di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Number Theory


Question: Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=151&width=891&top_left_y=453&top_left_x=644)

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
For each positive integer $m \geq 2$, determine a simplified expression for the sum of the terms in the $(m, 3)$-sawtooth sequence.

Solution:
Suppose that $m \geq 2$.

The $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .

Consider one of these teeth whose terms are

$$
2,3,4, \ldots, m-1, m, m-1, m-2, m-3, \ldots, 2,1
$$

When we write the ascending portion directly above the descending portion, we obtain

$$
\begin{aligned}
& 2, \quad 3, \quad 4, \quad \ldots, \quad m-1, \quad m \\
& m-1, \quad m-2, \quad m-3, \quad \ldots, \quad 2, \quad 1
\end{aligned}
$$

From this presentation, we can see $m-1$ pairs of terms, the sum of each of which is $m+1$. $($ Note that $2+(m-1)=3+(m-2)=4+(m-3)=\cdots=(m-1)+2=m+1$ and as we move from left to right, the terms on the top increase by 1 at each step and the terms on the bottom decrease by 1 at each step, so their sum is indeed constant.) Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$. This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\left(m^{2}-1\right)$, which equals $3 m^{2}-2$.
Suppose that $m \geq 2$.

The $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .

Consider one of these teeth whose terms are

$$
2,3,4, \ldots, m-1, m, m-1, m-2, m-3, \ldots, 2,1
$$

This tooth includes one 1 , two $2 \mathrm{~s}$, two $3 \mathrm{~s}$, and so on, until we reach two $(m-1) \mathrm{s}$, and one $m$.

The sum of these numbers is

$$
1(1)+2(2)+2(3)+\cdots+2(m-1)+m
$$

which can be rewritten as

$$
2(1+2+3+\cdots+(m-1)+m)-1-m=2 \cdot \frac{1}{2} m(m+1)-m-1=m^{2}+m-m-1=m^{2}-1
$$

Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$.

This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\left(m^{2}-1\right)$, which equals $3 m^{2}-2$.

Final answer: $3 m^{2}-2$
Subfield: Number Theory


Question: Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

<img_3354>

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
For each positive integer $m \geq 2$, determine a simplified expression for the sum of the terms in the $(m, 3)$-sawtooth sequence.

Solution:
Suppose that $m \geq 2$.

The $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .

Consider one of these teeth whose terms are

$$
2,3,4, \ldots, m-1, m, m-1, m-2, m-3, \ldots, 2,1
$$

When we write the ascending portion directly above the descending portion, we obtain

$$
\begin{aligned}
& 2, \quad 3, \quad 4, \quad \ldots, \quad m-1, \quad m \\
& m-1, \quad m-2, \quad m-3, \quad \ldots, \quad 2, \quad 1
\end{aligned}
$$

From this presentation, we can see $m-1$ pairs of terms, the sum of each of which is $m+1$. $($ Note that $2+(m-1)=3+(m-2)=4+(m-3)=\cdots=(m-1)+2=m+1$ and as we move from left to right, the terms on the top increase by 1 at each step and the terms on the bottom decrease by 1 at each step, so their sum is indeed constant.) Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$. This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\left(m^{2}-1\right)$, which equals $3 m^{2}-2$.
Suppose that $m \geq 2$.

The $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .

Consider one of these teeth whose terms are

$$
2,3,4, \ldots, m-1, m, m-1, m-2, m-3, \ldots, 2,1
$$

This tooth includes one 1 , two $2 \mathrm{~s}$, two $3 \mathrm{~s}$, and so on, until we reach two $(m-1) \mathrm{s}$, and one $m$.

The sum of these numbers is

$$
1(1)+2(2)+2(3)+\cdots+2(m-1)+m
$$

which can be rewritten as

$$
2(1+2+3+\cdots+(m-1)+m)-1-m=2 \cdot \frac{1}{2} m(m+1)-m-1=m^{2}+m-m-1=m^{2}-1
$$

Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$.

This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\left(m^{2}-1\right)$, which equals $3 m^{2}-2$.

Final answer: $3 m^{2}-2$
Subfield: Number Theory


Question: Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

<img_3354>

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer.

Solution:
In an $(m, n)$-sawtooth sequence, the sum of the terms is $n\left(m^{2}-1\right)+1$.

In each tooth, there are $(m-1)+(m-1)=2 m-2$ terms (from 2 to $m$, inclusive, and from $m-1$ to 1 , inclusive).

This means that there are $n(2 m-2)+1$ terms in the sequence.

