What is the remainder when 1723 is divided by 16?

Concept: if a base a leaves a remainder r when divided by m (i.e. a ≡ r mod m), then every positive-integer power of a leaves the same remainder as that power of r when divided by m — the remainder of a power depends only on the remainder of its base, not on the base's full value.

Application:

1. Reduce the base modulo the divisor first: dividing 17 by 16 gives quotient 1 and remainder 1, so 17 ≡ 1 (mod 16).

2. Raise both sides of that congruence to the required power: 1723 ≡ 123 (mod 16).

3. Simplify the right-hand side: 1 raised to any power is still 1, so 123 = 1.

4. Combine the steps: 1723 ≡ 1 (mod 16), which means the remainder when 1723 is divided by 16 is exactly 1.

Cross-check (binomial expansion):

Write 17 as 16 + 1, so 1723 = (16 + 1)23. Expanding this binomial gives a sum of terms of the form C(23, k)·16k·1(23−k). Every term with k ≥ 1 carries a factor of 16 and is therefore exactly divisible by 16; only the k = 0 term survives modulo 16, and it equals 1. This independently confirms the remainder is 1.

Result: the remainder when 1723 is divided by 16 is 1.