Count Like a Babylonian
JavaScript (ES6), 44 bytes

Takes input as an array of ASCII characters.



a=&gt;a.map(c=&gt;k+=c&lt;1?k*59:c&lt;'?'?10:c&lt;{},k=0)|k


Try it online!

How?

The Babylonian Numeral System can be seen as a 4-instruction language working with a single register -- let's call it the accumulator.

Starting with \$k=0\$, each character \$c\$ in the input array \$a\$ modifies the accumulator \$k\$ as follows:


space: multiply \$k\$ by \$60\$ (implemented as: add \$59k\$ to \$k\$)
&lt;: add \$10\$ to \$k\$
T: increment \$k\$
\: do nothing; this is the NOP instruction of this language (implemented as: add \$0\$ to \$k\$)