Last Friday’s puzzle was another deceptively simple one. The answers are below.

(1 + 1 + 1) ! = 6

2 + 2 + 2 = 6

3 * 3 - 3 = 6

$$ \sqrt{4} + \sqrt{4} +\sqrt{4} = 6 $$

$$ 5 + {{5}/{5}} = 6 $$

6 + 6 - 6 = 6

7 - (7 / 7) = 6

$$ \sqrt[3]{8} + \sqrt[3]{8} +\sqrt[3]{8} = 6 $$

$$ \sqrt{9} * \sqrt{9} -\sqrt{9} = 6 $$

Remembering factorial (!) is the tricky one.

Then as an additional puzzle we asked how many other ways are there of solving the “three twos” sum?

Here are some others, but we’re sure there must be more . . .

2 * 2 + 2 = 6

-2 * -2 +2 = 6

2 ^ 2 + 2 = 6

-2 ^ 2 + 2 = 6

(2 / 2 + 2)! = 6

And then we issued a challenge, can you extend the sequence further to solve “three tens”, “three elevens” and so on?

The “three tens” needs logarithms; common logarithms, or those based on 10 (as opposed to natural logarithms based on e). log (10) equals one, and so this can be substituted into the “three ones” above.

(log (10) + log (10) + log (10))! = 6

To solve the “three elevens” we need another device – the “integer part” or floor symbol └. The integer part of log (11) is one, and so . . .

(└log (11) + └log (11) + └log (11))! = 6

In fact the log of all numbers from 10 to 99 is one point something, so the integer part is one. Thus this method can be used all the way up to 99.

(└log (99) + └log (99) + └log (99))! = 6

And when we reach 100 the log is two and we can use └ and log all the way to 999 in the “three twos” equation.

By extension we can use └ and log all the way up to 9,999,999,999 as the integer part of log (9,999,999,999) is nine, and we can use each of the equations above as we reach a new power of ten.

Is that the end?

No it isn’t, as log (10,000,000,000) is ten and so log (log (10,000,000,000)) is one, and we can start again.

There is a solution for every positive integer.

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