The nine sets of six numbers had the property that the sum of the Nth powers of the members is the same for all the sets, for all (integer) values of N = 0 to N = 5. This is an extension of the Prouhet-Tarry-Escott Problem, in which only two sets are required.

This is one possible solution path.
All grid entries have either two or three digits,
so they’re in the range 10 to 999.
None of the letter values can be the stated power of
4, 8, 9 or any number that is itself a square or higher power,
because the result would be classed as a higher power (of a smaller number) than the one stated;
generally, (x^{y})^{z} = x^{yz} and would be listed as a (yz)th power, not a zth power.
For example, none of the cubes could be 4096 = 16^{3},
because that’s a 12th power (2^{12}).

D is an 8th power and 18ac = D + S means it’s < 1000;
3^{8} = 6561 is too big,
so **D = 256** = 2^{8}.
The cube S is then < 1000 − 256 = 744,
ie one of {8, 27, 125, 216, 343}
(not 64 = 2^{6}, 512 = 2^{9} or 729 = 3^{6})
and 18ac is one of {264, 283, 381, 472, 599}.
That means 18dn = FU + e + P is in the range [200, 599],
with U ≥ 4, e ≥ 8 and P ≥ 16,
so the maximum for the 5th power F is (599 − 16 − 8)/4 = 143.75,
ie **F = 32** = 2^{5}.
The other 5th power, I, is also < 1000 (from 6dn),
so **I = 243** = 3^{5}.

12dn = P has two digits, so P is one of the 4th powers {16, 81}.
19dn = H + 2P has three digits,
so the 6th power H < 1000 and must be one of {64, 729},
giving the values {96 (too small), 226, 761, 891} for 19dn.
But 19dn must start with one of {1, 2, 3, 4, 9} to match the last digit of 18ac,
so **19dn = 226**, **H = 64**
and **12dn = P = 81**,
which makes **18ac = 472** and **S = 216**.

Now 18dn = 32U + e + 81 matches 4__ in the grid,
where U is a square and e is a cube.
The maximum for U is (499 − 81 − 8)/32 ≈ 12.8,
so U is one of {4, 9}.
But from the grid 14dn = 2s + U matches _7, so U is odd,
ie **U = 9** and 18dn = 369 + e.
So e is between 400 − 369 = 31 and 499 − 369 = 130,
which can only be **e = 125**,
making **18dn = 494**.
As 14dn = 2s + 9 ends in 7, 2s must end in 8,
with s ending in 4 or 9;
the maximum for s is (97 − 9)/2 = 44,
so the only available square is **s = 4**,
making **14dn = 17**.

22ac = 96 − O, matching _9 in the grid,
so the cube O ends in 7 and can only be **O = 27**,
making **22ac = 69**.
3dn = 793 + W, and the square W must end in one of {0, 1, 4, 5, 6, 9},
so 3dn ends in one of {3, 4, 7, 8, 9, 2} respectively,
which is also the last digit of 11ac.
11ac = 341 + n, so the cube n is < 1000 − 341 = 659
and ends in one of {2, 3, 6, 7, 8, 1};
the only available values are {8, 343},
but if n is 8 then 11ac would be 349,
which doesn’t fit with the _8_ in the grid,
so **n = 343** and **11ac = 684**.
Now 3dn ends in 4, so W ends in 1 and must be < 1000 - 793 = 207;
the only available square is **W = 121**,
making **3dn = 914**.

Now 2dn = T − 343, matching __6 in the grid,
so the square T ends in 9 and is bounded by 106 + 343 = 449
and 999 (because 4dn = T has three digits),
ie **T = 529** (not 729, which is a 6th power)
which makes **2dn = 186**,
**4dn = 529** and **6dn = 897**.
The grid now has **9ac = 12** for the clue M − 4,
so **M = 16**.
Now 16dn = 291 + i (three digits), where i is a 4th power < 1000;
the only available value is **i = 625**,
making **16dn = 916**.

The grid has **13ac = 917** for the clue 881 + N,
so **N = 36**, which makes **7dn = 661**.
8ac = A, a square matching _8_ in the grid,
so it’s one of {289, 484, 784},
which makes 1dn = A + 88 one of {377, 572, 872};
the middle digit of 1dn thus has to be 7,
which forces **8ac = A = 784**,
making **1dn = 872** and also completing
**1ac = 816** and **23ac = 124**.

5ac = L + 45 is _86 in the grid,
so the square L matches _41 and is one of {441, 841}.
But 20dn = L − 499, so **L = 841**,
**5ac = 886** and **20dn = 342**.
Now 25ac = C + t − 216, with three digits;
for the 6th power C, 4^{6} = 4096
is too big (and also a 12th power),
so the only available value is
**C = 729** = 3^{6}.
That makes 25ac = t + 513, matching _62 in the grid,
so the square t must end in 49 and be < 962, which only allows
**t = 49**, making **25ac = 562**.

