Discussion
The digit at units place of the square of a number is 2 more than the digit at units place of that number. Which of the following cannot be the digit at units place of the square of that number?
Answer: B
The digit that cannot occur is 5.
Check each possible units digit \(d\) of the number and the units digit of \(d^{2}\); we need the square's digit to be \(d + 2\).
\[ \begin{aligned} d = 2 &: 2^{2} = 4 = 2 + 2 \\ d = 4 &: 4^{2} = 16, \text{ digit } 6 = 4 + 2 \\ d = 7 &: 7^{2} = 49, \text{ digit } 9 = 7 + 2 \end{aligned} \]
No other digit works (for example \(3^{2} = 9\), not 5, and \(5^{2} = 25\), not 7). So the units digit of the square can be 4, 6 or 9, but never 5, which is why 4, 6 and 9 are wrong answers.
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