Thus, the average of the terms in the sequence is $\frac{n\left(m^{2}-1\right)+1}{n(2 m-2)+1}$.

We need to prove that this is not an integer for all pairs of positive integers $(m, n)$ with $m \geq 2$.

Suppose that $\frac{n\left(m^{2}-1\right)+1}{n(2 m-2)+1}=k$ for some integer $k$. We will show, by contradiction, that this is not possible.

Since $\frac{n\left(m^{2}-1\right)+1}{n(2 m-2)+1}=k$, then

$$
\begin{aligned}
\frac{m^{2} n-n+1}{2 m n-2 n+1} & =k \\
m^{2} n-n+1 & =2 m n k-2 n k+k \\
m^{2} n-2 m n k+(2 n k-n-k+1) & =0
\end{aligned}
$$

We treat this as a quadratic equation in $m$.

Since $m$ is an integer, then this equation has integer roots, and so its discriminant must be a perfect square.

The discriminant of this quadratic equation is

$$
\begin{aligned}
\Delta & =(-2 n k)^{2}-4 n(2 n k-n-k+1) \\
& =4 n^{2} k^{2}-8 n^{2} k+4 n^{2}+4 n k-4 n \\
& =4 n^{2}\left(k^{2}-2 k+1\right)+4 n(k-1) \\
& =4 n^{2}(k-1)^{2}+4 n(k-1) \\
& =(2 n(k-1))^{2}+2(2 n(k-1))+1-1 \\
& =(2 n(k-1)+1)^{2}-1
\end{aligned}
$$

We note that $(2 n(k-1)+1)^{2}$ is a perfect square and $\Delta$ is supposed to be a perfect square. But these perfect squares differ by 1 , and the only two perfect squares that differ by 1 are



1 and 0.

(To justify this last fact, we could look at the equation $a^{2}-b^{2}=1$ where $a$ and $b$ are non-negative integers, and factor this to obtain $(a+b)(a-b)=1$ which would give $a+b=a-b=1$ from which we get $a=1$ and $b=0$.)

Since $(2 n(k-1)+1)^{2}=1$ and $2 n(k-1)+1$ is non-negative, then $2 n(k-1)+1=1$ and so $2 n(k-1)=0$.

Since $n$ is positive, then $k-1=0$ or $k=1$.

Therefore, the only possible way in which the average is an integer is if the average is 1.

In this case, we get

$$
\begin{aligned}
m^{2} n-2 m n+(2 n-n-1+1) & =0 \\
m^{2} n-2 m n+n & =0 \\
n\left(m^{2}-2 m+1\right) & =0 \\
n(m-1)^{2} & =0
\end{aligned}
$$

Since $n$ and $m$ are positive integers with $m \geq 2$, then $n(m-1)^{2} \neq 0$, which is a contradiction.

Therefore, the average of the terms in an $(m, n)$-sawtooth sequence cannot be an integer.
Subfield: Number Theory


Question: Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is

![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=151&width=891&top_left_y=453&top_left_x=644)

The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer.

Solution:
In an $(m, n)$-sawtooth sequence, the sum of the terms is $n\left(m^{2}-1\right)+1$.

In each tooth, there are $(m-1)+(m-1)=2 m-2$ terms (from 2 to $m$, inclusive, and from $m-1$ to 1 , inclusive).

This means that there are $n(2 m-2)+1$ terms in the sequence.

Thus, the average of the terms in the sequence is $\frac{n\left(m^{2}-1\right)+1}{n(2 m-2)+1}$.

We need to prove that this is not an integer for all pairs of positive integers $(m, n)$ with $m \geq 2$.

Suppose that $\frac{n\left(m^{2}-1\right)+1}{n(2 m-2)+1}=k$ for some integer $k$. We will show, by contradiction, that this is not possible.

Since $\frac{n\left(m^{2}-1\right)+1}{n(2 m-2)+1}=k$, then

$$
\begin{aligned}
\frac{m^{2} n-n+1}{2 m n-2 n+1} & =k \\
m^{2} n-n+1 & =2 m n k-2 n k+k \\
m^{2} n-2 m n k+(2 n k-n-k+1) & =0
\end{aligned}
$$

We treat this as a quadratic equation in $m$.

Since $m$ is an integer, then this equation has integer roots, and so its discriminant must be a perfect square.

