Talk:4-rings or 4-squares puzzle
Why 2860?
The equations are: a+b=X b+c+d=X d+e+f=X f+g=X which imply that d = a-c and d = g-e when d=9 the only values are a=9, c=0 when d=8 the values are a=9, c=1; a=8, c=0. which for d=0..9 sums to 55. So there are 55*55 cases to consider for a, c, d, e, and g. Fixing b fixes f so it should not be necessary to consider more than 55*55*10 (which is 25250) cases, which is rather less than the permutations that some solutions are testing! The task is to determine the number of solutions that there are, for which I need to go a little further. X must have a minimum value of d when b,c,e,f=0 and a maximum value of 18 when a,b=9. For d=0..9 I generate something Pascal's Triangle like for the count Z of solutions: d=9 Z9 = 10 ->10*1 d=8 Z8 = 38 -> 2*1 + 9*2*2 d=7 Z7 = 82 -> 2*1 + 2*2*2 + 8*3*3 d=6 Z6 =140 -> 2*1 + 2*2*2 + 2*3*3 + 7*4*4 d=5 Z5 =210 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 6*5*5 d=4 Z4 =290 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 5*6*6 d=3 Z3 =378 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 4*7*7 d=2 Z2 =472 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 2*7*7 + 3*8*8 d=1 Z1 =570 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 2*7*7 + 2*8*8 + 2*9*9 d=0 Z0 =670 -> 2*1 + 2*2*2 + 2*3*3 + 2*4*4 + 2*5*5 + 2*6*6 + 2*7*7 + 2*8*8 + 2*9*9 + 1*10*10 Sum of Z0 through Z9 is 2860
--Nigel Galloway (talk) 17:49, 25 January 2017 (UTC)