Big Holes by Elap
One approach was: 27dn had to be a four-digit cube and so T lay in the range 10 to 21. From 3dn, 2T = 3O and so T was a multiple of 3, ie, 12, 15, 18 or 21. The only factorial which fitted 14dn was 7! = 5040, and so 23ac had to end with a zero. From the clues to 3dn, 32dn, 14dn and 23ac the values of O, P, Q, T, U, V and W could be fixed. After further deductions, A=51, B=739, C=408, D=514, E=19, F=515, G=710, H=18, I=543, J=883, K=5317, L=400, M=7708, O=10, P=370, Q=5040, R=33, S=738, T=15, U=180, V=1807, W=45, X=771, Y=53 and Z=815. Sorting letters by ascending values gave OTHERWAYUPLCDFIGS…, and if the digits in the grid were read as if in LCD (liquid crystal display) segments, the grid revealed a crossword upside-down, reading 3 as E, 4 as h, 5 as S, etc, so that, eg, 38ac reads as LOESS. Individual letter values shared this property. The sum of all entries, 37819173, inverted made ELIGIBLE, to be written underneath the grid. It was not necessary for solvers to enter the digits as LCD segments.
One possible route is shown in full HERE. You will need to use the “BACK” button on your browser to return to this page.