14. Feifei separately counted the number of divisors of $\mathrm{A}$, the number of divisors of $\mathrm{A}$'s 2 times, the number of divisors of $\mathrm{A}$'s 3 times, ..., and the number of divisors of $\mathrm{A}$'s 10 times, and obtained the following table. If there is only one incorrect statistic in this table, then the integer $\mathbf{A}$ is
【Analysis】Considering the number of divisors of $9 \times \mathrm{A}$ is 49, then $9 \times \mathrm{A}$ is either the 48th power of some prime number, or the 6th power of one prime number multiplied by the 6th power of another prime number.
If this result is correct, then in the first case, $\mathrm{A}$ would be $3^{46}$, which clearly makes $\mathrm{A}$ and $3 \times \mathrm{A}$ incorrect, and it is impossible for there to be only one statistical error; if in the second case, then $\mathrm{A}$ would be $3^{4} \times p^{6}$, which clearly makes $\mathrm{A}$ and $2 \times \mathrm{A}$ incorrect, and it is impossible for there to be only one statistical error;
Therefore, the number of divisors of $9 \times \mathrm{A}$ being 49 is incorrect.
If $\mathrm{A}$ does not contain the prime factor 2, then the number of divisors of $2 \times \mathrm{A}$ should be twice the number of divisors of $\mathrm{A}$, but 30 is not twice 24, so $\mathrm{A}$ must contain the prime factor 2.
Similarly, $\mathrm{A}$ contains the prime factors $3$ and $7$.
It is also observed that the number of divisors of $9 \times \mathrm{A}$ is 42, and 42 can be written as the product of at most 3 integers greater than 1, i.e., $2 \times 3 \times 7$, so the number of prime factors of $\mathrm{A}$ cannot exceed 3.
In summary, $\mathrm{A}$ has exactly 3 prime factors: $2$, $3$, and $7$.
Considering the number of divisors of $2 \times \mathrm{A}$ is 30, and 30 can only be written as the product of 3 integers greater than 1 in one way, i.e., $2 \times 3 \times 5$, we examine $2 \times 2 \times 5=20$ and $2 \times 3 \times 4=24$, and conclude that the exponent of 2 in $\mathrm{A}$ is $4-1=3$.
Since the exponent of 2 in $\mathrm{A}$ is 3, considering the number of divisors of $3 \times \mathrm{A}$ is 36, and 36 can only be written as the product of 3 integers greater than 1 in one way, and it must include 4, i.e., $4 \times 3 \times 3$, and $4 \times 3 \times 2=24$, we conclude that the exponent of 3 in $\mathrm{A}$ is $2-1=1$.
Since the exponent of 2 in $\mathrm{A}$ is 3 and the exponent of 3 is 1, it is easy to see that the exponent of 7 is $24 \div (3+1) \div (1+1) - 1 = 2$.
Thus, the required $\mathrm{A}$ is $2^{3} \times 3 \times 7^{2} = 1176$.