10. (7 points) In the multiplication problem below, each box must be filled with a digit; each Chinese character represents a digit, different characters represent different digits, and the same character represents the same digit. Therefore, the final product of this multiplication problem is . $\qquad$

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【Solution】Solution: According to the problem, we know that,
, $1 \times 8=8$, the unit digit of the multiplicand is 8;
From the vertical calculation $9+9+8=26$, we can deduce that the digit represented by "年" is 6 or 7;
Assuming "年" $=7$, because
, there is no one-digit number that when multiplied by 8 ends with 7, which does not meet the problem's requirements;
Therefore, the digit represented by "年" can only be 6, and "新" + "年" < 10, i.e., "新" + 6 < 10, "新" < 4; , $48 \times 2=96, 28 \times 7=196$, we can deduce that "贺" is 2 or 7, when "贺" is 2, "恭" is 4, the hundred's digit of the multiplicand is greater than 4, because
, the result of the hundred's digit of the multiplicand multiplied by the unit's digit of the multiplier is
9 or 9 plus the carry, only $9 \times 1=9$, so the hundred's digit of the multiplicand is 9, and the unit's digit of the multiplier is 1, because $1 \times 948=9$ "新",
"新" $=4$, which does not match the problem's requirements; when "贺" is 7, "恭" is 2, the hundred's digit of the multiplicand is greater than
$1, 228 \times 4=912$, which meets the problem's requirements, the multiplicand is 228, the multiplier is 174, the vertical calculation is:
$$
228
$$
$\times 174$
912
1596
228
39672

Therefore, the final product of this multiplication problem is 39672.
The answer is: 39672.