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Math problem 3
Divided by 3, there are the following combinations: (1, 2,3), (1, 2,6), (1, 2,9), (2,3,4), (2,3,7), (3,4).

Each combination has six numbers, such as (1, 2,3)123,132,213,231,312,32/kloc-.

It can be seen that

The sum of hundreds is1* 2 * 3+2 * 5+4 * 2 * 5+5 * 2 * 4+6 * 2 * 4+7 * 2 * 4+8 * 2 * 3+9 * 2 * 3 = 654.

The sum of ten digits is 342.

The sum of the numbers is 342.

Then the sum of three digits divisible by 3 is 342 *100+342 *10+342 = 37962.