The number 2027 is prime. Let m = 242026 + 252027. Find the smallest positive integer n for which mn = 2027k + 1 for some integer k.
Since 2027 = p is prime, phi(p) = p-1 and any number coprime to p obeys x^(p-1) = 1 mod p
Write m = 24^2026 + 25^2027 = 24^(p-1) + 25 * 25^(p-1)
m = 1 + 25 * 1 mod p = 26 mod p. Then we seek n such that
26n = 1 mod p
Let's look at where 26n transitions from < p to > p:
26 * 100 = 2600 (> p)
26 * 20 = 520 so
26 * 80 = 2600 - 520 = 2080 (> p; 26*80 - p = 2080 - 2027 = 53 -- we can get closer)
26*2 = 52, so
26*78 = 2028 which, since it's p+1 is clearly = 1 mod p
then n = 78
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Posted by Paul
on 2025-04-19 23:06:40 |
analytic solution
