Sunday Times Teaser 3035 – Friend of the Devil
by Stephen Hogg
Published Sunday November 22 2020 (link)
My friend, “Skeleton” Rose, rambled on with me and my uncle (“The Devil” and “Candyman”) about Mr Charlie, who gave, between us, three identical boxes of rainbow drops.
Each identical box’s card template had a white, regular convex polygonal base section with under ten sides, from each of which a similar black triangular star point extended. All these dark star points folded up to an apex, making an enclosed box.
The number of sweets per box equalled the singlefigure sum of its own digits times the sum of the star points and the box’s faces and edges. If I told you how many of the “star point”, “face” and “edge” numbers were exactly divisible by the digit sum, you would know this number of sweets.
How many sweets were there in total?

John Z permalink123456789101112131415161718192021222324252627282930313233# function to return sum of digits for up to a three digit number# max value encountered will be 9 * ( 4 * 9 + 1) or 333sod3 = lambda x: sum(((a := divmod(x, 100))[0], *divmod(a[1], 10)))sols = [0] * 4 # number exactly divisiblesolns = [0] * 4 # number of sweets when exactly divisible# index over number of sides in polygonfor i in range(3, 10):faces = i + 1starpoints = iedges = i * 2# index over possible sums of digits of sweets in boxfor n in range(1, 10):sweets = n * (4 * i + 1) # i.e. n * (Faces + starpoints + edges)# test 'number of sweets per box equalled the singlefigure sum of its own# digits ...'if n == (sod := sod3(sweets)):p = (faces % sod == 0) + (starpoints % sod == 0) + (edges % sod == 0)sols[p] += 1 # 'number exactly divisble'# save number of sweets; ok to overwrite# as overwrite only occurs for nonunique solutionssolns[p] = sweets# print(i, n, sweets) # for debugging# unique solution when "number of exactly divisible" occurs once# there are 3 boxes so multiply by threeprint("Total number of sweets in the three boxes:", solns[sols.index(1)] * 3)