How many positive four-digit integers $N$ have the property that the three-digit number obtained by removing the leftmost digit is equal to $\frac{N}{17}?$

Answer 1

Given a four-digit integer \( N \) such that the three-digit number obtained by removing the leftmost digit equals \( \frac{N}{17} \), we denote \( N \) as \( 1000a + b \), where \( a \) is the leftmost digit (ranging from 1 to 9) and \( b \) is the rightmost three-digit part (ranging from 0 to 999). The problem states:

\[
b = \frac{1000a + b}{17}
\]

To find all possible values of \( N \), we start by solving for \( b \) from the provided equation. Multiplying both sides by 17, we get:

\[
17b = 1000a + b
\]

Subtract \( b \) from both sides to isolate terms involving \( b \):

\[
16b = 1000a
\]

Solving for \( b \), we find:

\[
b = \frac{1000a}{16} = 62.5a
\]

Since \( b \) must be an integer, \( 62.5a \) must also be an integer. This implies that \( a \) must be such that \( 62.5a \) is an integer.

Considering \( a \) to be within the range [1, 9], we need \( a \) to make \( 62.5a \) an integer. Given \( 62.5 = \frac{125}{2} \), \( a \) should be a multiple of 2 to ensure \( a / 2 \) is an integer since \( 125 \times \) (an integer) must be an integer. Hence, \( a \) must be even. The possible even values for \( a \) in this range are \( 2, 4, 6, 8 \).

To verify, calculate \( b \) for these \( a \):

- For \( a = 2 \):
\[
b = 62.5 \times 2 = 125
\]

- For \( a = 4 \):
\[
b = 62.5 \times 4 = 250
\]

- For \( a = 6 \):
\[
b = 62.5 \times 6 = 375
\]

- For \( a = 8 \):
\[
b = 62.5 \times 8 = 500
\]

These calculations confirm \( b \) values lead to valid four-digit numbers \( N \). Explicitly constructing these,
we get:

- For \( a = 2 \): \( N = 1000 \cdot 2 + 125 = 2125 \)
- For \( a = 4 \): \( N = 1000 \cdot 4 + 250 = 4250 \)
- For \( a = 6 \): \( N = 1000 \cdot 6 + 375 = 6375 \)
- For \( a = 8 \): \( N = 1000 \cdot 8 + 500 = 8500 \)

Thus, these configurations verify that we have four distinct four-digit \( N \) values satisfying the conditions. Therefore, the number of such four-digit integers \( N \) is:

\[
\boxed{4}
\]