Cupidinous Convivial Country

Here's my (somewhat tedious) solution:

Starting at $1a$ , we know we'll need at least two $1a$ coins, to make $1a$ and $2a$ . For $3a$ , we can just use one $3a$ coin.

So here's our coins so far:

• Two $1a$
• One $3a$

From these we can make $4a$ and $5a$ , but we'll need another $3a$ coin to make $6a$ . After we do this, we can make $7a$ and $8a$ , but we'll need another $3a$ coin for $9a$ . Values of $10a$ and $11a$ will then be covered, but we'll need to add one more $3a$ coin for $12a$ .

Here's our coins so far:

• Two $1a$
• Four $3a$

But don't worry, the whole process isn't so laborious!

For $13a$ , we can add a new $13a$ coin. But notice something here: we went from $1a$ to $12a$ just with our $1a$ and $3a$ coins, so using these, we can go from $13a$ all the way to $13a + 12a = 25a$ and hit every integer along the way!

Coins so far:

• Two $1a$
• Four $3a$
• One $13a$

For $26a$ , we'll need another $13a$ coin. As before, we can then easily go from $26a$ to $26a + 12a = 38a$ with all integers in between. Then another $13a$ coin for $39a$ and then we can go from $39a$ to $39a + 12a = 51a$ .

Coins so far:

• Two $1a$
• Four $3a$
• Three $13a$

Now at this point, we need a value of $52a$ , but notice that we're not using all of our available small change. In $1a$ s and $3a$ s, we can technically increase $14a$ . We only used $3a$ coins to increase $12a$ , so we can easily add another $1a$ coin to get $52a$ .

Coins so far:

• Two $1a$
• Four $3a$
• Three $13a$
• One $53a$

Now we get to use a $53a$ coin for $53a$ . With small change, we can go from there to $65a$ . Then, we could just use add a $13a$ coin to our $53a$ coin for $66a$ . Then from there to $78a$ . And with two $13a$ coins, $79a$ . Then from there to $91a$ . (Hopefully the pattern is beginning to emerge!) Then with three $13a$ coins, $92a$ . And then from there we don't even need a $93a$ coin. We just have to use our smaller change to get up to $99a$ !

Whew!

But there you have it: we can get every integer from $1a$ to $99a$ using $2 + 4 + 3 + 1 = \boxed{10}$ coins!

Nice problem Pi Han Goh!