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.
To find the remainder when (13^16 + 17^12) is divided by 221, we can start by simplifying the individual terms.
First, let's find the remainder of 13^16 when divided by 221.
To do this, we can use the property of modular arithmetic called Fermat's Little Theorem.
Fermat's Little Theorem states that if p is a prime number and a is any positive integer not divisible by p, then a^(p-1) ≡ 1 (mod p).
In this case, 221 is not a prime number, but we can find the prime factorization of 221 as 13 * 17.
Let's calculate 13^(16 mod (13-1)) = 13^4 ≡ 169 (mod 221)
Next, let's find the remainder of 17^12 when divided by 221.
Using the same property of Fermat's Little Theorem, we can calculate 17^(12 mod (17-1)) = 17^4 ≡ 113 (mod 221)
Now, let's simplify the expression by adding the remainders:
169 + 113 = 282
To find the remainder, we divide 282 by 221:
282 ÷ 221 = 1 remainder 61
Therefore, the remainder when (13^16 + 17^12) is divided by 221 is 61.