## Digimix by Oyler

### Detailed Solution

1. All Q start with either a 1 or a 2. The largest X_Y_Z is 987654321 and its square root is 31426.96805. So the smallest Q that we can use is 31245 and the smallest P associated is 6789. Squaring both of these and adding gives a 10 digit result. A smaller result can be obtained with larger values of Q and smaller values of P, but all such start with at least 998.
3. The way in is via the h, 5h in the same clue along with 6h. Only odd numbers need he considered for h since if h is even 5h ends in 0, so 5h ends in 5. Further, h ending in 5 can be discounted. 6h limits the calculation to odd numbers between 123 and 163 inclusive. Omitting repeat digits and those containing a 5 since 5h must have the 5 we have 123, 127, 129, 137, 139, 143, 147, 149, 163. Multiplying by 5 and eliminating digits in h already gives 127, 129 and 137 as possible choices for h. Multiplying these by 6 eliminates 129 and 137 so h = 127, 5h is 635, 6h is 762 and h′ is 721.
4. E must start with a 2 since D contains 1. As D ends in a 1 which when squared gives 1 we need the square of E to end in a 1 also so that the sum ends in a 2 from 6h = 762. So E ends in a 9.
5. m contains 4, 8 and 9. From 5G, G must be odd so G = 29 and 5G is 145.
6. From 6J² and 7J², J = 11. 6J² is 726 and 7J² is 847.
7. From jx, x ≤ 32 and from 5Gx, x is odd. In addition jx cannot have a 3 in it from 9K, which ends in 3. Using 5Gx, x can be 15, 17, 25 or 27. We can discount x as 15 by considering jx, 911 x 15 is 13*** and 919 x 15 is 13***.
8. 5Gx can be 2465, 3625 or 3915. 9T must be a multiple of 9 and so its digits must sum to a multiple of 9. 9T contains {1, 3, 7, 8, 9}, {1, 4, 7, 8, 9} or {2, 4, 6, 7, 8}. These digits sum to 28, 29 and 27 respectively. Only 27 is a multiple of 9 so 5Gx is 3915 and x = 27. So S starts with a 1. Jx is 297 and 9T starts with a 2.
9. 5(G′ − W) cannot contain 1, 2 or 7 and W must be odd and ≤ 29, else this starts 1 or 2 or contains 0. In fact W can be 13, 15, 19 or 23. From 4v we can eliminate W as 15 as 4v would end in 0. The remaining three values of 5(G′ − W) are all of the form 3?5 with ? 4, 6 or 9.
10. 3Jf is 33f so f starts with a 1 or 2. Taking the smallest and largest 9T of 24678 and 28764 with 3915, squaring and adding gives 3Jf 624 to 842. So f is 23, 24 or 25 (smaller values are ruled out by the effect on E) with 3Jf as 759, 792 or 825 respectively. The maximum for T starting with a 2 is 2999 so 9T would be 26991, squaring and adding 3915² gives 743 for 3Jf which is less than 759 so T starts with a 3. The minimum T starting with 3 is 3111 so 9T would be 27999. Applying the same procedure gives 3Jf as 799 which is more than 792 so we can conclude that f = 25 and 3Jf is 825.
11. We can now home in on T as 3162 or 3163. It must be the latter from 2p (which is even) and 5Gx (which is odd). T = 3163. 9T is 28467. R + k + y′ is 697 and 2p is 314 so p = 157 and 5p is 785.
12. P when squared cannot contain a 5, 7 or 8 or repeats so P can be 13, 14, 18, 19 or 31.
13. b and g do not end in 1, 2, 5 or 9 (from D and E, which occur as a P/Q pair).
14. v starts with a 1 or 2, since 4v is three digit.
15. f(M + s + z), M ends in 5 and s in 3 so z must be odd as f is 25, so z = 61 since W starts with 1 or 2 from above. f(M + s + z) ends in 5 as does its square, so the square of 3Gn must end in 6 to give 721 ending when summed, so 3Gn ends in 4 or 6. 3G is 87 so n ends in 2 or 8.
16. Consider CH′: the smallest C can be is 211 and the maximum for the product is 9876. Dividing gives 12 ≤ H′ ≤ 46 so n starts with a 1, 2, 3 or 4.
17. There are 8 possible values for n and testing each of these for 3Gn gives n as 28, 32 or 48. It cannot be 42 as that has a 5 in it and the Q part of the clue has that. 3Gn can be 2436, 2784 or 4176.
18. 5(G′ − W) starts with a 3. f(M + s + z) must start with a 1 since the smallest Q starting with 2 is 21345 which when squared is 455… and thus too big. 3Gn cannot contain a 1 which eliminates n as 48 so f(M + s + z) cannot contain a 2 or 4 so must end in 75 as it is a multiple of 25. Therefore n = 28 with 3Gn as 2436.
19. Taking the maximum for M with s starting with 1 then 2 etc, along with z as 61 shows that (185 + 193 + 61) and (285 + 293 + 61) when multiplied by 25, squared and 2436² added are not big enough whilst (485 + 413 + 61) is too big, so M = 385. Searching through the 9 possible values for s gives s = 313 and f(M + s + z) as 18975. Thus 5(G′ − W) is 365 and b(G + d) is 984. So W = 19 and v = 239 (139 doesn’t work) with 4v as 956. Since b(G + d) is 984 and its factorisation is 2.2.2.3.41 and G + d ends in a 1 it must be 41 so d = 12 and b = 24.
