Python(3) code to sieve solution(s) for “A quiz with Susi” n. 927

susi927.py

#! /usr/bin/python3
#
# italian magazine "La Settimana Enigmistica",
# quiz "Quesito con la Susi" n. 927
#
# R, B, S: prices of one bottle of red, white and sparkling wine
# a, b, c: number of sparkling (a), white (b) and red (c) wine in a
#          box of seven bottles
#
# R = 3B
# 7R = 3S
# a + b + c = 7
# aS + bB + cR = 119 (euro)
#
# we find:
#
# R(49 - 6b - 4c) = 3*119     (1)
#
# The code uses this one, plus the fact that the smallest
# euro coin is the cent, hence prices are multiplied by 100 and
# treated as integers
#
# We obtain 3 solutions. The original problem gives us another
# constraint (which I missed): prices (in €) are integer. So I've
# added a variable that makes the code consider this constraint too.
#
# To do that, it's enough we don't multiply by 100 as we did
# to count in eurocent.
#

integer_prices = True

B = lambda r: int(r/3)    # no reason to use lambda instead of def...
S = lambda r: int(7*r/3)

mult = 1   # prices are integer and in euro
if not integer_prices:
    mult = 100  # prices are integer and in eurocent

for a in range(1,7):
    sl = 7 - a      # b + c = 7 - a
    bc = [ (i, sl - i) for i in range(1,sl) ]
    n = 3*119*mult   # rhs of eq.(1), (*100)
    for pair in bc:
        b = pair[0]
        c = pair[1]
        d = 49 - 6*b - 4*c # lhs of eq.(1) /R
        if d <= 0:   # with these values, it won't happen
            continue
        r = int(n/d)
        v = a*S(r) + b*B(r) + c*r
        if v == 119*mult:
            print(a, b, c, "-", S(r), B(r), r)