# Puzzle 3: Solution

Q     Using each of digits 1, 2, 3 once, and only basic arithmetic operations, write a formula for 19.

A     19 = sqrt(((3!)!)/2 + 1).

I don't know that this amusing formula has any deeper significance. I do have a heuristic explanation that suggests that there are more such formulas than one might guess at first.

The number ((3!)!)/2=360 clearly has many factors, and so can be written as A·B in many ways. Often A,B are of the same parity, so can be written as x+y and x-y for some integers x,y, whence 360=x2-y2. Since 2y=A-B, we know that y will be small if A and B are close to each other. In particular, 360=18·20 yields 360=192-12, which is equivalent to our formula above.

This kind of thing happens a few more times; for instance, 71=sqrt(7!+1), and we can use three 9's to write 603 as the square root of 9!+9sqrt(9).

The formula 192-1=(3!)!/2 can be generalized in a different direction by writing (3!)!/2 as 6!/2! and noting that for all n>3 we have the identity

n!/(n-4)! = n(n-1)(n-2)(n-3) = n(n-3)·(n-1)(n-2) = (n2-3n)(n2-3n+2) = (n2-3n+1)2-1.

As it happens, the identity 71=sqrt(7!+1) above is a consequence of this formula together with the sporadic factorial identity 6!7!=10!.