Listener Crossword 4151: Solution Notes

Back to 2011 menu

Number or Nummer by Ruslan

Puzzle explanation

In the first grid, the values of letters are A = 11, B = 7, D = 19, E = 23, F = 13, I = 29, L = 3, S = 17, T = 5, Z = 2, and the answers transcribed in German are 13ac, 16ac, 1dn and 5dn. In the second grid, the letter values are A = 11, B = 17, D = 19, E = 7, F = 13, I = 29, L = 23, S = 5, T = 3, Z = 2, and the entries in German are 8ac and 9dn.

The footnote is intended to reveal that the claim in the preamble about the origins of the puzzle was nonsense.

The clues included references to German newspapers, such as Die ZEIT, FAZ (Frankfurter Allgemeine Zeitung), BILD, SDZ (Süddeutsche Zeitung) and BLATTES (journal).

Puzzle solution process

It’s useful to draw up a table to list the possibilities for cross-checking digits entered in English and German (G).

Digit Letter Cross-checking digit
0Z, N0, 2, 9if 0 crosses 2, 2 is G; if 0 crosses 9, 0 is G
1O, E1, 8 if 1 crosses 8, 1 is G
2T, Z0, 2, 3if 2 crosses 0, 2 is G
3T, D2, 3  
4F, V4, 5  
5F 4, 5  
6S 6, 7  
7S 6, 7  
8E, A1, 8 if 8 crosses 1, 1 is G
9N 0, 9 if 9 crosses 0, 0 is G

Any entry that contains O or T must be in English; any entry containing D, V or A must be in German.

One possible solution path is as follows.

Since 24ac = D² and 19dn = A² each have 3 digits, D ≥ 11 and A ≥ 11. All squares of odd primes end in 1 or 9 and those two digits can’t cross-check, so 24ac and 19dn must end with the same digit. 15dn = BD² = B × 24ac, which has 4 digits, and the letter for its last digit is the same as the letter for the middle digit of 24ac. This leads to the following possibilities for D, B and A:

D 11 13 17 19 23 29
24ac = D² 121 169 289 361 529 841
15dn ends 0, 2, 3 6, 7 1, 8 6, 7 0, 2, 3 4, 5
Possible B from 15dn 13, 23 23 19, 29 7, 17 7, 17 5
Possible A from 19dn 19, 29 17 13, 23 11, 29 13, 1711, 19

21ac = ABI has 4 digits and its last letter is the same as the middle letter of 24ac. Checking the possibilities for D, A, B and I, we get:

D A B 21ac ends AB Possible I BD² ABI
11 19 13 6, 7 247    
11 19 23 6, 7 437    
11 29 13 4, 5 377 5 1573 1885 (x)
11 29 23 4, 5 667 5 2783 3335 (x)
13 17 23 1, 8 391 11 3887 4301 (x)
17 13 19 6, 7 247 11 5491 2717 (x)
17 13 29 6, 7 377 11 8381 4147 (x)
17 23 19 0, 2, 3 437    
17 23 29 0, 2, 3 667    
19 11 7 0, 2, 3 77 29 2527 2233
19 11 17 0, 2, 3 187 29 6137 5423
19 29 7 4, 5 203 5 2527 1015 (x)
19 29 17 4, 5 493 5 6137 2465 (x)
23 13 7 6, 7 91 17 3703 1547 (x)
23 13 17 6, 7 221 7 8993 1547 (x)
23 17 7 1, 8 119 19, 29 3703 2261, 3451 (x)
29 11 5 0, 2, 3 55    
29 19 5 6, 7 95    

In all but two cases, the third digits of 21ac and 15dn are incompatible for cross-checking (marked with (x) above), so the A = 11, D = 19, I = 29 and B = 7 or 17. The two possible values for ABI are 2233 and 5423 and in both cases we have a 2 cross-checking a 3, which can only happen with both 21ac and 15dn entered in English. Since 21ac starts with a T in the first grid, we can enter TTTT for 2233 there, as well as TFTS (2527) for 15dn, and we know that B = 7 in that solution. In the other grid, we’re told that 21ac doesn’t start with T, so 21ac must be FFTT (5423) there, with 15dn as SOTS (6137) and B = 17 in that solution.

19dn = 121 has a middle letter of T in both grids, so it must be entered as OTO (English) in both places. 24ac = 361 is _SO in both grids, so it must be entered as TSO (English) in both places.

B = 7 (first grid) B = 17 (second grid)

22dn = B + A + L + Z = 18 + L + Z. It’s entered as TT, for one of {22, 23, 32, 33}, so L + Z is one of {4, 5, 14, 15}. The only available primes making those sums are {L, Z} = {2, 3} or {2, 13} in either order.

22dn = B + A + L + Z = 28 + L + Z. It’s entered as FT, for one of {42, 43, 52, 53}, so L + Z is one of {14, 15, 24, 25}. The only available primes making those sums are {L, Z} = {2, 13} or {2, 23} in either order.

18ac = I − B − L + ZA = 22 − L + 11Z. It’s entered as FO, for one of {41, 51}, so 11Z − L is one of {19, 29}.
Therefore Z = 2 and L = 3.

18ac = I − B − L + ZA = 12 − L + 11Z. It’s entered as OO, for 11, so 11Z − L = −1.
Therefore Z = 2 and L = 23.

14ac = 637 and its middle letter is T, so it’s entered in English as STS.