The discriminant of this quadratic equation is

$$
\begin{aligned}
\Delta & =(-2 n k)^{2}-4 n(2 n k-n-k+1) \\
& =4 n^{2} k^{2}-8 n^{2} k+4 n^{2}+4 n k-4 n \\
& =4 n^{2}\left(k^{2}-2 k+1\right)+4 n(k-1) \\
& =4 n^{2}(k-1)^{2}+4 n(k-1) \\
& =(2 n(k-1))^{2}+2(2 n(k-1))+1-1 \\
& =(2 n(k-1)+1)^{2}-1
\end{aligned}
$$

We note that $(2 n(k-1)+1)^{2}$ is a perfect square and $\Delta$ is supposed to be a perfect square. But these perfect squares differ by 1 , and the only two perfect squares that differ by 1 are



1 and 0.

(To justify this last fact, we could look at the equation $a^{2}-b^{2}=1$ where $a$ and $b$ are non-negative integers, and factor this to obtain $(a+b)(a-b)=1$ which would give $a+b=a-b=1$ from which we get $a=1$ and $b=0$.)

Since $(2 n(k-1)+1)^{2}=1$ and $2 n(k-1)+1$ is non-negative, then $2 n(k-1)+1=1$ and so $2 n(k-1)=0$.

Since $n$ is positive, then $k-1=0$ or $k=1$.

Therefore, the only possible way in which the average is an integer is if the average is 1.

In this case, we get

$$
\begin{aligned}
m^{2} n-2 m n+(2 n-n-1+1) & =0 \\
m^{2} n-2 m n+n & =0 \\
n\left(m^{2}-2 m+1\right) & =0 \\
n(m-1)^{2} & =0
\end{aligned}
$$

Since $n$ and $m$ are positive integers with $m \geq 2$, then $n(m-1)^{2} \neq 0$, which is a contradiction.

Therefore, the average of the terms in an $(m, n)$-sawtooth sequence cannot be an integer.

Final answer: 证明题，略
Subfield: Combinatorics


Question: At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected

<img_3522>
uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping.
For a 2-topping pizza, determine the probability that at least $\frac{1}{4}$ of the pizza is covered by both toppings.

Solution:
Assume that the first topping is placed on the top half of the pizza. (We can rotate the pizza so that this is the case.)

Assume that the second topping is placed on the half of the pizza that is above the horizontal diameter that makes an angle of $\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\theta$ to $\theta+180^{\circ}$.
<img_3295>

We may assume that $0^{\circ} \leq \theta \leq 360^{\circ}$.

When $0^{\circ} \leq \theta \leq 90^{\circ}$, the angle of the sector covered by both toppings is at least $90^{\circ}$ (and so is at least a quarter of the circle).

When $90^{\circ}<\theta \leq 180^{\circ}$, the angle of the sector covered by both toppings is less than $90^{\circ}$ (and so is less than a quarter of the circle).

When $\theta$ moves past $180^{\circ}$, the left-hand portion of the upper half circle starts to be covered with both toppings again. When $180^{\circ} \leq \theta<270^{\circ}$, the angle of the sector covered by both toppings is less than $90^{\circ}$ (and so is less than a quarter of the circle).

When $270^{\circ} \leq \theta \leq 360^{\circ}$, the angle of the sector covered by both toppings at least $90^{\circ}$ (and so is at least a quarter of the circle).

Therefore, if $\theta$ is chosen randomly between $0^{\circ}$ and $360^{\circ}$, the combined length of the intervals in which at least $\frac{1}{4}$ of the pizza is covered with both toppings is $180^{\circ}$.

Therefore, the probability is $\frac{180^{\circ}}{360^{\circ}}$, or $\frac{1}{2}$.

Final answer: $\frac{1}{2}$
Subfield: Combinatorics


Question: At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected

![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=288&width=525&top_left_y=1157&top_left_x=1193)
uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping.
For a 2-topping pizza, determine the probability that at least $\frac{1}{4}$ of the pizza is covered by both toppings.

Solution:
Assume that the first topping is placed on the top half of the pizza. (We can rotate the pizza so that this is the case.)

Assume that the second topping is placed on the half of the pizza that is above the horizontal diameter that makes an angle of $\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\theta$ to $\theta+180^{\circ}$.
![](https://cdn.mathpix.com/cropped/2023_12_21_2f637db46c3cc012c3e8g-1.jpg?height=300&width=616&top_left_y=435&top_left_x=856)

We may assume that $0^{\circ} \leq \theta \leq 360^{\circ}$.

When $0^{\circ} \leq \theta \leq 90^{\circ}$, the angle of the sector covered by both toppings is at least $90^{\circ}$ (and so is at least a quarter of the circle).

