Problem 3. Sasha's collection consists of coins and stickers, with fewer coins than stickers, but at least one coin is present. Sasha chose a positive number $t>1$ (not necessarily an integer). If he increases the number of coins by a factor of $t$ and leaves the number of stickers the same, his collection will have 100 items. If, instead, he increases the number of stickers by a factor of $t$ and leaves the number of coins the same, he will have 101 items. How many stickers could Sasha have? Find all possible answers and prove that there are no others.
Answer: 34 or 66.

Solution. Let $m$ be the number of coins, and $n$ be the number of stickers. Then the condition can be rewritten as the system of equations

$$
\left\{\begin{array}{l}
m t+n=100 \\
m+n t=101
\end{array}\right.
$$

Subtract the first equation from the second:

$$
1=101-100=(m+n t)-(m t+n)=(n-m)(t-1)
$$

Therefore, $t=1+\frac{1}{n-m}$. Now add the two original equations:

$$
201=101+100=(m t+n)+(m+n t)=(m+n)(t+1)
$$

Therefore, $t=\frac{201}{m+n}-1$. Introduce the notations $a=n-m, b=n+m$. Note that $a>0$, since $n>m$. Equate the two expressions for $t$:

$$
1+\frac{1}{n-m}=\frac{201}{m+n}-1 \Leftrightarrow 1+\frac{1}{a}=\frac{201}{b}-1 \Leftrightarrow \frac{2 a+1}{a}=\frac{201}{b}
$$

Note that the numbers $2 a+1$ and $a$ are coprime, so the fraction $\frac{2 a+1}{a}$ is irreducible. Therefore, 201 is divisible by $2 a+1$. The number $201=3 \cdot 67$ has only four divisors: $1,3,67$ and 201. Since $2 a+1>1$, we need to consider three cases.

- First case: $2 a+1=3$. Then $a=1$ and $\frac{201}{b}=\frac{3}{1}$. Therefore, $b=67$, from which

$$
m=\frac{1}{2}(b-a)=33, \quad n=\frac{1}{2}(a+b)=34
$$

Also, $t=2$. It is easy to check that this case works.

- Second case: $2 a+1=67$. Then $a=33$ and $\frac{201}{b}=\frac{67}{33}$. Therefore, $b=99$, from which

$$
m=\frac{1}{2}(b-a)=33, \quad n=\frac{1}{2}(a+b)=66
$$

Also, $t=\frac{34}{33}$. It is easy to check that this case works.

- Third case: $2 a+1=201$. Then $a=100$ and $\frac{201}{b}=\frac{201}{100}$. Therefore, $b=100$, from which

$$
m=\frac{1}{2}(b-a)=0, \quad n=\frac{1}{2}(a+b)=100
$$

This case does not work, as there must be at least one coin.