A Student (okay, my son Sammy, who is a 7th grade student), gave me this math puzzle:

“Use 6ix (sic) nines to equal 100, you can use +, -, x, ÷, ( ). [No exponents]. 9, 9, 9, 9, 9, 9. Good Luck, Sammy”

Welp, I was stumped and had to have him give me an answer.

This was his solution:

(9 x 9 + 9) + (9 ÷ 9 + 9)

which equals

90 + 10 = 100

I brought the problem to some colleagues of mine, and here are their answers:

Dennis wrote:

(999 – 99) ÷ 9

which equals

900 ÷ 9 = 100

Greg wrote:

(9 ÷ 9 + 9) x (9 ÷ 9 + 9)

which equals

10 x 10 = 100

I don’t remember who gave me this one:

99 + (9 ÷ 9) x (9 ÷ 9)

which equals

99 + 1 x 1

which equals

99 + 1 = 100

I finally came up with my own solution:

(99 ÷ 99) + 99

which equals

1 + 99 = 100

I happen to like my solution, since it uses the same numbers each time (99). However, Dennis’s is lovely, since it goes from three digits (999) to two digits (99) to one digit (9).

Which do you think is the most beautiful solution? The one that uses the most operations, or the one that uses the fewest?