Listener Crossword 4270: Solution Notes

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Alma Mater by Oyler

Puzzle explanation

This puzzle celebrated the 600th anniversary of the University of St Andrews, the setter’s alma mater. The puzzle was published on St Andrew’s day, and the clue letters spelt ST ANDREW in order of increasing numeric values. The unclued grid entries are the years 1413 and 2013, duplicated on the major diagonals as a representation St Andrew’s cross.

The University of St Andrews began as a school of higher studies in May 1410, which obtained a charter of incorporation and privileges from the Bishop of St Andrews in February 1411. Pope Benedict XIII conferred full University status in August 1413, in six papal bulls that arrived in St Andrews in February 1414 after a five-month journey from the pope’s exile in Spain. The university’s sexcentenary celebrations began in 2011 and culminated in November 2013.

Puzzle solution process

This is one possible logical path to a solution. The maximum sum of seven two-digit primes is 97 + 89 + 83 + 79 + 73 + 71 + 67 = 559, so none in the set of eight can be less than 600 − 559 = 41.

23ac = S and 15dn = AS end in the same digit, so A is a prime ending in 1, ie, one of {41, 61, 71}, and 12dn = AA is one of {1681, 3721, 5041}.

18ac is equivalent to 600 − W − 2E + D, for which the minimum is 600 − 89 − 2×97 + 41 = 358 and the maximum is 600 − 43 − 2×41 + 97 = 572. The only value for 12dn that fits is 5041, so A = 71.

12ac = W − S is now 5_, so W = 97 and S is one of {41, 43, 47}. 15dn = AS is then one of {2911, 3053, 3337}. But 14ac = E + A − T − S is even, so 15dn must be 2911 and S = 41 (which means the other values are the seven largest primes). We can also enter 23ac = S = 41, 12ac = W − S = 56 and 27dn = A − S = 30.

From 16ac and 3ac we know that E > D > A, and 21ac = W − D starts with 1, D is one of {79, 83} and E is one of {83, 89}. 17ac = S + E + E ends in 9, so E = 89 and we can enter 17ac = 219, 1dn = W + E = 186 and 19dn = 600 − W − E = 414. Then 5ac = T + E − A starts with 8, so T = 67 and we can enter 5ac = 85, 14ac = E + A − T − S = 52, 4dn = W + E + T = 253 and 26dn = 600 − T = 533.

3ac = (D − A)D now ends in 2, so D can’t be 83: D = 79 and we can enter 3ac = 632, 16ac = E − D = 10, 20ac = D = 79, 21ac = W − D = 18, 25ac = A + D − E = 61 and 13dn = E + A − S − D = 40. Also, 13ac = 600 − D − S = 480.

We have 28dn = (E − D)(N − A) = 10(N − 71). The only unassigned primes greater than 71 are 73 and 83; the latter gives 120, which has too may digits, so N = 73. That makes 27ac = 600 − N − S − T − W = 322, so R + E + A + D = 322 and R = 83. We now have all the letter assignments (spelling STANDREW in ascending order), and we can complete the grid except for the unclued 1ac and 31ac.

It’s not necessary to calculate the high powers in full in order to determine their last two digits. Any two integers being multiplied can be expressed as, say, 100a + b and 100c + d, where b and d are integers in the range [0, 99], ie, the final two digits of the numbers. The product is (100a + b)(100c + d) = 100(100ab + ad + bc) + bd, of which the final two digits come from bd only.

To calculate 10ac = 7371, we only need to consider the odd powers, ie, starting with 73 and successively multiplying by 732 = 5329. Or, as only the last two digits are significant, we can multiply by 29. This gives the sequence (for exponents of 1, 3, 5, 7, ...) as 73, 73×29=2117, 17×29=493, 93×29=2697, 97×29=2813, 13×29=377, 77×29=2233, 33×29=957, 57×29=1653, 53×29=1537, 37×29=1073, at which point the sequence of last digits repeats. Since the last two digits of 73n are the same as for 73(n+20), the last two digits of 7371 are the same as for 7311, ie, 10ac = 77 from the list above. (In fact, it’s enough to recognise that the last digit alternates between 3 and 7 in successive odd powers, so the unchecked digit of 73(4n−1) has to be 7.)

Similarly, for 2dn = 9767, the multiplier for odd powers is (972 modulo 100) = 9. We know the entry ends in 3, so we can skip alternate odd powers (which end in 7), using the multiplier (92 modulo 100) = 81. The sequence (for exponents of 3, 7, 11, 15, ...) is 97×9=873, 73×81=5913, 13×81=1053, 53×81=4293, 93×81=7533, 33×81=2673, at which point the sequence repeats. The last two digits of 97n are the same as for 97(n+20), so the last two digits of 9767 are the same as for 977, ie, 2dn = 13 from the list above.

The grid is now complete apart from 1ac = 1_1_ and 31ac = _0_3. From the letter values giving ST ANDREW and the “thematic shape” of the lines described in the preamble, a St Andrew’s cross may be deduced. Diagonals joining the centres of the corner cells pass through the centres of only four inner cells, and the digits along the two diagonals are 1413 and 2013 respectively. These and the letter sum of 600 refer to the sexcentenary of the University of St Andrews, celebrated in 2013.

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