This is one possible solution path. √66 = √2×√3×√11 ≈ 8.1 isn’t a rational number, meaning there’s no integer value of H that can make H√66 an integer. Multiplying out (B + C√66)(E + F√66) gives (BE + 66CF) + (BF + CE)√66 = G + H√66; the integer terms must be equal, so BE + 66CF = G. Having no zeros or repeated digits, all the BCEF and G values given by clues are at least 1234 and at most 9876.

The clues for y and t involve the square roots of I and D, so to give an integer (rational) result they must be square digits, ie two of {1, 4, 9}, and the product DI is one of {4, 9, 36}. Clue p = (DI − NO)(S + T)A + R means (DI − NO) = (p − R)/A/(S + T), for which the minimum is (1234 − 6)/9/(8 + 7) ≈ 9.1. So the minimum for DI is 10 + 1×2 = 12, which means DI = 36 with {D, I} = {4, 9} in some order, and NO ≤ 26.

The digit 1 can’t be T, because then the maximum for clue k would be (6 + 4 + 1^A + 8×7 − 2 − 3)9 = 558, too small; if A is 1 then k ≤ (6 + 4 + 5¹ + 8×7 − 2 − 3)9 = 594, still too small. If R is 1 then in clue o the term (1^O − √36) makes the answer negative; if O is 1 then o ≤ (6 + 5)7(8¹ − √36) = 154, also too small. If U is 1 then p ≤ 3×1^T + (7 + 6 + 9)8 + 5 = 184, too small. If S is 1 then u ≤ 8(1^T + 7×6 + 5) + 9 = 393, too small. The only letter left is N = 1, which leaves 235678 for the letters AORSTU in some order.

From clue n we have (1 + I)^T = n/(US) + (A + R)D. If I is 4, the maximum for the right-hand side is 9876/(2×3) + (7 + 8)9 = 1781 but that would leave T ≥ 5 and 5⁵ = 3125 is too big, so T is one of {2, 3}. (If I is 9, the maximum is a bit lower and we have 10^T, and 10⁵ = 100000 is even bigger.) If T is 2, then q < 8(8² + 8×8 + 1) + 9 = 1041, too small; therefore T = 3. Clue c contains T^I added and multiplied with other things; 3⁹ = 19683 would make c far too big, so I = 4 and D = 9.

Now one of the remaining letters AORSU is 2. If it’s A then k < (8 + 4 + 3² + 8×8 − 5 − 1)9 = 711, which is too small. If it’s S then u < 8(2³ + 8×8 + 1) + 9 = 593, also too small. If it’s U then p < 8×2³ + (8 + 8 + 4)8 + 1 = 225, too small again. If it’s O then o is (A + S)3(R² − 6), with ASR from the set 5678; if R is 5 the maximum possible is (8 + 7)3(5² − 6) = 855, too small; if R is 6 then 6² − 6 = 30 would make o end in 0; the remaining options are {(5 + 6)3(7² − 6) = 1419, (5 + 8)3(7² − 6) = 1677, (6 + 8)3(7² − 6) = 1806, (5 + 6)3(8² − 6) = 1914, (5 + 7)3(8² − 6) = 2088, (6 + 7)3(8² − 6) = 2262}, all of which have a repeated digit or 0. Therefore O isn’t 2, which makes R = 2.

We now have AOSU = 5678 in some order. For d we have 36(1^O×S + A + U)2 = 72(S + A + U) = 72(26 − O); the options are {72(26 − 5) = 1512, 72(26 − 6) = 1440, 72(26 − 7) = 1368, 72(26 − 8) = 1296}, and to avoid a repeated digit O must be one of {7, 8}. Also, e = (2AOU + 1)9 = (2×1680/S + 1)9 and the options are {(3360/5 + 1)9 = 6057, (3360/6 + 1)9 = 5049, (3360/7 + 1)9 = 4329, (3360/8 + 1)9 = 3789}, so S must also be one of {7, 8}. That leaves {A, U} = {5, 6}. Clue n is now U((1 + 4)³ − (A + 2)9)S = US(107 − 9A) and the four options for (A, U, S) are {(5, 6, 7), (5, 6, 8), (6, 5, 7), (6, 5, 8)}, giving the corresponding values {2604, 2976, 1855, 2120} for n. The only one without 0 or a repeat is 2976, making A = 5, U = 6 and S = 8, which only leaves O = 7.

