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?

We need to find two-digit numbers where the original number is more than three times its digit-reversed version.

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Let us represent any two-digit number as $10a + b$, where:

• $a$ is the tens digit (can be 1, 2, 3, ..., 9)
• $b$ is the units digit (can be 1, 2, 3, ..., 9, since it's non-zero)

Any two-digit number like 47 can be written as $10 \times 4 + 7 = 47$. This is just the place value system we use.

When we interchange the digits, we get $10b + a$. For example, if our original number is 47, then $a = 4$, $b = 7$, and the interchanged number is $10 \times 7 + 4 = 74$.

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The problem states: original number > 3 × (interchanged number)

This translates to: $10a + b > 3(10b + a)$

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$10a + b > 3(10b + a)$

$10a + b > 30b + 3a$

$10a - 3a > 30b - b$

$7a > 29b$

Key insight: This inequality $7a > 29b$ is much simpler to work with than the original condition.

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Since both $a$ and $b$ are single digits (1 through 9), we will check each possible value of $b$:

When $b = 1$:

$7a > 29 \times 1 = 29$

$a > \tfrac{29}{7} = 4.14...$

Since $a$ must be a whole number, $a \geq 5$.

Possible values: $a \in \{5, 6, 7, 8, 9\}$ → 5 pairs

When $b = 2$:

$7a > 29 \times 2 = 58$

$a > \tfrac{58}{7} = 8.28...$

Since $a$ must be a whole number, $a \geq 9$.

Possible values: $a \in \{9\}$ → 1 pair

When $b = 3$:

$7a > 29 \times 3 = 87$

$a > \tfrac{87}{7} = 12.43...$

Since $a$ must be a whole number, $a \geq 13$.

But $a$ can only be 1 through 9, so no valid pairs.

When $b \geq 4$:

For $b = 4$: $7a > 29 \times 4 = 116$, so $a > 16.57...$

This is impossible since $a \leq 9$.

Similarly, for larger values of $b$, the required value of $a$ becomes even larger.

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From $b = 1$: 5 pairs

From $b = 2$: 1 pair

From $b \geq 3$: 0 pairs

Total = 5 + 1 = 6

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Therefore, there are 6 such two-digit numbers.