UPSC CDS II 2025 Mathematics Paper

Concept: The last digit of a power depends on the cyclicity of the base’s last digit. For $D^{100}$ to end with 1, the digit $D$ must have a cycle that gives 1 when the exponent is a multiple of 4 (or 2 in some cases). Explanation: The number is $(54D)^{100}$. Its last digit is the same as $D^{100}$. Check all digits 0–9: $0 \to 0$, $1 \to 1$, $2 \to 6$, $3 \to 1$, $4 \to 6$, $5 \to 5$, $6 \to 6$, $7 \to 1$, $8 \to 6$, $9 \to 1$. Only $D = 1, 3, 7, 9$ give last digit 1. Four possibilities exist. Statement I: $D > 5$ gives $D = 6,7,8,9$. Among these, only $7$ and $9$ satisfy. Not unique. Statement II: $D$ is a multiple of $3$ gives $D = 0,3,6,9$. Among these, only $3$ and $9$ satisfy. Not unique. Using both statements together: $D > 5$ and $D$ multiple of $3$ gives $D = 6$ or $9$. Only $D=9$ works (since $6$ does not). So both statements are needed to get a unique answer. Answer: C. The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.