This is one possible solution path. For the special game described in the preamble, the cells landed on must contain different digits, so there can’t be more than ten of them. Trying repeated throws of each of the die numbers 1 to 6, the only one that reaches cell 100 within ten throws is 5, landing on cells {5, 10, 46, 51, 56, 40, 85, 90, 95, 100} in that order.

For 100ac we need a ten-digit value of n⁹, with the seventh digit > 0 because it’s the start of 94dn, so n is > 10 (because 1 000 000 000 has a 0 for 94dn) and n ≤ 9 999 999 999^(1/9) ≈ 12.9; 12⁹ = 5 159 780 352 would make 94dn start with 0, so n = 11 and 100ac = 2 357 947 691.

Each of 60ac, 61ac, 80ac, 81ac is a prime power of a two-digit prime, say p^r = N. Using natural logarithms, ln(p^r) = ln(N) is equivalent to r×ln(p) = ln(N), so r = ln(N)/ln(p) and since N has ten digits, r is between ln(1 000 000 000)/ln(97) ≈ 4.5 and ln(9 999 999 999)/ln(11) ≈ 9.6; the only primes in that range are 5 and 7, so these entries are all n⁵ or n⁷. For p⁵, the minimum for the prime p is 1 000 000 001^(1/5) ≈ 63.1; for p⁷, p is between 1 000 000 001^(1/7) ≈ 19.3 and 9 999 999 999^(1/7) ≈ 26.8, ie 23⁷ = 3 404 825 447 is the only possibility. Since all two-digit primes end in one of {1, 3, 7, 9}, their 5th and 7th powers must also end in one of those digits.

For 61dn we need a four-digit value of n⁴, ie one of {6⁴ = 1296, 7⁴ = 2401, 8⁴ = 4096, 9⁴ = 6561}, but all of its digits start across entries and therefore can’t be 0, leaving {1296, 6561}. Now 61ac starts with 1 or 6 and its 5th or 7th root must be a two-digit prime. If it starts with 6, the 5th root is between 6 000 000 000^(1/5) ≈ 90.3 and 6 999 999 999^(1/5) ≈ 93.1, but none of {91 = 7×13, 92 = 2×2×23, 93 = 3×31} is a prime; the 7th root is between 6 000 000 000^(1/7) ≈ 24.9 and 6 999 999 999^(1/7) ≈ 25.5, but 25 isn’t prime either. Therefore 61ac starts with 1 and 61dn = 1296. The 7th root of 61ac is between 1 000 000 000^(1/7) ≈ 19.3 and 1 999 999 999^(1/7) ≈ 21.3, but neither of {20, 21} is prime; the 5th root is between 1 000 000 000^(1/5) ≈ 63.1 and 1 999 999 999^(1/5) ≈ 72.5, ie one of {67, 71}, giving {1 350 125 107, 1 804 229 351} for 61ac. That means 68dn starts with 1 or 3, but the smallest product of five different primes is 2×3×5×7×11 = 2310, so 68dn starts with 3 and 61ac = 1 804 229 351.

Also, 60ac starts with 2 and its 7th root is between 2 000 000 000^(1/7) ≈ 21.3 and 2 999 999 999^(1/7) ≈ 22.6, but 22 isn’t prime; the 5th root is between 2 000 000 000^(1/5) ≈ 72.5 and 2 999 999 999^(1/5) ≈ 78.6 and the only prime in that range is 73, so 60ac = 2 073 071 593 = 73⁵. 40ac is a 6th power starting with 6, so its root is between 6 000 000 000^(1/6) ≈ 42.6 and 6 999 999 999^(1/6) ≈ 43.7, ie 40ac = 6 321 363 049 = 43⁶.

Now 68dn is 35_0 in the grid, so its prime factors include 2 and 5. They must also include 3 because 2×5×7×11×13 = 10010 is too big; to be a multiple of 3, the missing digit must be one of {1, 4, 7}. Similarly, 2×3×5×11×13 = 4290 is too big, so 7 must be a factor, making 68dn = 210p, where p is the remaining prime factor. Of the three possibilities, {3510/210 ≈ 16.7, 3540/210 ≈ 16.9, 3570/210 = 17}, only the last works, so 68dn = 3570 = 2×3×5×7×17.

