If a number is a multiple of 3(9), the sum of its digits must also be a multiple of 3(9).
1. Multiplier, multiplier and product are all multiples of 3. The product is a multiple of 9. The reason for this is the following：
When 1, one of the multiplicand and multiplier is a multiple of 3 and the other is not, the product should be a multiple of 3. But at this time, the sum of 8 numbers must not be a multiple of 3. Does not meet the problem.
2. When both the multiplicand and the multiplier are not multiples of 3: Let the multiplicand be 3a+ 1, the multiplier be 3b- 1, and the product after multiplication be 3ab+ 1b-3a- 1, but at this time the multiplicand+multiplier+product is not multiples of 3, which is inconsistent with the question. (A number is more than a multiple of 3 1, so the sum of its digits is also more than 3 1. ) Similarly, when the multiplicand is 3a+ 1 and the multiplier is 3b+ 1, it is also inconsistent with the problem.
Second, the product is a multiple of 9, and the sum of numbers can only be 18 or 27. But if it is 27 and the multiplier is 6, the sum of the multiplicand numbers is 36-27-6=3, which is impossible; When the multiplier is 3, the sum of the multiplicand numbers is 36-27-3=6, and the maximum three digits of the sum of numbers are 32 1, which is obviously unqualified. So the sum of products can only be 18.
3. When the multiplier is 3, the sum of the multiplicand numbers is 36- 18-3= 15, and the combination is 1 68,258,267,456, with 6 16 digits greater than 333 and/kloc-0. Like 6 18, 8 16, 528, 852, 672, 726, 762, they don't meet the requirements at first glance. Multiply the remaining six numbers. The answer is 582*3= 1746.
3. When the multiplier is 6, the sum of the multiplicand numbers is 36- 18-6= 12, and the combination is 138, 1 47,237,345, and the number of digits is not/.
The answer is 453*6=27 18.