How many two-digit numbers satisfy this property: The last digit (unit's digit) of the square of the two-digit number is 8?

### Concept & Logic Perfect squares can only end in specific digits. Let's evaluate the unit's digits of squares of all single digits: * $0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 6$ * $5^2 = 5, 6^2 = 6, 7^2 = 9, 8^2 = 4, 9^2 = 1$ Thus, any perfect square must end in $0, 1, 4, 5, 6,$ or $9$. A perfect square **never** ends in $2, 3, 7,$ or $8$. ### Step-by-Step Solution * **Given property:** The unit's digit of the square of a two-digit number is $8$. * **Deduction:** Based on the core logic of perfect squares, no integer squared can yield a number ending in $8$. Therefore, there are $0$ two-digit numbers that satisfy this condition. This matches option (d) 'None of these'. ### Exam Strategy & Shortcut Remember the **2-3-7-8 Rule**: No perfect square ever ends in the digits $2, 3, 7,$ or $8$. Since the question specifies a square ending in $8$, you can instantly mark 'None of these' without performing any calculations. ### Common Pitfall * **Mistake:** Confusing the rule and thinking that since $2^3 = 8$, some square might end in $8$, or listing numbers ending in $8$ instead of checking their squares. * **Tip:** Keep a firm mental checklist of impossible unit digits for squares ($2, 3, 7, 8$) and cubes ($0-9$ are all possible). ### Final Answer **Therefore, the correct answer is None of these.**