Leah had 32 chocolates and her sister had 42. So in total they had 32 + 42 = 74. After eating 35, they had 74 - 35 = 39 pieces left in total. The answer is 39.
###
Problem: Jason had 20 lollipops. He gave Denny some lollipops. Now Jason has 12 lollipops. How many lollipops did Jason give to Denny?
Answer: Jason had 20 lollipops originally. Then he had 12 after giving some to Denny. So he gave Denny 20 - 12 = 8 lollipops. The answer is 8.
###
Problem: Shawn has five toys. For Christmas, he got two toys each from his mom and dad. How many toys does he have now?
Answer: Shawn started with 5 toys. He then got 2 toys each from his mom and dad. So he got 2 * 2 = 4 more toys. Now he has 5 + 4 = 9 toys. The answer is 9.
###
Problem: There were nine computers in the server room. Five more computers were installed each day, from monday to thursday. How many computers are now in the server room?
Answer: There were originally 9 computers. For each day from monday to thursday, 5 more computers were installed. So 4 * 5 = 20 computers were added. Now 9 + 20 = 29 computers are now in the server room. The answer is 29.
###
Problem: Michael had 58 golf balls. On tuesday, he lost 23 golf balls. On wednesday, he lost 2 more. How many golf balls did he have at the end of wednesday?
Answer: Michael started with 58 golf balls. He lost 23 on Tuesday, and lost 2 more on wednesday. So he had 58 - 23 = 35 at the end of Tuesday, and 35 - 2 = 33 at the end of wednesday. The answer is 33.
###
Problem: Olivia has $23. She bought five bagels for $3 each. How much money does she have left? 
Answer: Olivia had 23 dollars. She bought 5 bagels for 3 dollars each. So she spent 5 * 3 = 15 dollars. Now she has 23 - 15 = 8 dollars left. The answer is 8."""
        else:  # complex
            return r"""Problem: Kevin Kangaroo begins hopping on a number line at 0. He wants to get to 1, but he can hop only $\frac{1}{3}$ of the distance. Each hop tires him out so that he continues to hop $\frac{1}{3}$ of the remaining distance. How far has he hopped after five hops? Express your answer as a common fraction.
Answer: Let's think step by step
Kevin hops $1/3$ of the remaining distance with every hop.
His first hop takes $1/3$ closer.
For his second hop, he has $2/3$ left to travel, so he hops forward $(2/3)(1/3)$.
For his third hop, he has $(2/3)^2$ left to travel, so he hops forward $(2/3)^2(1/3)$.
In general, Kevin hops forward $(2/3)^{k-1}(1/3)$ on his $k$th hop.
We want to find how far he has hopped after five hops.
This is a finite geometric series with first term $1/3$, common ratio $2/3$, and five terms.
Thus, Kevin has hopped $\frac{\frac{1}{3}\left(1-\left(\frac{2}{3}\right)^5\right)}{1-\frac{2}{3}} = \boxed{\frac{211}{243}}$.
The answer is \frac{211}{243}}

Problem: What is the area of the region defined by the equation $x^2+y^2 - 7 = 4y-14x+3$?
Answer: Let's think step by step
We rewrite the equation as $x^2 + 14x + y^2 - 4y = 10$ and then complete the square,
resulting in  $(x+7)^2-49 + (y-2)^2-4=10$,
or $(x+7)^2+(y-2)^2=63$.
This is the equation of a circle with center $(-7, 2)$ and radius $\sqrt{63},$
so the area of this region is $\pi r^2 = \boxed{63\pi}$.
The answer is 63\pi

Problem: If $x^2+y^2=1$, what is the largest possible value of $|x|+|y|$?
Answer: Let's think step by step
If $(x,y)$ lies on the circle,
so does $(x,-y),$ $(-x,-y),$ and $(-x,-y),$ (which all give the same value of $|x| + |y|$),
so we can assume that $x \ge 0$ and $y \ge 0.$
Then $|x| + |y| = x + y.$  Squaring, we get
\[(x + y)^2 = x^2 + 2xy + y^2 = 1 + 2xy.\]
Note that $(x - y)^2 \ge 0.$
Expanding, we get $x^2 - 2xy + y^2 \ge 0,$ so $2xy \le x^2 + y^2 = 1.$
Hence,\[1 + 2xy \le 2,\]which means $x + y \le \sqrt{2}.$
Equality occurs when $x = y = \frac{1}{\sqrt{2}},$
so the maximum value of $|x| + |y|$ is $\boxed{\sqrt{2}}.$
The answer is \sqrt{2}

