04/07/2008, 06:06 AM
I'm not sure what BO is saying, but I do know this, you don't need 7 trillion digits of accuracy. You just have to be clever.
Any integer \( 3^n \) can be easily calculated for small-ish n, so we can start with this and find cycles in the digits. The 1-place will always cycle between {1, 3, 9, 7}, and the 10-place will always cycle between {0, 0, 0, 2, 8, 4, 2, 8, 6, 8, 4, 4, 4, 2, 6, 0, 2, 6, 8, 6}, which means we can know for sure that if we wanted to find the first two digits (the 1-place and 10-place) of \( {}^{4}{3} = 3^{3^{27}} \) then we can find that (7625597484987 mod 4 = 3) so (counting from 0) the 3-rd element of {1, 3, 9, 7} is 7, so the 1-place of 3-tetra-4 is 7. Correspondingly, (7625597484987 mod 20 = 7) so the 10-place of 3-tetra-4 is 8. Going a little bit further, the 100-place of any integer \( 3^n \) will cycle between {0, 0, 0, 0, 0, 2, 7, 1, 5, 6, 0, 1, 4, 3, 9, 9, 7, 1, 4, 4, 4, 2, 6, 8, 4, 4, 3, 9, 9, 8, 6, 9, 8, 5, 5, 7, 1, 3, 0, 2, 8, 4, 2, 6, 8, 6, 9, 7, 3, 0, 2, 7, 2, 7, 1, 5, 5, 5, 6, 0, 2, 6, 8, 4, 2, 8, 5, 5, 7, 2, 8, 5, 6, 9, 7, 3, 9, 7, 2, 8, 6, 8, 4, 2, 6, 0, 1, 3, 1, 4, 4, 3, 0, 1, 3, 1, 3, 9, 8, 6} which means that given (7625597484987 mod 100 = 87), we know that the 100-place digit of 3-tetra-4 is going to be 3, so we know the final decimal value of 3-tetra-4 will be of the form ...387
I found the size of the cycles above by experimental means, I have no idea how to find the cycle size by other means. So really, what is more important than the digits themselves is finding the size (or period) of these cycles, so you can say, "I want the 10th digit of 3-tetra-4" and somehow know that the 10th digit is a period-50000 cycle or something (I just made up the 50000). Knowing this information allows one to calculate the 10th digit of \( 3^n \) by finding the 10th digit of \( 3^{(n\text{ mod }50000)} \).
Andrew Robbins
Any integer \( 3^n \) can be easily calculated for small-ish n, so we can start with this and find cycles in the digits. The 1-place will always cycle between {1, 3, 9, 7}, and the 10-place will always cycle between {0, 0, 0, 2, 8, 4, 2, 8, 6, 8, 4, 4, 4, 2, 6, 0, 2, 6, 8, 6}, which means we can know for sure that if we wanted to find the first two digits (the 1-place and 10-place) of \( {}^{4}{3} = 3^{3^{27}} \) then we can find that (7625597484987 mod 4 = 3) so (counting from 0) the 3-rd element of {1, 3, 9, 7} is 7, so the 1-place of 3-tetra-4 is 7. Correspondingly, (7625597484987 mod 20 = 7) so the 10-place of 3-tetra-4 is 8. Going a little bit further, the 100-place of any integer \( 3^n \) will cycle between {0, 0, 0, 0, 0, 2, 7, 1, 5, 6, 0, 1, 4, 3, 9, 9, 7, 1, 4, 4, 4, 2, 6, 8, 4, 4, 3, 9, 9, 8, 6, 9, 8, 5, 5, 7, 1, 3, 0, 2, 8, 4, 2, 6, 8, 6, 9, 7, 3, 0, 2, 7, 2, 7, 1, 5, 5, 5, 6, 0, 2, 6, 8, 4, 2, 8, 5, 5, 7, 2, 8, 5, 6, 9, 7, 3, 9, 7, 2, 8, 6, 8, 4, 2, 6, 0, 1, 3, 1, 4, 4, 3, 0, 1, 3, 1, 3, 9, 8, 6} which means that given (7625597484987 mod 100 = 87), we know that the 100-place digit of 3-tetra-4 is going to be 3, so we know the final decimal value of 3-tetra-4 will be of the form ...387
I found the size of the cycles above by experimental means, I have no idea how to find the cycle size by other means. So really, what is more important than the digits themselves is finding the size (or period) of these cycles, so you can say, "I want the 10th digit of 3-tetra-4" and somehow know that the 10th digit is a period-50000 cycle or something (I just made up the 50000). Knowing this information allows one to calculate the 10th digit of \( 3^n \) by finding the 10th digit of \( 3^{(n\text{ mod }50000)} \).
Andrew Robbins