What percentage of the numbers from 1 to 50 have squares that end in the digit 1?

### Concept & Formula A number's square ends in the digit $1$ if and only if the number itself ends in either $1$ or $9$ (since $1^2 = 1$ and $9^2 = 81$). $$\text{Percentage} = \left( \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} \right) \times 100$$ ### Step-by-Step Solution * **Given:** Total count of numbers from $1$ to $50 = 50$. * **Identification:** Numbers between $1$ and $50$ ending in $1$ or $9$: * Ending in 1: $1, 11, 21, 31, 41$ (5 numbers) * Ending in 9: $9, 19, 29, 39, 49$ (5 numbers) Total favorable numbers = $5 + 5 = 10$. * **Calculation:** $$\text{Percentage} = \left( \frac{10}{50} \right) \times 100 = 20\%$$ ### Exam Strategy & Shortcut In every block of $10$ consecutive numbers (e.g., $1-10$, $11-20$), exactly $2$ numbers end in $1$ or $9$. Since $2$ out of $10$ numbers satisfy the condition, the proportion is always $\frac{2}{10} = 20\%$, regardless of whether the range goes up to $10, 20, 50,$ or $100$. ### Common Pitfall * **Mistake:** Forgetting to count the numbers ending in $9$, or missing the boundary value like $49$, leading to an incorrect count of $5$ numbers ($10\%$). * **Tip:** Always write out the unit possibilities ($1^2=1$ and $9^2=81$) explicitly before counting. ### Final Answer **Therefore, the correct answer is 20.**