41ac is (11up)³ matching 9____8_7__; since 11up and its cube end with the same digit, that digit must be one of {0, 1, 4, 5, 6, 9}. So 11up is between 9 000 080 700^(1/3) ≈ 2080.1 and 9 999 989 799^(1/3) ≈ 2154.4. Its second digit is the end of 20ac = 1 + (a number ending in one of {1, 3, 7, 9}), so it must be even, ie 11up matches 209_. Of the cubes of {2090, 2091, 2094, 2095, 2096, 2099}, the only one that fits the grid entry is 41ac = 9 208 180 736, with 11up = 2096. Now 21ac = 1 + p^r ends in 0, so p^r ends in 9; it can’t be 23⁷ (calculated above), so it’s p⁵ and p is a prime > 63 ending in 9, ie one of {79, 89}, making 21ac one of {3 077 056 400, 5 584 059 450}; whichever it is, we can enter the digits ____05_4_0. Now 15up = 2n² matches _568, so n² ends in 284 or 784 and is ≤ 4784; the only such square is 28² = 784, so 15up = 1568.

For 45up = n² we have 102__ in the grid, so n ≤ √10299 ≈ 101.5 and > 100 (whose square is 10000), ie, it’s 11 and 45up = 10201. Now the 5th digits of 80ac and 81ac are 0 and 1 respectively, so neither can be 23⁷ = 3 404 825 447 (calculated above) and both must be 5th powers. 60ac and 61ac have already used 73⁵ and 71⁵, so 80ac and 81ac are from {67⁵ = 1 350 125 107, 71⁵ = 1 804 229 351, 73⁵ = 2 073 071 593, 79⁵ = 3 077 056 399, 83⁵ = 3 939 040 643, 89⁵ = 5 584 059 449, 97⁵ = 8 587 340 257}. The only one with 1 as the 5th digit is 67⁵, so 81ac = 1 350 125 107.

Now the lowest across answer 20ac has to be < 1 350 125 107, and as a cube ending in 2 its cube root must end in 8. The root is ≤ 1 350 125 112^(1/3) ≈ 1105.2, so it’s one of {1008, 1018, 1028, …, 1098}. Of those, the only one with a cube that has 1 as the 6th digit (to fit with 15up) is 20ac = 1 287 913 472 = 1088².

Of the two options for 21ac, if it’s 3 077 056 400 then 42dn is 2302, and we would have 36 even digits and 48 odd digits (including the 5 and 9 not yet placed in the game path cells 5 and 10) in the grid so far, putting the possible total number of even digits in the range [36, 52], so we’d need 2302/2 = 1151 to be divisible by something in the range [18, 26], which it isn’t. Therefore 21ac = 5 584 059 450 and 42dn = 2352 = 2⁴×3×7². The possible count of even digits is now in the range [35, 51] and the only factors of 2352 in that range are {42 = 2×3×7, 48 = 2⁴×3, 49 = 7²}.

94dn is 75_ in the grid and must be a multiple of one of {42, 48, 49}. If it’s 49n then n is between 750/49 ≈ 15.3 and 759/49 ≈ 15.5, not possible. If it’s 48n then n is between 750/48 ≈ 15.6 and 759/48 ≈ 15.8, again not possible. So it’s 42n, with n between 750/42 ≈ 17.9 and 759/42 ≈ 18.1: 42×18 = 756 works, so 94dn = 756 and there should be 42 even digits in the completed grid.

For 80ac we now have ____0_6___; 23⁷ = 3 404 825 447 doesn’t fit, so it’s a 5th power of a prime > 63. Excluding the ones already used, the options are {79⁵ = 3 077 056 399, 83⁵ = 3 939 040 643, 89⁵ = 5 584 059 449, 97⁵ = 8 587 340 257} and the only one that fits the grid is 80ac = 3 077 056 399.

Finally, the highest entry 1ac is a cube n³ > 41ac = 2096³ and has to match either 9___5____9 or 9___9____5. Its root n < 9 999 999 999^(1/3) ≈ 2154.4, so n is in the range [2097, 2154]. If 1ac is 9___9____5 then n ends in 5 and the options are {2105³ = 9 327 307 625, 2115³ = 9 460 870 875, 2125³ = 9 595 703 125, 2135³ = 9 731 810 375, 2145³ = 9 869 198 625}, but none of those has a 9 in 5th place. So 1ac is 9___5____9, n ends in 9 and the options are {2099³ = 9 247 776 299, 2109³ = 9 380 581 029, 2119³ = 9 514 651 159, 2129³ = 9 649 992 689, 2139³ = 9 786 611 619, 2149³ = 9 924 513 949}, of which the 5th digit is 5 only in {9 380 581 029, 9 924 513 949}. Without 1ac completed, we have 37 even digits in the grid, so we need 1ac to contain five even digits, which is true only for 1ac = 9 380 581 029. As a final check, the sum of all digits in the grid is 414, twice the 207 at 38up, as expected.