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.

AArgh I should never have found this problem, because I will never have any free time again! Anyway, I'm trying to compose my list without using such functions as floor(), ceiling(), or trig functions.
For now, I'll offer a short proof:
If n! is defined as the product of all positive integers from 1 to n, then:
1! = 1*1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
n! = 1*2*3*...*(n-2)*(n-1)*n
You could also say that
n! = n(n-1)!
1! = 1(1-1)! = 1(0!) = 1
:. 0!=1
I've just begun my list, I'll post the finished product.

Posted by DJ on 2003-03-05 05:48:42