A naive approach of appending multiples of \(X\) to the end of \(S\) would exceed the length limit of \(S\), so we need to choose \((X, S)\) such that the multiples of \(X\) are efficiently embedded in \(S\).

One way to construct such an integer exploits the structure of repeating decimals. For example, consider \(1/7 = 0.142857142857\dots\). If we take the repeating segment to form the integer \(142857\) as \(X\), and let \(S\) be the integer formed by writing \(142857\) twice in a row, i.e. \(142857142857\), then we have:

• \(142857 \times 1 = 142857\) : \(\underline{\textbf{142857}}142857\)
• \(142857 \times 2 = 285714\) : \(14\underline{\textbf{285714}}2857\)
• \(142857 \times 3 = 428571\) : \(1\underline{\textbf{428571}}42857\)
• \(142857 \times 4 = 571428\) : \(1428\underline{\textbf{571428}}57\)
• \(142857 \times 5 = 714285\) : \(14285\underline{\textbf{714285}}7\)
• \(142857 \times 6 = 857142\) : \(142\underline{\textbf{857142}}857\)

Here, the products \(X \times 1\) through \(X \times 6\) are all embedded in \(S\).

In general, for a prime \(p\) such that the length of the repeating section of \(1/p\) is \(p-1\), one can apply the above idea to obtain \(X\) along with \(S\) in which \(X \times 1, X \times 2, \dots, X \times (p - 1)\) are all embedded.

It can be proved that the smallest positive integer \(a\) satisfying \(10^a \equiv 1 \pmod{p}\) is \(p-1\) if and only if the length of the repeating section of \(1/p\) is \(p-1\), so we can find such a prime \(p\) using a bit of straightforward computation. (Alternatively, you may directly compute \(1/p\) and check the length of its repeating segment.)

If we use such a prime \(p\) with \(1001 \leq p \leq 2500\) (for example, \(p=1019\)) to obtain \((X, S)\), we can satisfy all conditions in the problem statement.