Galochkina A.I.

Sasha's collection consists of coins and stickers, and there are fewer coins than stickers, but there is at least one. Sasha chose some positive number $\$ \mathrm{t}>1 \$$ (not necessarily an integer). If he increases the number of coins by $\$ \mathrm{t} \$$ times, without changing the number of stickers, then his collection will have $\$ 100 \$$ items. If instead he increases the number of stickers by $\$ t \$$ times, without changing the number of coins, then he will have $\$ 101 \$$ items. How many stickers could Sasha have? Find all possible answers and prove that there are no others.
Let's denote the number of coins by $\$ \mathrm{~m} \$$ and the number of stickers by $\$ \mathrm{n} \$$, then the condition can be rewritten as a system of equations $\$ \$$ begin $\{$ cases $\} \mathrm{mt}+\mathrm{n}=100, \ \mathrm{~m}+\mathrm{nt}=101$. \end } \{ cases \} \$ \$ Subtract the first equation from the second: \$ $1=101-100=(\mathrm{m}+\mathrm{nt})-(\mathrm{mt}+\mathrm{n})=(\mathrm{n}-\mathrm{m})(\mathrm{t}-1) . \$ \$$ Therefore, $\$ \mathrm{t}=1+(\mathrm{frac}\{1\}\{\mathrm{n}-\mathrm{m}\} \$$. Now add the two initial equations: \$ $201=101+100=(\mathrm{mt}+\mathrm{n})+(\mathrm{m}+\mathrm{nt})=(\mathrm{m}+\mathrm{n})(\mathrm{t}+1) . \$ \$$ Therefore, $\$ \mathrm{t}=\mid$ frac $\{201\}\{\mathrm{m}+\mathrm{n}\}-1 \$$. Introduce the notations $\$ a=n-m \$, \$ b=n+m \$$. Note that \$a>0 \$, since $\$ n>m \$$. Equate the two expressions for \$t\$: $\$ \$ 1+\mid$ frac $\{1\}\{n-m\}=\backslash$ frac $\{201\}\{m+n\}-1 \backslash \backslash$ Leftrightarrow $\backslash 1+\mid$ frac $\{1\}\{a\}=\mid$ frac $\{201\}\{b\}-1 \backslash \backslash$ Leftrightarrow $\backslash$ frac $\{2 a+1\}\{a\}=\mid f r a c\{201\}\{b\} . \$ \$$ Note that the numbers $\$ 2 a+1 \$$ and \$a are coprime, so the fraction $\$$ frac $\{2 a+1\}$

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$\$ 67$ and $\$ 201 \$$. Since $\$ 2 a+1>1 \$$, we need to consider three cases.

1) $\$ 2 a+1=3 \$$. Then $\$ a=1 \$$ and $\$$ frac $\{201\}\{b\}=\mid f r a c\{3\}\{1\} \$ .3$ so, $\$ b=67 \$$, from which $\$ \$ m=\mid f r a c\{1\}\{2\}(b-a)=33$, \quad $\mathrm{n}=\mid \operatorname{frac}\{1\}\{2\}(\mathrm{a}+\mathrm{b})=34 . \$ \$$ Also, $\$ \mathrm{t}=2 \$$. It is easy to verify that this case works.
2) $\$ 2 \mathrm{a}+1=67 \$$. Then $\$ \mathrm{a}=33 \$$ and $\$|\operatorname{frac}\{201\}\{\mathrm{b}\}=|$ frac $\{67\}\{33\} \$$. So, $\$ \mathrm{~b}=99 \$$, from which $\$ \$ \mathrm{~m}=\mid$ frac $\{1\}\{2\}(\mathrm{b}-\mathrm{a})=33$, Iquad $\mathrm{n}=\mid$ frac $\{1\}\{2\}(\mathrm{a}+\mathrm{b})=66 . \$ \$$ Also, $\$ \mathrm{t}=\mid$ frac $\{34\}\{33\} \$$. It is easy to verify that this case works.
3) $\$ 2 a+1=201 \$$. Then $\$ a=100 \$$ and $\$$ frac $\{201\}\{b\}=\mid f r a c\{201\}\{100\} \$$. So, $\$ b=100 \$$, from which $\$ \$ m=\mid f r a c\{1\}\{2\}(b-$ a)=0, |quad n=|frac $\{1\}\{2\}(a+b)=100 . \$ \$$ This case does not work, as there must be at least one coin.

## Answer

34 or 66 stickers.

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