Take the digits 2, 0, 0 and 3. Make equations equating to all the integers from 1 to 150 using these digits according to the following rules:-

a) The above digits are the only digits to be used and no other digits should appear anywhere in the equation (except on the side where the answer will be).

b) Use of any mathematical symbols are allowed.

c) The digits 2, 0, 0 and 3 should appear in the given order in the equation. e.g - 0 + 2 + 3 + 0 = 5 is not acceptable.

d) When using the mathematical symbols try using the most simplest forms as much as possible.

This puzzle got me thinking, but for some of the trouble spots in my list I used the trig identity sec (arctan (sqrt x)) = sqrt (x + 1).
Using that identity I came up with a system to generate every number 6 to infinity.

(sec (arctan (sqrt 2)))^(0!+0!) + 3 = 6
(sec (arctan (sec (arctan (sqrt 2)))))^(0!+0!) + 3 = 7
(sec (arctan (sec (arctan (sec (arctan (sqrt 2)))))))^(0!+0!) + 3 = 8
(sec (arctan (sec (arctan (sec (arctan (sec (arctan (sqrt 2)))))))))^(0!+0!) + 3 = 8

Repeatedly applying sec (arctan (x)) increases the value of the expression by 1.

Posted by Brian Smith on 2003-04-15 04:22:56