When $90^{\circ}<\theta \leq 180^{\circ}$, the angle of the sector covered by both toppings is less than $90^{\circ}$ (and so is less than a quarter of the circle).

When $\theta$ moves past $180^{\circ}$, the left-hand portion of the upper half circle starts to be covered with both toppings again. When $180^{\circ} \leq \theta<270^{\circ}$, the angle of the sector covered by both toppings is less than $90^{\circ}$ (and so is less than a quarter of the circle).

When $270^{\circ} \leq \theta \leq 360^{\circ}$, the angle of the sector covered by both toppings at least $90^{\circ}$ (and so is at least a quarter of the circle).

Therefore, if $\theta$ is chosen randomly between $0^{\circ}$ and $360^{\circ}$, the combined length of the intervals in which at least $\frac{1}{4}$ of the pizza is covered with both toppings is $180^{\circ}$.

Therefore, the probability is $\frac{180^{\circ}}{360^{\circ}}$, or $\frac{1}{2}$.

Final answer: $\frac{1}{2}$
Subfield: Number Theory


Question: What is the smallest positive integer $x$ for which $\frac{1}{32}=\frac{x}{10^{y}}$ for some positive integer $y$ ?

Solution:
Since $10^{y} \neq 0$, the equation $\frac{1}{32}=\frac{x}{10^{y}}$ is equivalent to $10^{y}=32 x$.

So the given question is equivalent to asking for the smallest positive integer $x$ for which $32 x$ equals a positive integer power of 10 .

Now $32=2^{5}$ and so $32 x=2^{5} x$.

For $32 x$ to equal a power of 10, each factor of 2 must be matched with a factor of 5 .

Therefore, $x$ must be divisible by $5^{5}$ (that is, $x$ must include at least 5 powers of 5 ), and so $x \geq 5^{5}=3125$.

But $32\left(5^{5}\right)=2^{5} 5^{5}=10^{5}$, and so if $x=5^{5}=3125$, then $32 x$ is indeed a power of 10 , namely $10^{5}$.

This tells us that the smallest positive integer $x$ for which $\frac{1}{32}=\frac{x}{10^{y}}$ for some positive integer $y$ is $x=5^{5}=3125$.

Final answer: 3125
Subfield: Algebra


Question: Determine all pairs $(x, y)$ of real numbers that satisfy the system of equations

$$
\begin{aligned}
x\left(\frac{1}{2}+y-2 x^{2}\right) & =0 \\
y\left(\frac{5}{2}+x-y\right) & =0
\end{aligned}
$$

Solution:
From the first equation, $x\left(\frac{1}{2}+y-2 x^{2}\right)=0$, we obtain $x=0$ or $\frac{1}{2}+y-2 x^{2}=0$.

From the second equation, $y\left(\frac{5}{2}+x-y\right)=0$, we obtain $y=0$ or $\frac{5}{2}+x-y=0$.

If $x=0$, the first equation is satisified.

For the second equation to be true in this case, we need $y=0$ (giving the solution $(0,0)$ ) or $\frac{5}{2}+0-y=0$. The second equation gives $y=\frac{5}{2}$ (giving the solution $\left(0, \frac{5}{2}\right)$ ).

If $y=0$, the second equation is satisified.

For the first equation to be true in this case, we need $x=0$ (giving the solution $(0,0)$ ) or $\frac{1}{2}+0-2 x^{2}=0$. The second equation gives $x^{2}=\frac{1}{4}$ or $x= \pm \frac{1}{2}$ (giving the solutions $\left(\frac{1}{2}, 0\right)$ and $\left.\left(-\frac{1}{2}, 0\right)\right)$.

So far, we have accounted for all solutions with $x=0$ or $y=0$.

If $x \neq 0$ and $y \neq 0$, then for both equations to be true, we need $\frac{1}{2}+y-2 x^{2}=0$ (or $1+2 y-4 x^{2}=0$ ) and $\frac{5}{2}+x-y=0$ ( or $5+2 x-2 y=0$ ).

Adding these two equations, we obtain $6+2 x-4 x^{2}=0$.

This is equivalent to $2 x^{2}-x-3=0$ or $(2 x-3)(x+1)=0$, whose solutions are $x=\frac{3}{2}$ and $x=-1$.

The equation $\frac{5}{2}+x-y=0$ tells us that $y=x+\frac{5}{2}$.

If $x=\frac{3}{2}$, then $y=4$; if $x=-1$, then $y=\frac{3}{2}$.

