+6248 \(\text{Well conceptually it's simple enough}\\ n + d + q = 30\\ 5n+10d+25q=500 \\ n+2d+5q=100\)
\(\text{we can take the difference of these two equations to obtain}\\ d + 4q=70\\ q = \dfrac{70-d}{4},~q \in \mathbb{Z^+}+\{0\}\\ \text{and we can now read off the possible $q$'s}\\ (d,q)=(2,17), (6,16),(10,15),(14,14),(18,13),(22,12), (26,11),(30,10)\\ \text{and of these we select those who's sum is 30 or less}\\ (d,q)=(2,17), (6,16),(10,15),(14,14)\\ \text{and we can then fill in $n$}\\ (n,d,q) = (11,2,17), (8,6,16), (5,10,15), (2,14,14) \)
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+26367 OSL4#13
\(\text{Let the number of nickels $=n$, $1$ nickel $= 5 $ cent } \\ \text{Let the number of dimes $=d$, $1$ dime $= 10 $ cent } \\ \text{Let the number of quaters $=q$, $1$ quater $= 25$ cent }\)
\(\begin{array}{|lrcll|} \hline (1) & \mathbf{ n+d+q } &=& \mathbf{30} \\\\ & 5n+10d+25q &=& 500 \qquad \text{in cents}\quad &| \quad : 5 \\ (2) & \mathbf{ n+2d+ 5q } &=& \mathbf{100} \\ \hline (2)-(1): & n+2d+ 5q -(n+d+q) &=& 100-30 \\ & n+2d+ 5q -n-d-q &=& 70 \\ & d+ 4q &=& 70 \quad &| \quad -4q \\ (3)& \mathbf{ d } &=& \mathbf{70- 4q} \\ \hline & n+d+q &=& 30 \\ &n&=& 30-d-q \quad &| \quad d =70- 4q \\ &n&=& 30-(70- 4q)-q \\ &n&=& 30-70+ 4q-q \\ (4)& \mathbf{ n } &=& \mathbf{-40+ 3q} \\ \hline \end{array} \)
\(\mathbf{d \geq 0}\)
\(\begin{array}{|rcll|} \hline 70- 4q &\geq& 0\quad &| \quad + 4q \\ 70 &\geq& 4q \quad &| \quad : 4 \\ \dfrac{70}{4} &\geq& q \\ q &\leq& \dfrac{70}{4} \\ q &\leq& 17.50 \\ \mathbf{q} &\leq& \mathbf{17} \\ \hline \end{array}\)
\(\mathbf{n \geq 0}\)
\(\begin{array}{|rcll|} \hline -40+ 3q &\geq& 0 \quad &| \quad +40 \\ 3q &\geq& 40 \quad &| \quad : 3 \\ q &\geq& \dfrac{40}{3} \\ q &\geq& 13.3333333333 \\ \mathbf{q} &\geq& \mathbf{14} \\ \hline \end{array} \)
\(\mathbf{14\leq q \leq 17 } \qquad q=\{14,\ 15,\ 16,\ 17 \} \)
There are 4 different combinations.
The combinations:
\(\begin{array}{|c|c|c|} \hline \text{q (Quaters)} & d=70- 4q \text{ (dimes)} & n=3q-40 \text{ (nickels)} \\ \hline 14 & 14 & 2 \\ 15 & 10 & 5 \\ 16 & 6 & 8 \\ 17 & 2 &11 \\ \hline \end{array}\)