14ac = 767 and its middle letter is S, so it’s entered as SSS (in either language).

10ac = 65, entered as SF (in either language).

10ac = 75, entered as SF (in either language).

6dn = 873 − 6(F + T), entered as _S_ in both grids, so its middle digit is 6 or 7 (in either language). The available prime values give:

{F, T}F + T6dn
{5, 13} 18 765
{5, 17} 22 741
{5, 23} 28 705
{13, 17}30 693
{13, 23}36 657
{17, 23}40 633

{F, T} = {5, 13} in either order; 6dn = 765, entered as SSF.

{F, T}F + T6dn
{3, 5} 8 825
{3, 7} 10 813
{3, 13} 16 777
{5, 7} 12 801
{5, 13} 18 765
{7, 13} 20 753

{F, T} = {3, 13} or {5, 13} in either order; 6dn = 777 (SSS) or 765 (SSF).

9dn = 13021 − (F + T) = 13003, to end with T, so it must be entered as OTZZT (English).

9dn = 13011 − (F + T) = 12995 or 12993, to end with F, so it must be 12995 (OTNNF or EZNNF), {F, T} = {3, 13} in either order and 6dn = 777 (SSS).

8ac = 10471 − 219(B + L) = 8281, to end with O, so it must be entered as ETEO (English).

8ac = 10471 − 219(B + L) = 1711, entered as OSOO (English) or ESEE (German).

7dn = 40(F + T) + 3826 = 4546, entered as FFFS (English) or VFVS (German).

7dn = 40(F + T) + 3836 = 4476, entered as FFSS (English).

4ac = (58 + F + T)² − 2 = 5774 (FSSF or FSSV).

4ac = (58 + F + T)² − 2 = 5474 (FFSF).

5dn = (22(F + T) − 115)² + 7 = 78968 (SENSE or SANSA).

5dn = (22(F + T) − 115)² + 7 = 56176 (FSOSS or FSESS).

12ac = ((F + T)(F + T - 2))² = 82944 to match _TNF_, so it’s entered in English as ETNFF. That makes 7dn = _FFS, so it must be FFFS, making 4ac FSSF (both English).

12ac = ((F + T)(F + T - 2))² = 50176 (FZOSS or FNESS). The second letter is T or Z from 9dn, so 12ac = FZOSS (English), 9dn = EZNNF (German) and 8ac = ESEE (German).

12dn = (28(F + T) − 206)² = 88804, starting with E, so it’s entered as EEEZF (English).

12dn = (28(F + T) − 206)² = 58564, entered as FEFSF (English) or FAFSV (German).

3dn = A + B = 18, ending in E, so it’s entered as OE (English).

3dn = A + B = 28, ending in E, so it’s entered as TE (English).

1ac = 939 − (F + T) = 921, ending in O, so it’s entered as NTO (English).

1ac = 939 − (F + T) = 923, ending in T, so it’s entered as NTT (English).

1dn = 870 + 7B = 919, with E as the middle letter, so it’s entered as NEN (German).

1dn = 870 + 7B = 989, with E as the middle letter, so it’s entered as NEN (English).

2dn = 2358, entered as TT__, so it must be TTFE (English).

2dn = 2658, entered as TS__, so it must be TSFE (English).

11ac is NF in both grids, so it must be either 94 (English) or 95 (English or German); since it’s the sum of 7 odd numbers, it must be 95. It’s the sum of all ten of the primes except D, F and Z, so 95 = 108 - F, making F = 13.

We now have T = 5, which leaves {E, S} = {17, 23} in some order.

We now have T = 3, which leaves {E, S} = {5, 7} in some order.

13ac = 632 − 11(S + 5)/2, entered as _EE. If S is 23 we get 478, which doesn’t fit, so the entry is FEE (German) for 511, with S = 17. Then the last letter assignment is E = 23.

13ac = 632 − 11(S + 3)/2, entered as _E_. If S is 7 we get 577, which doesn’t fit, so the entry is FEE (English) for 588, with S = 5. Then the last letter assignment is E = 7. 12dn is now fixed as FEFSF (English).

23ac = (F + 2I)² = 5041, entered as FZFO (English) in both grids, because 12dn ends with F, so 13dn is entered as F__F and 17dn ends with Z in both grids.

13dn = 5044, entered as FZFF (English).

13dn = 4784, entered as FSEF (English).

17dn = 980, entered as NEZ (English).

17dn = 780, entered as SEZ (English). The second grid is now complete.

20ac = 58, which agrees with FE (English).

20ac = 88, which agrees with EE (English).

16ac = 29128; to match ZNEZ_ it must be entered as ZNEZA (German). That means 5dn = 78968 must be SANSA (German). The first grid is now complete.

16ac = 66496, which agrees with SSFNS (English) in the grid.

We now have two completed grids, in which the letter values are as follows:

A11
B7
D19
E23
F13
I29
L3
S17
T5
Z2
A11
B17
D19
E7
F13
I29
L23
S5
T3
Z2

The entries transcribed in German and their English counterparts (ignoring entries which would be the same in English and German) are as follows:

Clue Value German English
13ac 511 FEE FOO
16ac 29128 ZNEZA TNOTE
1dn 919 NEN NON
5dn 78968 SANSA SENSE

The English versions make FOOTNOTE and NONSENSE.

Clue Value German English
8ac 1711 ESEE OSOO
9dn 12995 EZNNF OTNNF

The English versions don’t make real words.

Back to 2011 menu