Discussion
If
\[ \begin{array}{r} 2\,A \\ \times\, A \\ \hline B\,2\,A \end{array} \]
where A and B are digits, then
Answer: D
The digits are A = 5, B = 1.
The two-digit number 2A times A ends in A, so \(A \times A\) must end in the digit A. Only 0, 1, 5 and 6 have that property.
\[ \begin{aligned} A = 5 &: 25 \times 5 = 125 \\ A = 6 &: 26 \times 6 = 156 \\ A = 1 &: 21 \times 1 = 21 \end{aligned} \]
Only 125 has the form B2A, with middle digit 2 and last digit 5, giving \(B = 1\). The product 156 has middle digit 5, so A = 6 fails, and A = 1 gives a two-digit product, so no option with A = 1 can work.
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