29 - 1 = 30 how?

I suppose you could always round the answer to the nearest multiple of 5, although that has nothing to do with Athena.

$$begin{align} 29-1&=28 xrightarrow{text{rounds to }}30 14-1&=13 xrightarrow{phantom{text{________}}} 15 11-1&=10 xrightarrow{phantom{text{________}}} 10 9-1 &= phantom08 xrightarrow{phantom{text{________}}} 10 end{align}$$

$$9 - 1 = 10 = 11 - 1,$$ thus

$$29 - 1 = 20 + 9 - 1 = = 20 + 11 - 1 = 31 - 1 = 30.$$

Edit

Or just using that $9-1 = 10$

$$ 29 - 1 = 20 + 9-1 = 20 + 10 = 30.$$

Here's another approach that doesn't require redefining ‘$-$’ as a string operation, rather than a mathematical one:

$$begin{align} 29 - 1 (text{base} 11) &= 30 (text{base} 10) 14 - 1 (text{base} 12) &= 15 (text{base} 10) 11 - 1 (text{base} 10) &= 10 (text{base} 10) 9 - 1 (text{base} 10) &= 10 (text{base} 8) end{align}$$

I like the OP's intent better, though. It's a clever puzzle.

This is mathematically true in $mathbb{Z}/2mathbb{Z}$, i.e. $bmod 2$:

$$begin{align} 29 - 1 equiv 30 equiv 0 pmod 2 14 - 1 equiv 15 equiv 1 pmod 2 11 - 1 equiv 10 equiv 0 pmod 2 9 - 1 equiv 10 equiv 0 pmod 2 end{align}$$

This is inspired by/alternative to Olba12's nice answer:

$$begin{align} 30 &= 10 + 10 + 10 &= (11 - 1) + (11 - 1) + (9 - 1) &= 31 - 3 &= 31 - 2 - 1 &= 29 - 1 end{align}$$

Hence: $29 - 1 = 30$.

It is just the '-' operator has changed its semantics to compute sum of its operands.

This is a cheeky answer.

Assume the usual rules of arithmetic and logic.

We are told that $9 - 1 = 10$, that is, $8 = 10$, which is a contradiction.

By the rules of logic, since we have derived a contradiction, we can now derive anything else (like 'magic', cf the storybook character Minerva), such as, that

$29 - 1 = 30$.

QED