2021 AMC Fall 10A Problem 12

Write $N$ in expanded form: with digits $d_{10} d_9 \dots d_1 d_0 = 2,7,0,0,6,0,0,0,0,5,2$,
$$N = \sum_{i=0}^{10} d_i \cdot 9^i.$$

Since $9 \equiv -1 \pmod 5$, we have $9^i \equiv (-1)^i \pmod 5$. So modulo $5$, $N$ is the alternating sum of its base-nine digits, with the units digit positive:
$$N \equiv d_0 - d_1 + d_2 - \cdots + d_{10} \pmod 5.$$

Only five digits are nonzero: $d_0 = 2$, $d_1 = 5$, $d_6 = 6$, $d_9 = 7$, $d_{10} = 2$. Even positions get $+$, odd positions get $-$:
$$N \equiv 2 - 5 + 6 - 7 + 2 = -2 \equiv 3 \pmod 5.$$

The answer is $\boxed{\textbf{(D)}\ 3}$.

The familiar divisibility test for $11$ in base ten (alternating digit sum) works because $10 \equiv -1 \pmod{11}$. The same idea applies whenever the base is one less than a multiple of the modulus: here $9 \equiv -1 \pmod 5$, so base-nine numbers reduce mod $5$ by an alternating digit sum. Look for base $\equiv \pm 1$ relations before converting anything to base ten.

• Using the plain digit sum (as if 9 were 1 mod 5) or misplacing which digits sit in even and odd positions.
• Converting the whole eleven-digit number to base ten, which is slow and invites arithmetic errors.