How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
- 5
- 6
- 8
- 7
The Correct Option is B
Solution and Explanation
The correct answer is (B):
Let the original two-digit number be \( ab \), where \( a \) is the tens digit and \( b \neq 0 \) is the units digit.
The value of the number is \( 10a + b \). After interchanging the digits, the new number becomes \( ba = 10b + a \).
Given condition: \[ 10a + b > 3(10b + a) \]
Expand the inequality: \[ 10a + b > 30b + 3a \] \[ 10a - 3a > 30b - b \Rightarrow 7a > 29b \]
Now we test integer values of \( b \) from 1 to 9 (since \( b \neq 0 \)), and find integer values of \( a \) from 1 to 9 that satisfy the inequality.
- For \( b = 1 \): \( 7a > 29 \Rightarrow a > \frac{29}{7} \approx 4.14 \Rightarrow a = 5, 6, 7, 8, 9 \) ⇒ 5 values
- For \( b = 2 \): \( 7a > 58 \Rightarrow a > \frac{58}{7} \approx 8.29 \Rightarrow a = 9 \) ⇒ 1 value
- For \( b \geq 3 \): \( 7a > 87 \) and higher ⇒ No valid \( a \leq 9 \)
Hence, total valid combinations = \( 5 + 1 = \boxed{6} \)
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