We can now calculate all the clue values, as shown here. For the grid entries, they need to be converted from BCEF values to G values or vice versa, using G = BE + 66CF, where the highest possible value for BE or CF is 72 = 8×9.

Starting with the lowest value, d = 1368, if we try converting it from BCEF to G we get 1×6 + 66×3×8 = 1590, which is invalid because it contains 0; so 1368 must be a G. Its CF = (1368 − BE)/66 must be between (1368 − 8×9)/66 ≈ 19.6 and (1368 − 1×2)/66 ≈ 20.7, which only allows 20, so {C, F} = {4, 5} (in either order). Then BE = 1368 − 66×4×5 = 48 and {B, E} = {6, 8}, so the entry for d is one of {6485, 6584, 8465, 8564}.

If b = 1538 is a G value then its CF is between (1538 − 8×9)/66 ≈ 22.2 and (1538 − 1×2)/66 ≈ 23.3, but 23 isn’t the product of any two digits. So 1538 is a BCEF value and we can enter b = 2643 = 1×3 + 66×5×8 in the grid, which narrows the d entry to {6485, 6584}.

There are six other clue answers that can’t be G values, as follows: j = 1947 would need CF between 28.4 and 29.5 but 29 isn’t the product of two digits, f = 3659 would need CF = 55, c = 3687 would need CF = 55, e = 3789 would need CF = 57, i = 5987 would need CF = 90; finally, s = 1597 needs CF = 24, but then BE = 1597 − 66×24 = 13, not the product of two digits. These are therefore all BCEF values, so we can calculate their G values and enter c = 2796, e = 4182, f = 3579, i = 4198, j = 4162 and s = 2319 in the grid. With b, these are the seven BCEF answers mentioned in the puzzle’s preamble, so the rest are all G values.

For the remaining clue answers, the lower and upper bounds of CF are (G − 8×9)/66 and (G − 1×2)/66 respectively, and in each case they allow either a single integer or a consecutive pair of which only one can be the product of two digits. Knowing the CF product for each answer, the BE products can also be calculated; these are listed in the second table here.

From that, we can see that n’s BE = 6, which agrees with the first and third digits 1 and 6 we have in the grid; the other two digits have a product (CF) of 45, so they’re 5 and 9; from above (and the table) we know that d’s second and fourth digits are 4 and 5, so 5 goes where n and d cross, giving entries of d = 6485 and n = 1965.

From the grid we have k = 87__ and from the table its BE = 48 = 6×8 and CF = 35 = 5×7, so k = 8765. Similarly, o = 29__ with BE = 6 and CF = 72, so o = 2938.

The entry for m is 589_ with CF = 32, so m = 5894. Now w ends in 4 with CF = 24, so its second digit is 6, and to make BE = 40 it’s one of {5684, 8654}. u = 36__ with BE = 12 and CF = 42, so u = 3647. t starts with 2 and has BE = 2 and CF = 30, so it’s 2_1_ with 5 and 6 as the other digits; but the first digit of w can’t be 6, so t = 2516 and w = 5684.

Now y = __67 with BE = 48 and CF = 28, so y = 8467. p starts with 6 and has BE = 12 and CF = 27, so it’s one of {6329, 6923}. Then for q we have 2__8 with BE = 10 and CF = 24, so q = 2358. r ends in 4 and has CF = 28, so its second digit is 7; that makes v = 2_37 and to make its C = 35 it must be v = 2537.

For h we have __7415__, which represents G = 15__ = 7B + 66×4C. If C is 5, the second term is 66×485 = 1320, too far away to reach 1500 by adding the maximum 7B value of 63; if C is 7, we get 1848, too far away from 1599 on the high side. Therefore C = 6 and G is at least 7×1 + 66×4×6 = 1591, but that has a repeated digit; if B is 3 then 7×3 + 66×4×6 = 1605, which doesn’t match the second digit of G, so B is 2, making h = 2674 1598.

Now g is _96318__, giving G = 18__ = 6B + 66×9×3. The minimum for B is (1823 − 1782)/6 ≈ 6.8, so B is one of {7, 8} (not 9 because that would repeat a digit in the BCEF value). It can’t be 8 because 6×8 + 66×9×3 = 1830 has a 0, so B = 7 and g = 7963 1824.