Problem: If $f(x)=\frac{ax+b}{cx+d}, abcd\not=0$ and $f(f(x))=x$ for all $x$ in the domain of $f$, what is the value of $a+d$?
Answer: Let's think step by step
The condition $f(f(x))$ means that $f$ is the inverse of itself,
so its graph is symmetrical about the line $y = x$.
With a rational function of this form, we will have two asymptotes:
a vertical one at $x=-d/c$ if $cx+d$ does not divide $ax+b$,
and a horizontal one at $y=a/c$,
if we take the limit of $f(x)$ as $x$ goes to $\pm\infty$.
In order for $f$ to be its own inverse, the intersection of the asymptotes must lie on the line $y=x$
so that it and its asymptotes reflect onto themselves.
This means that $-d/c=a/c$,
and therefore $-d=a$ and $a+d=\boxed{0}$.
The answer is 0

Problem: A math teacher requires Noelle to do one homework assignment for each of the first five homework points she wants to earn; for each of the next five homework points, she needs to do two homework assignments; and so on, so that to earn the $n^{\text{th}}$ homework point, she has to do $n\div5$ (rounded up) homework assignments. For example, when she has 11 points, it will take $12\div5=2.4\rightarrow3$ homework assignments to earn her $12^{\text{th}}$ point. What is the smallest number of homework assignments necessary to earn a total of 25 homework points?
Answer: Let's think step by step
Noelle only has to do 1 homework assignment to earn her first point,
and the same is true for each of her first five points.
She must then do 2 homework assignments to earn her sixth point, seventh point, and so on, up to her tenth point.
Continuing, we see that Noelle must do a total of \[1+1+1+1+1+2+2+2+2+2+\dots+5+5+5+5+5\] homework assignments to earn 25 points.
This sum may be rewritten as $5(1+2+3+4+5)=5(15)=\boxed{75}$.
The answer is 75

Problem: The quadratic equation $x^2+mx+n=0$ has roots that are twice those of $x^2+px+m=0,$ and none of $m,$ $n,$ and $p$ is zero. What is the value of $n/p?$
Answer: Let's think step by step
Let $r_1$ and $r_2$ be the roots of $x^2+px+m=0.$
Since the roots of $x^2+mx+n=0$ are $2r_1$ and $2r_2,$ we have the following relationships: \[
m=r_1 r_2,\quad n=4r_1 r_2,\quad p=-(r_1+r_2), \quad\text{and}\quad
m=-2(r_1+r_2).
\] So \[
n = 4m, \quad p = \frac{1}{2}m,
\quad\text{and}\quad
\frac{n}{p}=\frac{4m}{\frac{1}{2}m}=\boxed{8}.
\]
Alternatively, the roots of \[
\left(\frac{x}{2}\right)^2 + p\left(\frac{x}{2}\right) + m = 0
\] are twice those of $x^2 + px + m = 0.$
Since the first equation is equivalent to $x^2 + 2px + 4m = 0,$
we have \[m = 2p \quad\text{and}\quad n = 4m, \quad\text{so}\quad \frac{n}{p} = \boxed{8}.\]
The answer is 8

Problem: Expand $(2z^2 + 5z - 6)(3z^3 - 2z + 1)$.
Answer: Let's think step by step
$$\begin{array}{crrrrrrr}
& & & 3z^3 & & -2z & + 1 & \\
\times & & & & 2z^2 & +5z & -6 \\
\cline{1-7}\rule{0pt}{0.17in}
& & & -18z^3 & & +12z & -6 & \\
& & +15z^4 & & -10z^2 & +5z & & \\
+ & 6z^5 & & -4z^3 & +2z^2 & & & \\
\cline{1-7}\rule{0pt}{0.17in}
& 6z^5 & +15z^4 & -22z^3 & - 8z^2 &+17z & -6 &
\end{array}$$
The answer is 6z^5+15z^4-22z^3-8z^2+17z-6}.

Problem: Find the mean of all solutions for $x$ when $x^3 + 3x^2 - 10x = 0$.
Answer: Let's think step by step
First, we factor the equation as $x(x^2 +3x - 10) = 0$.
So, one solution is $x=0$ and the other two solutions are the solutions to $x^2 + 3x-10=0$.
We could either factor the quadratic, or note that the sum of the solutions to this quadratic is $-(3/1)=-3$,
so the mean of the three solutions to the original equation is $-3/3=\boxed{-1}$.
The answer is -1"""