Hosted by The Math Forum

# Easy as 123

A classic puzzle is to use four 4s to make various integers. I recall how, over four decades ago when I was about 12 or 13 years old, I would work out as much as I could of this puzzle, which I saw in the newspaper in the J. A. H. Hunter column "Fun With Figures." Getting 88 is easy as 4 (4!) − 4 − 4 or 4(4! − 4/√4).

Can you get 88 using only the integers 1, 2, and 3, using each of them only once, and using the standard arithmetic operations, including factorials and repeating decimals?

To be more precise, use only 1, 2, and 3, each used once, and use any of the following functions in whatever order and however many times you wish ...

 plus a + b times a*b subtract a − b negate −a divide a/b square root √a factorial a! exponentiation ab roots a(1/b)

... and the following three operations, which work on raw digits only, not on more complicated expressions ...

concatenation of digits (as in 23 from 2 and 3)
the decimal point (as in .2)
the repeating decimal point (as in .2^. which means 0.2222222..., or 1.23^. which means 1.2323232323...)

... and of course as many parentheses as you wish.

Source: Dick Hess, All-Star Mathlete Puzzles, Sterling, New York, \$6.95, 2009.