Now 17dn = 962 − o, matching _6_ in the grid,
so the square o has a tens digit of 0 or 9
and is between 962 − 869 = 93 and 962 − 160 = 802.
The available values are {100, 196, 400},
but 15dn = 377 − o, so we can rule out 400,
and 15dn is one of {181, 277}.
21ac = E + 425, matching _1_ in the grid,
and must start with 7 or 8 from 15dn,
so the square E is between 710 − 425 = 285 and 819 − 425 = 394,
and has a tens digit of 8 or 9;
the only available value is **E = 289**,
making **21ac = 714**,
which forces **15dn = 277** and **o = 100**,
in turn completing **17dn = 862**
and the unclued **16ac = 918**.

Finally, 24ac = 965 − R matches 76_ in the grid,
so the square R is limited by 965 − 769 = 196 and 965 − 760 = 205,
ie **R = 196**,
**24ac = 769** and then **10ac = 596**,
completing the grid.

The sets can now be completed as {816, 886, 529, 102, 32, 389}, {784, 17, 342, 134, 901, 576}, {12, 596, 769, 906, 322, 149}, {684, 917, 226, 234, 1, 692}, {472, 69, 862, 446, 849, 56}, {714, 661, 916, 204, 257, 2}, {124, 562, 897, 794, 356, 21}, {872, 81, 494, 46, 837, 424}, {186, 914, 277, 732, 4, 641}.

In ascending value order, the clue letters spell “sUM OF NtH PoWeRS IDEnTiCAL”, so we need to find the sum of the Nth powers of the members of each set, for various values of N.

Obviously, for N = 0, x^{0} = 1 for every member,
so the sum for each set will always be 6.
Also, for N = 1,
the sum of x^{1} is just the sum of the members themselves,
and as each set is defined as
{x, y, z, 918 − x, 918 − y, 918 − z}
the sum will be 3×918 = 2754, for all of the sets.
The sum of squares for the first set is
(816^{2} + 886^{2} + 529^{2}
+ 102^{2} + 32^{2} + 389^{2}) = 1893442
and for the second set it’s
(784^{2} + 17^{2} + 342^{2}
+ 134^{2} + 901^{2} + 576^{2}) = 1893442,
so the result works for N = 2.

Higher powers are more laborious to calculate,
especially on smaller calculators,
but to avoid overflow one can divide each number by 100
before taking the power
(being wary of rounding errors in the fractional part)
and multiply the sum by 100^{N}
(ie, moving the decimal point 2N places to the right)
to make it an integer.
For example, the sum of cubes for the first set can be represented as
(8.16^{3} + 8.86^{3} + 5.29^{3}
+ 1.02^{3} + 0.32^{3} + 3.89^{3}) = 1446.838686,
and the second set also gives
(7.84^{3} + 0.17^{3} + 3.42^{3}
+ 1.34^{3} + 9.01^{3} + 5.76^{3}) = 1446.838686,
and so on.
The actual sum for N = 3 is 1446838686,
which is the fractional version multiplied by 100^{3}.

For a bigger example,
the largest number in all the sets is 917,
and its 5th power can be calculated on a smaller calculator as follows.
9.17^{5} = 64840.5482564357,
which on a 10-digit calculator would be rounded to 64840.54826,
and shifting the decimal point 10 places gives the approximation 648405482600000.
To get the correct last five digits,
calculate 917^{2} = 840889 and 917^{3} = 771095213,
then multiply the last five digits of those to get 40889×95213 = 3893164357,
and add the last five digits of that to the approximation,
giving 648405482664357.
In case the last digit (6) in the approximation has been rounded,
check the digit sum (remainder when divided by 9) as follows.
The digit sum of 917 is 17, which in turn has a digit sum of 8,
and raising that to the 5th power gives 32768,
with a digit sum of 26 → 8,
so the digit sum of 917^{5} should be 8.
But the sum of digits in 648405482664357 is 72 → 9,
one greater than it should be,
so the correct value of 917^{5} is 648405482**5**64357.

However the calculations are done, the sum of 4th powers is 1160901254866 for all the sets, and the sum of 5th powers is always 958099908623814. For N = 6, using fractional calculations we get sums of 804326.6583… for the first set and 805338.3330… for the second set, differing even in the non-fractional part, so the property doesn’t hold for N = 6. Checking the lower bound, N = -1, ie the sum of the reciprocals of set members, gives approximately 0.0479 for the first set and 0.0733 for the second, so that’s excluded. Therefore, the sums of Nth powers are equal for all the sets for N = 0 to 5 only.