I'm a sucker for those math problems that go around Facebook.

x = 6 \div 2(1 + 2) \\ x = \frac{6}{2(1 + 2)} \\ x = \frac{6}{2 + 4} \\ x = \frac{6}{6} \\ x = 1 \\ y = 6 \div 2(1 + 2) \\ y = \frac{6}{2(1 + 2)} \\ y = \frac{2 \cdot 3}{2(1 + 2)} \\ y = \frac{2 (1 + 2)}{2(1 + 2)} \\ y = 1
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Added what appear to be the original form of the problem.

x = 6 \div 2(1 + 2) \\x = \frac{6}{2(1 + 2)} \\x = \frac{6}{2 + 4} \\x = \frac{6}{6} \\x = 1 \\y = 6 \div 2(1 + 2) \\y = \frac{6}{2(1 + 2)} \\y = \frac{2 \cdot 3}{2(1 + 2)} \\y = \frac{2 (1 + 2)}{2(1 + 2)} \\y = 1 \\z = 6/2 \space \space (1 + 2) \\z = \frac{6}{2}(1 + 2) \\z = \frac{\tfrac{6}{2}}{\tfrac{12}{4}} \\z = \frac{3}{3} \\z = 1

The internet claims that the correct answer is actually 9. In this situation it's one of the reasons I hate the "division" operator when writing things out mathematically. Algebra uses the fractional notation for a reason: It's unambiguous.

Interestingly the form it was presented in best translates to way a line in a program might write it. It's a bit funny to me that I see the algebraic translation first before I see the pure programmatical version.

x = 6 \div 2(1 + 2) \\x = 6 \div 2 \cdot (1 + 2) \\x = \frac{6}{2} \cdot (1 + 2) \\x = 3 \cdot (1 + 2) \\x = 3 \cdot 3 \\x = 9

6 \div 2(1 + 2) \\x \div y(j + k) \\\frac{x}{y(j + k)} \Leftrightarrow x/(y(j+k))