2024 CSAT — Q17

Strip the tens: $30^{30} = (3 \times 10)^{30} = 3^{30} \times 10^{30}$. The factor $10^{30}$ is exactly thirty trailing zeros and nothing else, so the digit just before the zeros is the units digit of $3^{30}$.

Units digit of $3^n$ cycles with period $4$:

Find $30 \bmod 4 = 2$. The $2$nd entry of the cycle is $9$.

Answer: (d) 9.

The whole template is “factor out the $10$s, then ride the unit-digit cycle of what’s left.” Any base ending in $0$ ($20,30,40,\dots$) splits cleanly into (digit)$\times 10$; the rightmost non-zero digit is just the unit-digit power of that leading digit. Memorise the four cycles you actually need: $2\!:[2,4,8,6]$, $3\!:[3,9,7,1]$, $7\!:[7,9,3,1]$, $8\!:[8,4,2,6]$ (all period $4$); $4,9$ have period $2$; $0,1,5,6$ are fixed. The one place people bleed marks is the indexing: I read the cycle as positions $1,2,3,4$ matching $n \bmod 4 = 1,2,3,0$ — so a remainder of $0$ means the last slot, not the first. Get that mapping once and every unit-digit item in the paper is a 20-second kill.