All solutions were successfully tested on 18 April 2013.

**make_bricks:**

def make_bricks(small, big, goal): return goal%5 >= 0 and goal%5 - small <= 0 and small + 5*big >= goal

**lone_sum:**

def lone_sum(a, b, c): if a == b == c: return 0 if b == c: return a if a == c: return b if a == b: return c return a + b + c

**lucky_sum:**

def lucky_sum(a, b, c): if a == 13: return 0 if b == 13: return a if c == 13: return a + b return a + b + c

**no_teen_sum:**

def no_teen_sum(a, b, c): return fix_teen(a) + fix_teen(b) + fix_teen(c) def fix_teen(n): #if 13 <= n <= 14 or 17 <= n <= 19: if n in [13, 14, 17, 18, 19]: return 0 return n

I consider checking for list membership to be more elegant than multiple comparison operations.

**round_sum:**

def round_sum(a, b, c): return round10(a) + round10(b) + round10(c) def round10(n): if n % 10 >= 5: return n + 10 - (n % 10) return n - (n % 10)

**close_far:**

def close_far(a, b, c): cond1 = abs(a-b) <= 1 and abs(b-c) >=2 and abs(a-c) >= 2 cond2 = abs(a-c) <= 1 and abs(a-b) >=2 and abs(c-b) >= 2 return cond1 or cond2

**make_chocolate:**

def make_chocolate(small, big, goal): maxBig = goal / 5 if big >= maxBig: if small >= (goal - maxBig * 5): return goal - maxBig * 5 if big < maxBig: if small >= (goal - big * 5): return goal - big * 5 return -1

Tom TabakHi! I’m just starting to learn coding(for about 5 days now) so I’m still not clear if all working solutions are acceptable, or should I follow some unwritten rules. If so, I would like to know which ones ðŸ™‚

For example, here is my take on the def make_chocolate problem:

Thank you for your effort! I believe I learned a lot from your solutions.

def make_chocolate(small, big, goal):

x = goal%5

if x>small or small+big*5big:

return (goal-big*5)

return x

Tom Tabak[CODE]

def make_chocolate(small, big, goal):

x = goal%5

if x>small or small+big*5big:

return (goal-big*5)

return x

[/CODE]

Tom TabakSorry, can’t find a way to post code,feel free to delete all comments ðŸ™‚

Johngood job on the make_bricks problem

you can simplify it a bit:

def make_bricks(small, big, goal):

return small >= goal%5 and small + 5*big >= goal

your “goal%5 >= 0” clause is a tautology, because a%b is always >= 0

TaniaI appreciate the solutions you have posted. Very helpful. Here is an alternative for lone_sum

def lone_sum(a, b, c):

sum = 0

values = [a, b, c]

for v in values:

if values.count(v) == 1:

sum += v

return sum

DarkBergamotteHere is an alternative for make_chocolate :

def make_chocolate(small, big, goal):

if goal>big*5 + small or small<goal % 5:

return -1

remaining = goal – big*5

while remaining<0:

remaining += 5

return remaining

Solomon Lidef no_teen_sum(a, b, c):

return sum(filter(fix_teen, [a, b, c]))

def fix_teen(n):

my_range = list(range(13, 15)) + list(range(17, 20))

return n not in my_range

Leo LiangFor make_chocolate, I have

def make_chocolate(small, big, goal):

if goal / 5 <= big:

amount_big = goal / 5

else:

amount_big = big

if goal – (amount_big * 5) <= small:

amount_small = goal – (amount_big * 5)

else:

amount_small = -1

return amount_small

———- or simply ————–

def make_chocolate(small, big, goal):

amount_big = (goal / 5) if (goal / 5 <= big) else big

amount_small = (goal – (amount_big * 5)) if (goal – (amount_big * 5) <= small) else -1

return amount_small

Josiahdef make_chocolate(small, big, goal):

while big > 0 and goal >= 5:

goal = goal – 5

big = big – 1

if goal <= small:

return goal

return -1

There's my solution for make chocolate. Hope this helps!