Christmas Puzzle – The First by Matrix

Solution Notes

10d is a cube with a multiple at 3d and linking them is triangular 16a. 10d is one of 125, 216 or 343 (512 does not allow multiples). 343 has the multiple 686 but then there is no valid triangular 11a. 216 has multiples 432, 648 or 864 but none allow a triangular 16a. 10d=125 allows 3d=250 or 750, 16ac=210 and 11a=55.

8d is a square divisor of 27d which in turn is a divisor of 28a and this forces 8d=16. 14ais then one of 678, 679 or 689 but only 678 allows for a valid 6a=311. With 14a=678, 15d=841. 6d is a square starting 3. If 6d=361 this leads to a duplicate 16 at 19a leaving 6d=324 and 19a=49. 12a is a palindrome with digit sum 5d so 12a=282, 5d=12. 17a is the reverse of 7d and cannot start 0 so 7d=15, 17a=51. 4a possibilities as a multiple of 17a are 510, 612, 714, 816, 918 and the only one whose digit sum divides 4d is 816 making 4d=855.

26a is a multiple of 311 that allows 18d to be triangular and this forces 26ac=933, 18dn=190, 27d=32, 28a=64 or 96. This allows just one 20d/32a/28a combination namely 994/124. Possible 1a values are 421 less 144 or 324 with the only valid combination 421 less 144=277. This confirms 1a=277, 3d=750, 32a=124, 21a=144, 21d=163 or 167, 13d=853.

33a is an anagram of a square with an odd middle digit. The options are 316 or 792. 316 has no 2-digit divisor ending 1 to occupy 29d but 792 has 99.

9a/9d combinations are 19/17, 22/27, 82/87, 99/97. 22a has its digital product ending 0 at 31a and so must contain an even digit. The candidates with ascending digits are 568, 569, 578, 589. Only 589 is divisible by one of the 9a options 19 confirming 22a=589, 9a=19, 9d=17, 31a=360, 24d=93. 25d is a multiple of 7 ending 6 hence 2d=78, 25d=56.

Possible 30a values that combine with 93 to form a square are 28, 51 and 76. The corresponding values for 23d+reverse(30a) are 170, 96 and 153 with only the latter triangular confirming 23dn=86, 30a=76 to complete the solution.


Christmas Puzzle The Second by MatriX

Solution Notes

The grid comprises 55 squares using 11 of the 12 shapes. The total of all the digits in the puzzle is 253 and so the digits in each shape must total 253/11=23.

There are many ways of proceeding, all essentially involving some trial and error. In the following, grid locations are given as co-ordinate pairs (row, column); rows numbered from 1 at the bottom to 5 at the top and columns from 1 at the left to 11 at the right.

Sometimes working from the corner possibilities establishes a way in. One systematic approach is to determine possible locations for the more symmetrical pentominos such as I&X that have fewer possibilities for rotation or reflection. I can appear in column 6 but then the sum of the digits to either side is not divisible by 23 and so no fill is possible. I can also appear in the middle row starting (3,6) but after V covering (3,11) no further progress is possible. Another I possibility is starting (4,5) then after L covering (5,4) to (5,7) no further progress is possible. This leaves I starting (1,6) which forces N to cover the bottom right corner then V to cover the top right corner. Working to the left allows only X with its centre at (4,8) with U beneath it. Then T fits to the left of the X into (5,7). The stem of the P fits into (2,6) in both orientations however it has to cover the bottom row to allow progress. W fits next around (5,5) then Z into (3,4). The rest follows straightforwardly with F into (4,3) and T into (2,2). L is not used.

There is also the possibility that I does not appear, in which case X must. X can be centred at (3,7) or (4,8). The latter produces the same fill as that already found. The former allows L running (5,4) to (5,7) and then the hollow in the U can be at (3,6). However counting the unallocated digits in the last 4 columns totals 94 and no effective multiple of 23 can be found that fits below the U&X. Another option with X centred at (3,7) is with the hollow of the U at (4,7) and again totalling the digits on the right shows that no multiple of 23 fits around the bottom of the X. No other pattern works so we are left with the single solution:


You will have noticed that there is no L and with a seasonal puzzle calling for an appropriate title to follow ‘The First’ the answer of course is Noel.