Form the smallest seven-digit number, completely divisible by 99, by using the digits 0, 2, 4, 6, 8 each at least once. Give the detailed justification.

The number must be divisible by both 9 and 11. The digits 0, 2, 4, 6, 8 add up to 20, and the two extra digits are also even, so the digit sum is an even multiple of 9, that is 36, which forces the extra digits to be 8 and 8; the digits are 0, 2, 4, 6, 8, 8, 8. For divisibility by 11, the sums of the digits in odd places $$(O)$$ and even places $$(E)$$ satisfy $$O + E = 36$$ with $$O - E$$ a multiple of 11, and since both sums are even, $$O = E = 18$$. The three even-place digits must add to 18, and the only such choice from $$\{0, 4, 6, 8, 8, 8\}$$ (after using 2 as the smallest possible leading digit) is $$\{4, 6, 8\}$$, leaving $$2, 0, 8, 8$$ for the odd places. Arranging each group in increasing order gives the smallest number 2406888, and indeed $$2406888 = 99 \times 24312$$.