Discussion
What is the remainder when \(15^{37}\) is divided by 17?
Answer: B
We can write 15 as \(17-2\). So we need \((17-2)^{37} \pmod{17}\), which is equivalent to \((-2)^{37} \pmod{17}\). Since the power is odd, this is \(-(2^{37}) \pmod{17}\). By Fermat's Little Theorem, since 17 is prime, \(2^{16} \equiv 1 \pmod{17}\). We can write \(2^{37} = 2^{16} \times 2^{16} \times 2^5 \equiv 1 \times 1 \times 32 \pmod{17}\). The remainder of \(32 \div 17\) is 15. So we have \(-(15) \pmod{17}\), which is \(-15 \equiv 2 \pmod{17}\). The remainder is 2.
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