Take a four digit number (abcd) and repeat it to make an eight digit number (abcdabcd). This eight-digit number always has at least three different prime factors, unless you start with a certain four-digit number. Which number is it?

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To solve this problem, we need to analyze the requirements and properties of the eight-digit number formed by repeating a four-digit number.

Let's split the eight-digit number "abcdabcd" into two groups: the first four digits, "abcd," and the last four digits, "abcd."

To determine the prime factors of this eight-digit number, we can first consider the prime factors of the four-digit number "abcd." Let's assume "abcd" has prime factors P1, P2, P3, ...

Since we have repeated the four-digit number, "abcd" appears twice consecutively, so the prime factors will also appear twice in the resulting eight-digit number.

Now, let's consider the special case where the four-digit number "abcd" is a perfect square. In this case, its prime factorization will have some prime factors appearing twice.

For example, if "abcd" is a square of a three-digit number, the prime factors of "abcd" will be those of the three-digit number repeated twice. This means that the eight-digit number will have prime factors repeated four times, making the number only possess two distinct prime factors.

To solve for this special case, we need to find a perfect square four-digit number "abcd" such that its prime factorization only contains prime factors appearing twice.

The only perfect square four-digit number with this property is 7744, which is the square of 88. Its prime factorization consists of only two different prime factors, 2 and 11, which appear twice.

Therefore, the number "abcd" is 7744.