My number has 2 hundreds. The tens digit is 9 more than the ones digit
The Correct Answer and Explanation is:
To solve the problem, consider the clues provided:
- The number has 2 hundreds.
This means the hundreds digit is 2, so the number is in the 200s. It can be written in the form:
2 _ _ - The tens digit is 9 more than the ones digit.
Denote the ones digit as x.
Then the tens digit becomes x + 9.
Digits must be between 0 and 9 (inclusive). Therefore, the value of x + 9 must also be a single digit (less than or equal to 9). This limits the possible values for x.
Let’s test values for x:
- If x = 0, then x + 9 = 9 → valid
- If x = 1, then x + 9 = 10 → invalid (not a single digit)
- Any value of x greater than 0 results in x + 9 exceeding 9 → invalid
Therefore, x = 0 is the only possible value for the ones digit.
Now calculate the tens digit:
- Tens digit = x + 9 = 0 + 9 = 9
Now assemble the digits:
- Hundreds digit = 2
- Tens digit = 9
- Ones digit = 0
The number is 290.
Explanation:
The problem involves simple digit constraints and number construction. The key was understanding the relationship between the digits and ensuring all digits stayed within the allowable range of 0–9. By letting the unknown ones digit be x and using the condition that the tens digit is 9 more than the ones digit, a simple inequality shows the valid value of x must be 0. Substituting back confirms the tens digit becomes 9. Combining all digits in place-value order forms the number 290, which satisfies both conditions in the problem.
