Answer: The last remaining coin will be a dime.
To see this, we observe that no matter what two coins are selected from the jug, we will always end up with one less coin in the jug after each selection as one coin is always added to the jug after two are removed. Additionally, in each selection, the number of dimes in the jug is either increased by 1 or decreased by 1, while the number of nickels either remains the same or is decreased by 2. This is summarized in the table below.
| Case (coins chosen) |
Net Change in the # of Dimes | Net Change in the # of Nickels | Net Change in the Total # of Coins |
|---|---|---|---|
| i. both dimes | -1 | 0 | -1 |
| ii. both nickels | +1 | -2 | -1 |
| iii. 1 dime and 1 nickel | -1 | 0 | -1 |
Since the total number of coins decreases by 1 after each selection, at some point we must have one coin remaining in the jug. However, the number of nickels is initially even (2018) and always decreases by an even amount (either 2 or 0), so it is not possible for there to ever be one nickel in the jug. Thus we conclude that the last coin in the jar must be a dime.