We were at dinner last night and Slade set us a challenge. He claimed that by using any mathematical operation you can make all of the following true:

1 ? 1 ? 1 = 6

2 ? 2 ? 2 = 6

9 ? 9 ? 9 = 6

for all digits, 1 through 9. Now most of these are trivial, eg.

$2 + 2 + 2 = 6$

$\sqrt[3]{9} \times \sqrt[3]{9} - \sqrt[3]{9} = 6$

But I think that the first one presents a whole lot of really interesting possibilities, depending on how you choose to view it. The text book answer is (1 + 1 + 1)! = 3! = 6. Which is really quite nice, but what other ways can you think about it?

A BIG cheat, would be to apply an expression on each term. For example:

$f(x) = x + 1 => f(1) + f(1) + f(1) = 6$

You could probably argue this, given that this is really what you’re doing with say the 8’s:

$\sqrt[3]{8} + \sqrt[3]{8} + \sqrt[3]{8} = 6$
$\equiv f(x) : \sqrt[3]{ x} \Rightarrow f(8) + f(8) + f(8) = 6$

But, it seems like the easy way out. I came up with another solution, that I would argue is completely valid – the trick is to read the statement in a non-obvious way. The naive interpretation would be:

“One and One and One gives six”

where ‘and’ is some operation. But what if we read it entirely differently:

1 ? 1 ? 1 => “give me the first, move forwards and give me the next, move forwards and give me the next”

From this, what we are really getting is a sequence and it opens  many more possibilities. For example, using the fibonacci series (beginning at 1) we can show that the statement is true:

fibonacci series= (1, 2, 3, 5, 8, 13, 21, …)

read:      Give me One + next + next = 1 + 2 + 3 = 6   =>   True

I like this 🙂 So often, we look for the obvious answer and that is enough to satisfy us. But, sometimes it is far more interesting to look beyond and see what else there is to discover. If anyone else has another way to interpret this, I would be really keen to hear (leave me a comment below :p)