6. In the number puzzle below, each letter represents a number. Different letters represent different numbers, and the same letter represents the same number. What is $\mathrm{T}$?



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【Analysis】Since the ten-thousands place $\mathrm{F}$ and $\mathrm{S}$ are different, it is easy to see that there is a carry-over to the thousands place.
Since the units place $\mathrm{Y}$ remains unchanged, $\mathrm{N}$ must be 0 or 5.
If $\mathrm{N}$ is 5, then the tens place $\mathrm{T}+2\mathrm{E}+1$ would still be $\mathrm{T}$, which is clearly impossible.
Therefore, $\mathrm{N}$ is 0, and thus there is no carry-over from the units place.
Observing the tens place, similarly, since $\mathrm{T}$ remains unchanged, $\mathrm{E}$ must be 0 or 5. Since $\mathrm{N}$ is already 0, $\mathrm{E}$ must be 5.
Observing the hundreds place, since the thousands place digit has changed, there must be a carry-over from the hundreds place, which could be 1 or 2.
If the hundreds place carries over 1, and the thousands place also has a carry-over, then $\mathrm{O}$ must be 9, making $\mathrm{I}$ 0, which is a repetition.
Therefore, the hundreds place carries over 2, and $\mathrm{O}$ cannot be 8, otherwise $\mathrm{I}$ would be 0. So $\mathrm{O}$ is 9, and $\mathrm{I}$ is 1.
Since $\mathrm{R}+2\mathrm{T}+1$ exceeds 20, and 9 has already been used,
$\mathrm{T}$ must be at least 6.
If $\mathrm{T}$ is 6, then $\mathrm{R}$ is 7 or 8. If $\mathrm{R}$ is 7, then $\mathrm{X}$ is 0, which is a repetition; if $\mathrm{R}$ is 8, then $\mathrm{X}$ is 1, which is also a repetition. Therefore, $\mathrm{T}$ is not 6.
If $\mathrm{T}$ is 7, then $\mathrm{R}$ is 6 or 8 (5 and 7 are already used). If $\mathrm{R}$ is 6, then $\mathrm{X}$ is 1, which is a repetition; if $\mathrm{R}$ is 8, then $\mathrm{X}$ is 3. Since 0, 1, 3, 5, 7, 8, and 9 are already used, only 2, 4, and 6 remain, and $\mathrm{S}$ is only 1 more than $\mathrm{F}$, which cannot be satisfied. Therefore, $\mathrm{T}$ is not 7.
If $\mathrm{T}$ is 8, then $\mathrm{R}$ is 3, 4, 6, or 7. If $\mathrm{R}$ is 3, then $\mathrm{X}$ is 0, which is a repetition; if $\mathrm{R}$ is 4, then $\mathrm{X}$ is 1, which is a repetition; if $\mathrm{R}$ is 6, then $\mathrm{X}$ is 3, and only 2, 4, and 7 remain, which cannot satisfy $\mathrm{S}$ and $\mathrm{F}$. Therefore, $\mathrm{R}$ is 7, and $\mathrm{X}$ is 4. Only 2, 3, and 6 remain, so $\mathrm{F}$ is 2, $\mathrm{S}$ is 3, and $\mathrm{Y}$ is 6.
In summary, $\mathrm{T}$ is 8.