Question

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?

• A. 5
• B. 6
• C. 8
• D. 7

Answer 1

The Correct Option is B

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 = \( 5 + 1 = \boxed{6} \)