is the answer 1287??

No, the set $ \{1001, 1008, 1053, 1100 \}$ fullfils that condition but probably there is another optimal one. Another "small" set is $ \{1001, 1008, 1014, 1053, 1056 \}$

mszew's example of $ \{1001, 1008, 1014, 1053, 1056\}$ is optimal. Since $ 1001 = 7 \cdot 11 \cdot 13$, $ A$ must contain another multiple of 7, another multiple of 11, and another multiple of 13. We list all multiple of 7, 11, and 13 between 1002 and 1056, and their prime factorizations: \begin{align*} 1001 &= 7 \cdot 11 \cdot 13, \\ 1008 &= 7 \cdot 2^4 \cdot 3^2, \\ 1015 &= 7 \cdot 5 \cdot 29, \\ 1022 &= 7 \cdot 2 \cdot 73, \\ 1029 &= 7^3 \cdot 3, \\ 1036 &= 7 \cdot 2^2 \cdot 37, \\ 1043 &= 7 \cdot 149, \\ 1050 &= 7 \cdot 2 \cdot 3 \cdot 5^2, \\ 1057 &= 7 \cdot 151, \\ 1012 &= 11 \cdot 2^2 \cdot 23, \\ 1023 &= 11 \cdot 3 \cdot 31, \\ 1034 &= 11 \cdot 2 \cdot 47, \\ 1045 &= 11 \cdot 5 \cdot 19, \\ 1056 &= 11 \cdot 2^5 \cdot 3, \\ 1014 &= 13^2 \cdot 2 \cdot 3, \\ 1027 &= 13 \cdot 79, \\ 1040 &= 13 \cdot 2^4 \cdot 5, \\ 1053 &= 13 \cdot 3^4. \end{align*} We see that no multiple of 11 smaller than 1056 will lead to a better bound on $ A$. For example, the number 1012 contains the prime factor 23, and no other number in the list above contains another prime factor of 23. Therefore, the bound of 1056 is optimal.

How about $ \{1001,1008,1012,1035,1040\}$?

nsato wrote: mszew's example of $ \{1001, 1008, 1014, 1053, 1056\}$ is optimal. Since $ 1001 = 7 \cdot 11 \cdot 13$, $ A$ must contain another multiple of 7, another multiple of 11, and another multiple of 13. We list all multiple of 7, 11, and 13 between 1002 and 1056, and their prime factorizations: \begin{align*} 1001 & = 7 \cdot 11 \cdot 13, \\ 1008 & = 7 \cdot 2^4 \cdot 3^2, \\ 1015 & = 7 \cdot 5 \cdot 29, \\ 1022 & = 7 \cdot 2 \cdot 73, \\ 1029 & = 7^3 \cdot 3, \\ 1036 & = 7 \cdot 2^2 \cdot 37, \\ 1043 & = 7 \cdot 149, \\ 1050 & = 7 \cdot 2 \cdot 3 \cdot 5^2, \\ 1057 & = 7 \cdot 151, \\ 1012 & = 11 \cdot 2^2 \cdot 23, \\ 1023 & = 11 \cdot 3 \cdot 31, \\ 1034 & = 11 \cdot 2 \cdot 47, \\ 1045 & = 11 \cdot 5 \cdot 19, \\ 1056 & = 11 \cdot 2^5 \cdot 3, \\ 1014 & = 13^2 \cdot 2 \cdot 3, \\ 1027 & = 13 \cdot 79, \\ 1040 & = 13 \cdot 2^4 \cdot 5, \\ 1053 & = 13 \cdot 3^4. \end{align*} We see that no multiple of 11 smaller than 1056 will lead to a better bound on $ A$. For example, the number 1012 contains the prime factor 23, and no other number in the list above contains another prime factor of 23. Therefore, the bound of 1056 is optimal. Who says that every number in the set must be a multiple of either 7, 11 or 13 ?

The number should be 1040. We need another multiple of 13. $ 1014=2*3*13*13$ does not help. $ 1027=13*79$ will ask for another multiple of $ 79$, a number greater than 1040.

Ok, thanks for the correction.