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}=\midfrac\{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 {2a+1}
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\ 67and\ 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}=\midfrac\{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 \,fromwhich\ \ m=\mid f r a c\{1\}\{2\}(b-a)=0,∣quadn=∣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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