Posted by **Adam** on Monday, July 30, 2012 at 12:43pm.

find the remainder when 13^16 + 17^12 is divided by 221.

- math -
**Steve**, Monday, July 30, 2012 at 2:19pm
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.

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