In my previous post, we looked at the largest known prime, P = 257885161 – 1, and how many digits it has in various bases. This post looks at how to find the last digit of P in base b. We assume b < P. (If b = P then the last digit is 0, and if b > P the last digit is P.)
If b is a power of 2, we showed in the previous post that the last digit of P is b-1.
If b is odd, we can find the last digit using Euler’s totient theorem. Let φ(b) be the number of positive integers less than b and relatively prime to b. Then Euler’s theorem tells us that 2φ(b) ≡ 1 mod b since b is odd. So if r is the remainder when 57885161 is divided by φ(b), then the last digit of Q = P+1 in base b is the same as the last digit of 2r in base b.
For example, suppose we wanted to know the last digit of P in base 15. Since φ(15) = 8, and the remainder when 57885161 is divided by 8 is 1, the last digit of Q base 15 is 2. So the last digit of P is 1.
So we know how to compute the last digit in base b if b is a power or 2 or odd. Factor out all the powers of 2 from b so that b = 2k d and d is odd. We can find the last digit base 2k, and we can find the last digit base d, so can we combine these to find the last digit base b? Yes, that’s exactly what the Chinese Remainder Theorem is for.
To illustrate, suppose we want to find the last digit of P in base 12. P ≡ 3 mod 4 and P ≡ 1 mod 3, so P ≡ 7 mod 12. (The numbers are small enough here to guess. For a systematic approach, see the post mentioned above.) So the last digit of P is 7 in base 12.
If you’d like to write a program to play around with this, you need to be able to compute φ(n). You may have an implementation of this function in your favorite environment. For example, it’s
sympy.ntheory.totient in Python. If not, it’s easy to compute φ(n) if you can factor n. You just need to know two things about φ. First, it’s a multiplicative function, meaning that if a and b are relatively prime, then φ(ab) = φ(a) φ(b). Second, if p is a prime, then φ(pk) = pk – pk-1.