The digit in the unit's place of the product $(2464)^{1793} \times (615)^{317} \times (131)^{491}$ is

### Concept & Strategy The unit digit of a product is entirely dependent on the unit digits of its multipliers. Furthermore, multiplying any even number by a number ending in $5$ will always yield a product that ends in $0$. ### Step-by-Step Solution **Given:** Find the unit digit of $(2464)^{1793} \times (615)^{317} \times (131)^{491}$. **Calculation:** * Look at the base of the middle term: $615$. Since it ends in $5$, any power of it will also end in $5$. * $(615)^{317} \rightarrow$ unit digit is $5$. * Look at the base of the first term: $2464$. Since it ends in an even number ($4$), any integer power of it will result in an even unit digit. * $(2464)^{1793} \rightarrow 4^{\text{odd}} \rightarrow$ unit digit is $4$ (an even number). * Look at the base of the last term: $131$. It ends in $1$, so its power will end in $1$. * Multiply the resulting unit digits: $$4 \times 5 \times 1 = 20$$ * The unit digit of $20$ is $0$. ### Exam Strategy & Shortcut Do not waste time calculating the exact unit digit for every term using cyclicity rules. The moment you spot a factor that will end in $5$ (like $615^{317}$) and *any* factor that will be even (like $2464^{1793}$), the unit digit of the entire product instantly becomes $0$. ### Common Pitfall Students often go into "autopilot" mode, meticulously calculating $1793 \pmod 4$ to find the power of $4$, taking unnecessary time. Always do a quick holistic scan of the expression before applying brute-force rules. ### Final Answer **Therefore, the correct answer is 0.**