Let the four-digit dividend be \(x\).

According to the problem, we have the following congruences:

• \(x \equiv 14 \pmod{21}\)
• \(x \equiv 35 \pmod{42}\)
• \(x \equiv 140 \pmod{147}\)
• \(x \equiv 98 \pmod{105}\)

Notice that the remainders are related to the divisors as follows:

• \(14 = 21 - 7\)
• \(35 = 42 - 7\)
• \(140 = 147 - 7\)
• \(98 = 105 - 7\)

Therefore, we can rewrite the congruences as:

• \(x \equiv -7 \pmod{21}\)
• \(x \equiv -7 \pmod{42}\)
• \(x \equiv -7 \pmod{147}\)
• \(x \equiv -7 \pmod{105}\)

This means \(x + 7\) is divisible by 21, 42, 147, and 105. We need to find the least common multiple (LCM) of these numbers.

First, find the prime factorization of each number:

• 21 = 3 x 7
• 42 = 2 x 3 x 7
• 147 = 3 x 7²
• 105 = 3 x 5 x 7

The LCM is \(2 \times 3 \times 5 \times 7^2 = 1470\).

So, \(x + 7\) is a multiple of 1470. Let \(x + 7 = 1470k\) for some integer \(k\). Then \(x = 1470k - 7\).

Since \(x\) is a four-digit number, we test values of \(k\):

• If \(k = 1\), \(x = 1463\)
• If \(k = 2\), \(x = 2933\)
• If \(k = 3\), \(x = 4403\)
• If \(k = 4\), \(x = 5873\)
• If \(k = 5\), \(x = 7343\)
• If \(k = 6\), \(x = 8813\)

Therefore, the four-digit dividend is 8813.