Gonitzoggoβ
This is a proof based problem added for learning purposes and does not accept submissions.
You can view the solution by clicking on the solution tab.
You can view the solution by clicking on the solution tab.
Disclaimer: The solutions we've shared are just one exciting approach, and
there are surely many other wonderful methods out there. We’d love to hear
your alternative solutions in the community thread below, so let's keep the
creativity
flowing!
Let $a_k$ be the number of $k$-cent coins used to make $A$ cents with $B$ coins.
\[a_1+a_2+a_5+a_{10}+a_{20}+a_{50}+a_{100}=B\]
\[a_1+2a_2+5a_5+10a_{10}+20a_{20}+50a_{50}+100a_{100}=A\]
Now, we use $ka_k$ $\frac{100}{k}$-cent coins for each possible value of $k$.
\[\text{The number of coins used}=a_1+2a_2+5a_5+10a_{10}+20a_{20}+50a_{50}+100a_{100}=A\]
\[\text{The amount of cents}=100(a_1+a_2+a_5+a_{10}+a_{20}+a_{50}+a_{100})=100B=B\text{ dollars}\]
$\therefore$ We can use $A$ coins to form $B$ dollars.