- #1
Jimmy Snyder
- 1,127
- 21
This one is just for fun, I do not have the answer myself. I was reminded of it by BicycleTree's procedure. The goal is to get each integer as the result of using any of the four operations, and exponentiation, operating on four fours. For instance:
1 = 4 - 4 + 4/4
2 = 4/4 + 4/4
3 = (4 + 4 + 4) / 4
4 = 4 + 4 * (4 - 4)
5 = 4 + 4 ^ (4 - 4)
6 = 4 + (4 + 4) / 4
7 = 4 + 4 - 4/4
8 = 4 * (4 + 4 ) / 4
9 = 4 + 4 + 4/4
I worked on this a few years back and got most numbers, but not all. I do not remember which ones I got and which ones I didn't. I think I was allowing two fours to be used as 44.
1 = 4 - 4 + 4/4
2 = 4/4 + 4/4
3 = (4 + 4 + 4) / 4
4 = 4 + 4 * (4 - 4)
5 = 4 + 4 ^ (4 - 4)
6 = 4 + (4 + 4) / 4
7 = 4 + 4 - 4/4
8 = 4 * (4 + 4 ) / 4
9 = 4 + 4 + 4/4
I worked on this a few years back and got most numbers, but not all. I do not remember which ones I got and which ones I didn't. I think I was allowing two fours to be used as 44.