20. 3fs is 23475 and m is 948 or 984 with m′ as 849 or 489. We can check both of these to determine the **** since the Q, X and Y are known. This gives 9168 with m′ as 849 so m = 948.
21. Q = 81 so 3Q is 243.
22. P(A + J). As the smallest A + J is 33 (d is 12, so A cannot be), P cannot now be 31 or 19 which is W so P is 13, 14 or 18. As q contains a 1, P(A + J) cannot. Testing possible values for A with P gives A as 22 or 42. Using 8Ar with r taking its least value of 21 eliminates A as 42 so A = 22 and P(A + J) must contain a 4.
23. a doesn’t contain 3, 5 or 8 so D starts with 6 or 7.
24. F contains 1, 7 and 8. It cannot start with 8 as a doesn’t contain 8 and it doesn’t end in 1 since c cannot contain a 1 because q does. R + k + y′ is 697 so 1 is the middle digit of F (the other possibilities are too large), hence F = 718. F′ is 817 and D starts with a 6. a = 267.
25. c ends in 8 and 9K ends in 3. The square of 9K ends in 9. The square of jx must also end in 9 in order to give an 8 as the last digit of c. Since 9K has the 3, jx must end in 7 so j = 911 and jx is 24597.
26. From 8Ar, r is 41 or 51.
27. b + e + p contains 1, 4 and 9 so can be 149, 194, 419, 491, 914 or 941. b + p is 181 which rules out the first two. Subtracting 181 from 491 or 941 give an entry containing a 0 which is forbidden. So e is 238 or 733. E can be 23579 or 23589. Testing these with the corresponding value for D gives D = 6471 and E = 23589. w(J + d)/2 is 598 and u′ is 314 so u = 413, w = 52 and R = 472.
28. Neither Pg nor P² can contain 5, 7 or 8. Testing the possible value for P (13, 14, 18) with those for g yields only one solution P = 13 and g = 18. So Pg is 234, P² is 169 and P(A + J) is 429.
29. Testing the possible values for 9K (K is ?47 or ?57) and eliminating those that contain a 2, 4, 5, 7 or 9 yields 757 or 957 as possible values for K so r = 51 and 8Ar is 8976. With 9K as 6813 or 8613 we can quickly see which gives a zero-less pandigital when squared. It is the former so K = 757, 9K is 6813, q = 651 and c = 378.
30. 3Nz′ is 48N and N is 57 or 58 (other values are ruled out, eg, w=52 already) giving 2736 or 2784. Both of which when squared end in 6. Using the R + k + y′ total of 697 with k 177 or 178 gives y′ of 47 or 48. U ends in 7 so the Uy′ can end in 9 or 6 which when squared gives 1 and 6. The sum of the squares ends in 7 so we want Uy′ that ends in 9 so y′ is 47 which means N = 58, k = 178 and y = 74. 3Nz′ is 2784 and knowing all the rest of the clue allows us to find Uy′ which is 15369 so U = 327.
31. 9H′y/2 will give H as the X_Y_Z is known and testing the 2 possible values for H (using the possible endings for e) which when squared and subtracted will yield the answer. We get 7659 and 9324 for 23 and 28 respectively. Only the former works so H′ is 23 and the ***** is 13248. H = 32, e = 733 and b + e + p is 914.
32. de is 8796.
33. CH′ yields C = 216 with CH′ as 4968 and S = 15237. The unknown bits in this clue can now be calculated to be 256 and 193 so B = 391.
34. V(g + x) ends in 5 as does its square so the square of 9(t − y) must end in 9 so 9(t − y) ends in 3 or 7 so t = 337 and 9(t − y) is 2367. So V = 433 with V(g + x) as 19485.
35. The unknown bits in the 8Ar clue are found most easily by looking at the Y which contains 3, 8 and 9. Trying the 6 possible combinations yields 389 as the answer coupled with 25413.
36. Rather than work through 720 possible combinations for the X_Y_Z part of the last clue it is more sensible to look at the 5 digit part. 8796 when squared ends in a 6 so the square of the 5 digit part must end in 1 to give the 297 ending. This can only come from a 1 or 9 and since the 9 is already present from de it must end in 1 and so must start with 2 So it is 2???1 with the missing digits 3, 4 and 5 in some order. Testing each of the 6 possibilities in turn yields only one solution 23541 with 631 and 548.
37. The missing entries are 9168, 25413, 389, 13248, 256, 23541, 631 and 548 which when summed give 73194. This number has no significance whatsoever (apart from having five different digits). As implied in the preamble, it merely serves to show solvers have deduced the asterisked items.

The values of the grid entries are thus as follows:

Entry Value Entry Value Entry Value Entry Value
A 22 M 385 a 267 p 157
B 391 N 58 b 24 q 651
C 216 P 13 c 378 r 51
D 6471 Q 81 d 12 s 313
E 23589 R 472 e 733 t 337
F 718 S 15237 f 25 u 413
G 29 T 3163 g 18 v 239
H 32 U 327 h 127 w 52
J 11 V 433 j 911 x 27
K 757 W 19 k 178 y 74
m 948 z 61
n 28

The P, Q, X, Y and Z values are as follows:

P Q X Y Z
9168 23475 635 127 849
2436 18975 365 984 721
2784 15369 243 956 817
3915 28467 825 697 314
8976 25413 726 389 145
2367 19485 385 267 914
7659 13248 234 169 785
6813 24597 651 429 378
4968 15237 256 847 193
6471 23589 598 314 762
8796 23541 631 548 297