2021 CSAT — Q5

Quant Number theory 2.5 marks Easy

If $3^{2019}$ is divided by $10$ , then what is the remainder?

A 1
B 3
C 7 Answer
D 9

Worked rationale

The remainder on division by $10$ is just the units digit. Powers of $3$ cycle with period $4$ :

3^1=3,\quad 3^2=9,\quad 3^3=27,\quad 3^4=81,\quad 3^5=243,\dots

so the units digits run $3,9,7,1,\,3,9,7,1,\dots$

Find $2019 \bmod 4$ : $2019 = 4\times 504 + 3$ , so $2019 \equiv 3 \pmod 4$ . The third entry of the cycle $(3,9,7,1)$ is $7$ .

3^{2019} \equiv 3^3 \equiv 7 \pmod{10}.

Answer: (c) 7.

Why the other options miss

A

counted one slot off: takes $2019\equiv 0\pmod 4$ (the $3^4\equiv 1$ slot), miscounting the cycle position.
B

counted one slot off: reads $2019\equiv 1\pmod 4$ , landing on $3^1=3$ .
D

counted one slot off: reads $2019\equiv 2\pmod 4$ , landing on $3^2=9$ .

Specialist insight

Every “remainder mod $10$ of a power” question is a cyclicity question. The only real work is the exponent reduced modulo the cycle length ( $4$ for $3$ ), and the one place students bleed marks is the fencepost: the cycle is $(3,9,7,1)$ for exponents $(1,2,3,4)$ , so a remainder of $3$ in $2019\bmod 4$ means the 3rd entry, not " $3$ ". Map the residue to the slot, not to the digit.

The trap, in one line

$2019\equiv 3\pmod 4$ picks the 3rd unit-digit in $3,9,7,1$ , which is $7 \Rightarrow$ (c).

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