# Paint numbers from 1 to 23 with three colours

Can you paint every number from 1 to 23 with three colours, such that there are no distinct numbers $$𝑎,𝑏,𝑐$$ of the same colour with $$𝑎+𝑏=𝑐$$? For example, you cannot have 2, 3 and 5 of the same colour since 2+3=5. You may need to use a computer to solve this.

Here is a similar puzzle for numbers 1 to 8 painted with two colours: Paint numbers from 1 to 8 with two colours

Good luck!

• This is a job for... S̶u̶p̶e̶r̶m̶a̶n̶ SAT solvers! Oct 20, 2019 at 6:56

I have found that solution (thanks Python !) :

Red : 1, 2, 4, 8, 11, 16, 22
Green : 3, 5, 6, 7, 19, 21, 23
Blue : 9, 10, 12, 13, 14, 15, 17, 18, 20

Here is my code if you're curious !

 # Tells for each color if it is available for this number
def getColorsAvailabilities(n):
av = [True] * 3
for i in range(1, (n+1) // 2): # Only iterate up to half the number
if (colors[i] == colors[n-i]):
# If i and (n-i) have same color, n cannot be of that color
av[colors[i]] = False
return av

# Main recursive function
def test(n):
if (n >= 24):
return True
colorsAvailabilities = getColorsAvailabilities(n)
# Try each available color, in order
for color in range(3):
if colorsAvailabilities[color]:
colors[n] = color
if test(n+1):
return True  # Stop when we found a solution
# If every available color results in a failure, backtrack
return False

# First try : 1 and 2 have same colors
colors = [-1] * 24
colors = 0
colors = 0
print(test(3)) # It works !
print(colors) # My solution (with Red = 0, Green = 1, Blue = 2)
# Second try : 1 and 2 have different colors
colors = [-1] * 24
colors = 0
colors = 1
print(test(3)) # Doesn't work (stops at 22)
print(colors)


• I saw someone else answer in the while I was editing my own answer, but since I have added my code, it is not a total duplicate, so here you go ^^ Oct 18, 2019 at 13:39

I used a computer, and found the following solution:

First group: $$1,2,4,8,11,22$$
Second group: $$3,5,6,7,19,21,23$$
Third group: $$9,10,12,13,14,15,16,17,18,20$$

You can move $$16$$ or $$17$$ to the first group, but not both. Those are the only three solutions according to my program, other than permuting the groups. Note that $$24$$ cannot be added to any group.