For x we now have __5_142_, giving G = 142_ = 5B + 66CF. CF is between (1423 − 5×9)/66 ≈ 20.9 and (1429 − 5×1)/66 ≈ 21.6, ie CF = 21 and the second and fourth digits are {3, 7}. But r has BE = 27 with its third digit being one of {3, 9}, so now we know r = 9734, which resolves p = 6329. x is now narrowed to _753142_, with B between (1423 − 66×21)/5 = 7.4 and (1429 − 66×21)/5 = 8.6; the only possibility is B = 8, making x = 8753 1426.

The only remaining grid entry is a, matching _619357_, giving G = 357_ = 1B + 66×6×9 = B + 3564. B is at least 3571 − 3564 = 7 and it can’t be 9 because that would repeat a digit in the BCEF part. So B is one of {7, 8}, giving {76193571, 86193572} respectively for a.

Looking for a common feature shared by a, g, h and x, we may notice that the latter three have no repeated digits, and only one of the options for a shares that feature. If that’s what’s required, for the final step we need to find a fifth BCEF-G pair with eight different digits. The preamble says that the G part is greater than the ones we have so far, ie it’s > 3572. But the upper bound for G values is 7×6 + 66×9×8 = 4794. The CF we need is therefore between (3574 − 8×9)/66 ≈ 53.1 and (4793 − 1×2)/66 ≈ 72.6. The numbers in that range that are the product of two digits are {54 = 6×9, 56 = 7×8, 63 = 7×9, 72 = 8×9}. If CF is 54 then G = BE + 66×54 = BE + 3564; because 6 and 9 are the digits CF, the maximum for BE is 7×8 = 56, so G ≤ 3564 + 56 = 3620 and, to avoid repeating 6 or 9, G is in the range [3574, 3587], with BE correspondingly in the range [10, 23], being the product of two digits not including 3 or 5 (which are in G) or 6 or 9 (which are C and F). The only such values are {14 = 2×7, 16 = 2×8}, but 2×7 + 66×6×9 = 3578 repeats 7 and 2×8 + 66×6×9 = 3580 repeats 8 (and has a disallowed 0); therefore CF isn’t 54.

If CF is 56 then G = BE + 3696; because 7 and 8 are the digits CF, the maximum for BE is 9×6 = 54, so G ≤ 3696 + 54 = 3750. G is ≥ 3696 + 1×2 = 3698 but in the range [3698, 3750] the only value that doesn’t contain 7 or 8 or is 3699, with a repeated 9; therefore CF isn’t 56.

If CF is 72 then G = BE + 4752; because 8 and 9 are the digits CF, the maximum for BE is 7×6 = 42, so G ≤ 4752 + 42 = 4794, but because it can’t repeat 9, 8 or 7 its upper bound is 4765. G is ≥ 4752 + 1×2 = 4754, so it’s in the range [4754, 4765]; the only values in that range that don’t contain 8, 9, 0 or a repeated digit are {4756, 4761, 4762, 4763, 4765} with corresponding BE values of {4, 9, 10, 11, 13}, from which we can eliminate 4 = 1×4 (we have a 4 in G), 9 = 1×9 (either C or F is 9) and the primes 11 and 13. If BE is 10 = 2×5 then G = 4762, which repeats a 2; therefore CF isn’t 72.

So we’re left with CF = 63 = 7×9, making G = BE + 4158. The maximum available for BE is 6×8 = 48, so G ≤ 4158 + 48 = 4206, but it’s not allowed to contain 0 (or 7 or 9), so the maximum comes down to 4186; BE can’t include 1 (or 4) because that would be repeated in G, so the minimum for G is 4158 + 2×3 = 4164. The only values in the range [4164, 4186] without 0, 7, 9 or repeats are {4165, 4168, 4182, 4183, 4185, 4186}, with corresponding BE values of {7, 10, 24, 25, 27, 28}. We can eliminate 7 = 1×7, 25 = 5×5, 27 = 3×9 and 28 = 4×7 because they repeat digits. If BE is 24 we have G = 24 + 4158 = 4182, but then either 3×8 or 4×6 for BE would repeat a digit in G. Therefore BE = 10 = 2×5 and G = 4168. There are four permutations of the BE and CF digits, namely {2759, 2957, 5729, 5927}, but the preamble says we want the smallest one, so the required numbers are 2759 and 4168. Converting those to the letters used in clues gives ROAD INUS (“in US”), a description of the famous Route 66 hinted at by the puzzle’s title.