------------------------------------------------------ ------------------------------------------------------- Solution: 1522343 4657981 584x187 1231842 3318768 Alan Lemm ------------------------------------------------------- jdsmith sent a picture of a solution. ------------------------------------------------------- Solve for "SIMPLE ADDITION, by John Gowland" A=152 B=234 C=657 D=981 E=584 F=187 G=123 H=184 J=318 K=768 a=14 b=568 c=27 d=48 e=317 f=543 g=918 h=513 j=846 k=23 m=18 n=28 Marcis Petersons ------------------------------------------------------- 1522343 4657981 584x187 1231842 3318768 Triads 152,784,936 196,542,738 146,583,729 317,529,846 184,392,576 186,543,729 192,384,576 342,576,918 324,657,981 Dwight Kidder ------------------------------------------------------- Hello! Solution to XmasPuzzle: 1522343 4657981 584.187 1231842 3318768 First I thought it would be impossible by hand, but the key for me was: K/4 + K/2 = (d/2)^2 Best Xmas, Juha -------------------------------------------------------1522343 4657981 584 187 1231842 2318768 I didn't understand the problem for the longest time. I assumed x,y,z were fixed. Once I understood it, it became easier... I assume the rubric means that each row in the table consists of a sequence of 3 numbers, the last of which is the sum of the first 2, taking all digits 1-9 once and never taking the digit 0. Then it may seem that there are three ways to fit the numbers together: ???+???=??? (1) ??+???=???? (2) ?+????=???? (3) However, (3) provides no solutions. The last three digits of the left-hand 4-digit number must be at most 987. The 1-digit number must be at most 9. Adding these together, we get at most 996, and are unable to carry anything over to give different first digits in the 2 4-digit numbers. Similarly, (2) provides no solutions, as the left hand side is less than 100+1000=1100, and so the right hand side must have a 0 digit in it somewhere. Therefore all solutions are of form (1). Let x+y=z be such that x,y,z are 3-digit numbers which between them include all the digits 1-9. Then x+y+z is a multiple of 9, but x+y+z=2z. Therefore z is a multiple of 9. Also z>=135+246=381 and z has 3 different digits, none of which is 0. Going through all such z, finding valid x,y (noting that each row contains a square) we find that each triple is either in the following sequence or the sequence you get by swapping the first two terms of each element: [ 194, 382, 576 ], [ 192, 384, 576 ], [ 184, 392, 576 ], [ 182, 394, 576 ], [ 186, 543, 729 ], [ 183, 546, 729 ], [ 146, 583, 729 ], [ 143, 586, 729 ], [ 196, 542, 738 ], [ 243, 576, 819 ], [ 317, 529, 846 ], [ 135, 729, 864 ], [ 324, 567, 891 ], [ 342, 576, 918 ], [ 152, 784, 936 ], [ 324, 657, 981 ] We now take each row of the matrix, one at a time. 3) E/4,11J/6,c^2 E is 3-digits long, and therefore the smallest number in the triple must be less than 250. The second number must be a multiple of 11. The largest number is a square and so 576 or 729. Therefore the triple is 146,583,729: E=584,J=318,c=27 ...2... ...7... 584 ... ....... .318... 5) H,2a^2,(d/2)^2 The second number is twice a square and the largest number is a square (576 or 729). Therefore the triple is 184,392,576:H=184,a=14,d=48 1..2.4. 4..7.8. 584 ... ...184. .318... 7) K/4,K/2,(d/2)^2 The second number is twice the first. Therefore the triple is 192,384,576:K=768,d=48 1..2.4. 4..7.8. 584 ... ...184. .318768 1) A,n^2,4*B The largest number must be a multiple of 4 and therefore either 576,864 or 936. Then given that the second number must be a square, we have that the triple is either 135,729,864: A=135, n=27, B=216 or 152,784,936: A=152, n=28, B=234 Furthermore, looking at the grid, n ends with 8. Therefore A=152,n=28,B=234 152234. 4..7.8. 584 ... ...1842 .318768 2) a^2,b-B/9,6*G The largest number must be a multiple of 6 and therefore either 576,738,846,864,918,938. Given that the first number must be 14^2=196, the triple is 196,542,738: a=14, b-B/9=542, G=123 Since B=234, B/9=26 and therefore b=568. 152234. 46.7.8. 584 ... 1231842 .318768 4) e,k^2,j Note from the grid that k=23. Therefore, the triple is 317,529,846:e=317,k=23,j=846 1522343 46.7.81 584 .87 1231842 .318768 9) m^2,C,D From the grid, we have m=18, C=6?7, D=?81. Solving 324+6?7=?81 gives 324+657=981 Therefore C=657,D=981. 1522343 4657981 584 .87 1231842 .318768 6) h-D/3,f,c^2 From the grid, f=543, c^2=27^2=729. Now 729-543=186. Therefore h-D/3=186. and h=186+D/3=186+(981/3)=186+327=513 1522343 4657981 584 .87 1231842 3318768 8) k^2-F,(d/2)^2,g Letting the missing digit be x, we see from the grid that k^2-F=529-(100x+87)=442-100x. (d/2)^2=576. g=908+10x. Therefore 1018-100x=908+10x. Solving for x, we find 110=110x, and x=1. Therefore k^2-F=342, (d/2)^2=576, g=918, and the grid is 1522343 4657981 584 187 1231842 3318768 Luke Pebody ------------------------------------------------------------------ 1522343 4657981 584X187 1231842 3318768 Sorry about the lack of separators, but typing in text-only does that to you. Good problem, and I love the fiendish way the clues end up interacting with each other. I've never solved one of these before, either. Cheers to you all! Darrel C Jones