2023 CSAT — Q8

Quant Number theory 2.5 marks Easy

What is the unit digit in the expansion of $(57242)^{9 \times 7 \times 5 \times 3 \times 1}$ ?

A 2 Answer
B 4
C 6
D 8

Worked rationale

Only two things matter: the units digit of the base and the exponent mod $4$ .

Units digit of $57242$ is $2$ , and $2$ has the $4$ -cycle $2, 4, 8, 6$ (for exponents $1,2,3,4$ , then repeating).

The exponent is $9 \times 7 \times 5 \times 3 \times 1 = 945$ . Reduce mod $4$ : $945 = 4 \times 236 + 1$ , so $945 \equiv 1 \pmod 4$ .

Cycle position $1$ gives units digit $2^1 = 2$ .

Answer: (a) 2.

Why the other options miss

B

an arithmetic slip: takes the exponent as $\equiv 2 \pmod 4$ (e.g. a slip in $945 \bmod 4$ ), landing on $2^2 = 4$ .
C

wrong formula: uses cycle position $4$ (i.e. treats the exponent as a multiple of $4$ ), which is what a product of consecutive even–odd factors would not give here.
D

off by one: reads the cycle one place too far ( $2^3 = 8$ ), mismatching $945 \bmod 4$ .

Specialist insight

The exponent is a product of odd numbers, so it is odd — and an odd number is never $\equiv 0 \pmod 4$ ; since $945$ ends in $5$ , $945 = 940 + 5 \equiv 1 \pmod 4$ instantly. The danger is computing $945$ and then fumbling the mod- $4$ step; the disciplined move is “units digit $2$ , $4$ -cycle, exponent $\bmod 4$ ” and nothing else — never expand the power.

The trap, in one line

Base ends in $2$ (cycle $2,4,8,6$ ); exponent $945\equiv 1\pmod 4$ , so the units digit is $2^1=2$ .

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