We can write the problem as the pair of simultaneous equations
\[2x+10y+50z=600 \rightarrow x+5y+25z=300\]
\[x+y+z=100\]
where
\[x, \; y, \; z\]
are the number of 2 pence, 10 pence and 50 pence coins respectively and only positive integer solutions are required.Let
\[z=k\]
then we can write the above equations as\[x+5y=300-25k\]
(1)\[x+y=100-k\]
(2)(2)-(1) gives
\[4y=200-24k \rightarrow y=50-6k\]
then from (2) \[x=100-k=100-(k)=50+5k\]
.\[z=k, \; y \ge 0 \rightarrow 0 \le k \le 8 \]
.Possible values of
\[x, \; y, \; z\]
are given in the table.| \[k\] |
\[x\] |
\[y\] |
\[z\] |
| 0 | 50 | 50 | 0 |
| 1 | 55 | 44 | 1 |
| 2 | 60 | 38 | 2 |
| 3 | 65 | 32 | 3 |
| 4 | 70 | 26 | 4 |
| 5 | 75 | 20 | 5 |
| 6 | 80 | 14 | 6 |
| 7 | 85 | 8 | 7 |
| 8 | 90 | 2 | 8 |