# Arrange numbers and operators to the magic triangle

Note : A # B = A*10 + B

Arrange the numbers 1 to 9 to green triangles, and arrange operator (+,-,x,/,^, and #) to the white triangles, so the math operations below are equals.

• (((A op1 B) op2 D) op3 F) = constant
• (((F op4 G) op5 H) op6 I) = constant
• (((I op6 E) op3 C) op1 A) = constant
• A x F x I = constant

A bit mistake in question :

The first must be : (((A op1 B) op2 D) op4 F) = constant.

I will accept both answer. (initial question and corrected question)

NOTE: this solutions is for the puzzle before OP found the mistake. I posted the real solutions in the comments of this answer.

First solution I found:

numbers:

A = 1, B = 7, C = 5, D = 8, E = 3, F = 4, G = 2, H = 9, I = 6

operators:

[-, +, #, ^, *, /]

proof:

(1-7+8)#4 = 24
((4^2) * 9) / 6 = 24
(6/3)#5 -1 = 24
1 * 4 * 6 = 24

And here's two more solutions (they should be all):

[5, 3, 4, 8, 7, 1, 6, 2, 9]
[#, -, /, *, ^, +]
constant = 45

and

[3, 9, 8, 7, 4, 5, 6, 2, 1]
[-, +, #, *, /, ^]
constant = 15

• Can you provide a clue of how you solved it, or was it brute force? – IanF1 Nov 13 '16 at 15:57
• Brute force. I wrote a thing in Java. – FrodCube Nov 13 '16 at 17:29
• After checking my answer and your answer, I see different result. I was a bit wrong in question. The first should be : (((A op1 B) op2 D) op4 F) = constant. But whatever your answer is right. – Jamal Senjaya Nov 13 '16 at 17:36
• So that's what it is! It seemed weird to me that I didn't find 3 (or multiples of 3) equal solutions that you get by rotating the triangle by 120°! I ran the program again and I have found those 3 solutions with the same constant. Here they are if anyone is interested: pastebin.com/RcN7QgGe – FrodCube Nov 13 '16 at 17:45