2. (8 points) In the vertical multiplication shown in the figure, the sum of the two multipliers that makes the product the smallest is $\qquad$

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【Solution】Solution:
(1) The highest digit of the product is 2, which indicates that the highest digit of the products calculated in the first two steps is 1. From this, we can deduce that the first digit of the first multiplicand is 1, and the last digit is 3;
(2) Based on the unit digit of the product being 1, we can determine that the unit digit of the product of the unit digits of the two multiplicands is 1. Combining this with the fact that the unit digit of the first multiplicand is 3, we can determine that the unit digit of the second multiplicand is 7;
(3) Since the product of the first multiplicand and the tens digit of the second multiplicand is over one hundred, we can determine that the tens digit of the second multiplicand is 1;
(4) Based on the product of the first multiplicand and 7 being one thousand and a few, we can deduce that the tens digit of the first multiplicand is 4. Therefore, the two multiplicands in this problem are 143 and 17, the first calculated product is 1001, the second product is 143, and the final product is 2431.

Thus, the sum of these two multiplicands is $143+17=160$.