math
posted by Adam on .
find the remainder when 13^16 + 17^12 is divided by 221.

13 mod 221 = 13
13^2 mod 221 = 169
13^3 mod 221 = 208
13^4 mod 221 = 52
13^16 mod 221 = 52^4 mod 221 = 52
17 mod 221 = 17
17^2 mod 221 = 68
17^3 mod 221 = 51
17^4 mod 221 = 204
17^12 mod 221 = 204^3 mod 221 = 170
52 + 170 = 222 mod 221 = 1
Hmmm. That's so simple, there must be an easier way to get it.
Since 13 and 17 are relatively prime, we can find a,b such that
13a + 17b = 1
In fact, 13*21  17*16 = 1
Now, letting n=13^16 + 17^12,
n = 17^12 mod 13 = 1
n = 13^16 mod 17 = 1
so, n = 13*21  16*17 mod 221 = 1 mod 221
Not so sure it's simpler, but it relies more on number theory.