Match List I with List II and select the correct answer: List I (Product) A. $(1827)^{16}$ B. $(2153)^{19}$ C. $(5129)^{21}$ List II (Digit in the unit's place) (1) $1$ (2) $3$ (3) $5$ (4) $7$ (5) $9$

### Concept & Logic The unit digit of any number with a large exponent $x^y$ depends exclusively on the unit digit of its base $x$ and the cyclicity (repeating pattern) of that unit digit's powers. ### Step-by-Step Solution **Calculation:** We will evaluate the unit digit for each expression in List I: * **A. $(1827)^{16}$** * The unit digit of the base is $7$. The cyclicity of $7$ is $4$ ($7, 9, 3, 1$). * Divide the exponent by the cyclicity: $16 \pmod 4 = 0$. * A remainder of $0$ maps to the $4$th power in the cycle ($7^4$), which ends in $1$. * Therefore, A matches with (1). * **B. $(2153)^{19}$** * The unit digit of the base is $3$. The cyclicity of $3$ is $4$ ($3, 9, 7, 1$). * Divide the exponent by the cyclicity: $19 \pmod 4 = 3$. * A remainder of $3$ maps to the $3$rd power ($3^3$), which ends in $7$. * Therefore, B matches with (4). * **C. $(5129)^{21}$** * The unit digit of the base is $9$. The cyclicity of $9$ is $2$ (odd powers end in $9$, even powers end in $1$). * The exponent $21$ is an odd number, so it behaves like $9^1$. * Therefore, C matches with (5). * **Resulting Sequence:** A $\rightarrow$ 1, B $\rightarrow$ 4, C $\rightarrow$ 5. This forms the sequence $1 \ 4 \ 5$. ### Exam Strategy & Shortcut For matching questions, you often don't need to solve every item. Solve the easiest one first. For instance, $C$ has a base of $9$ and an odd exponent, which instantly gives a unit digit of $9$ (List item 5). Looking at the options, only (c) and (d) end in $5$. Then solve A: $16$ is perfectly divisible by $4$, so base $7$ gives unit digit $1$. This immediately isolates the correct sequence as $1 \ 4 \ 5$. ### Common Pitfall Students sometimes get confused when the remainder is $0$ (like in A) and mistakenly assume the unit digit is $x^0 = 1$ for all bases. While $7^0$ happens to be $1$, this logic fails for even bases (e.g., $2^0=1$, but the $4$th step of $2$'s cycle is $6$). Always remember remainder $0$ means the *last* step of the cycle. ### Final Answer **Therefore, the correct answer is 1 4 5.**