Therefore, the complete list of pairs that satisfy the given system of equations is

$$
(x, y)=(0,0),\left(0, \frac{5}{2}\right),\left(\frac{1}{2}, 0\right),\left(-\frac{1}{2}, 0\right),\left(\frac{3}{2}, 4\right),\left(-1, \frac{3}{2}\right)
$$

Final answer: $(0,0),(0, \frac{5}{2}),(\frac{1}{2}, 0),(-\frac{1}{2}, 0),(\frac{3}{2}, 4),(-1, \frac{3}{2})$
Subfield: Algebra


Question: Determine all real numbers $x>0$ for which

$$
\log _{4} x-\log _{x} 16=\frac{7}{6}-\log _{x} 8
$$

Solution:
Note that $x \neq 1$ since 1 cannot be the base of a logarithm. This tells us that $\log x \neq 0$. Using the fact that $\log _{a} b=\frac{\log b}{\log a}$ and then using other logarithm laws, we obtain the following equivalent equations:

$$
\begin{aligned}
\log _{4} x-\log _{x} 16 & =\frac{7}{6}-\log _{x} 8 \\
\frac{\log x}{\log 4}-\frac{\log 16}{\log x} & =\frac{7}{6}-\frac{\log 8}{\log x} \quad(\text { note that } x \neq 1, \text { so } \log x \neq 0) \\
\frac{\log x}{\log 4} & =\frac{7}{6}+\frac{\log 16-\log 8}{\log x} \\
\frac{\log x}{\log \left(2^{2}\right)} & =\frac{7}{6}+\frac{\log \left(\frac{16}{8}\right)}{\log x} \\
\frac{\log x}{2 \log 2} & =\frac{7}{6}+\frac{\log 2}{\log x} \\
\frac{1}{2}\left(\frac{\log x}{\log 2}\right) & =\frac{7}{6}+\frac{\log 2}{\log x}
\end{aligned}
$$

Letting $t=\frac{\log x}{\log 2}=\log _{2} x$ and noting that $t \neq 0$ since $x \neq 1$, we obtain the following equations equivalent to the previous ones:

$$
\begin{aligned}
\frac{t}{2} & =\frac{7}{6}+\frac{1}{t} \\
3 t^{2} & =7 t+6 \quad(\text { multiplying both sides by } 6 t) \\
3 t^{2}-7 t-6 & =0 \\
(3 t+2)(t-3) & =0
\end{aligned}
$$

Therefore, the original equation is equivalent to $t=-\frac{2}{3}$ or $t=3$.

Converting back to the variable $x$, we obtain $\log _{2} x=-\frac{2}{3}$ or $\log _{2} x=3$, which gives $x=2^{-2 / 3}$ or $x=2^{3}=8$.

Final answer: $2^{-2 / 3}$, $8$
Subfield: Combinatorics


Question: Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$.
Calculate $P(10)$.

Solution:
Here, $k=10$ and so there are 10 balls in each bag.

Since there are 10 balls in each bag, there are $10 \cdot 10=100$ pairs of balls that can be chosen.

Let $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen. To determine $P(10)$, we count the number of pairs $(a, b)$ for which $a b$ is divisible by 10 .

If the number of pairs is $m$, then $P(10)=\frac{m}{100}$.

For $a b$ to be divisible by 10, at least one of $a$ and $b$ must be a multiple of 5 and at least one of $a$ and $b$ must be even.

If $a=10$ or $b=10$, then the pair $(a, b)$ gives a product $a b$ divisible by 10 .

In this case, we obtain the 19 pairs

$$
(a, b)=(1,10),(2,10), \ldots,(9,10),(10,10),(10,9), \ldots,(10,2),(10,1)
$$

If neither $a$ nor $b$ equals 10 , then either $a=5$ or $b=5$ in order for $a$ or $b$ to be divisible by 5 .

In this case, the other of $a$ and $b$ must be even and not equal to 10. (We have already counted the pairs where $a=10$ or $b=10$.)

In this case, we obtain the 8 pairs

$$
(a, b)=(5,2),(5,4),(5,6),(5,8),(2,5),(4,5),(6,5),(8,5)
$$

From our work above, there are no additional pairs for which $a b$ is divisible by 10 .

Thus, there are $19+8=27$ pairs $(a, b)$ for which $a b$ is divisible by 10 , which means that $P(10)=\frac{27}{100}$.

(We note that we could have made a 10 by 10 table that listed all possible combinations of $a$ and $b$ and their product, from which we could obtain $P(10)$.)

Final answer: $\frac{27}{100}$