Hi, Beulah, yes, D=6, but A is not 7.
For the last few years, I've been in Australia in late February and haven't been able to do the February numerical puzzle. Unlike last week, where the gridfill was quick but the endgame was long and tedious, this week I luckily saw the likely cipher for converting digits to letters almost immediately, even before filling in Grid A. As already noted, the way into the puzzle is first to note that various numbers are divisors of other numbers, and this limits the values that certain letters can take. One of these is E. This means that there are only a limited number of possible combinations of E, T and L (see 21dn), and feeding these into 11ac gives a single value for D, and hence for E, T and L. Using the fact that O is an exponent in 13ac and that 13ac has three digits narrows down the possible values for O, and substituting back into 21dn shows that there is only one possible value that O can take. Working on 22ac and 4dn gives M, I, and K, and the rest is then fairly plain sailing.
Grid B was straightforward as the gridfill of Grid A was confirming my suspicions about the cipher and what needed to be done. I was amused by the nice touch in the fifth row of Grid B!