3. Between the digits of the number 987654321, signs + should be placed so that the sum is 99. In how many ways can this be achieved?
3. Answer: in two ways.

It is clear that there cannot be three-digit addends, and there must be at least one two-digit addend, since the sum of all digits is 45. Let there be one two-digit addend, and we group the digits \(a+1\) and \(a\). Then the two-digit addend is \(10(a+1) + a = 11a + 10\), and the sum of the remaining digits is \(45 - (a+1) - a = 44 - 2a\). In total, we get \(11a + 10 + 44 - 2a = 99\), which means \(a = 5\). This gives the solution \(9 + 8 + 7 + 65 + 4 + 3 + 2 + 1 = 99\).

Now let there be two two-digit addends. We will assume that in one case we group the digits \(b+1\) and \(b\), and in the other \(a+1\) and \(a\), where \(b > a+1\). The two-digit numbers are \(11b + 10\) and \(11a + 10\) respectively, and the sum of the single-digit addends is \(45 - (2b + 1) - (2a + 1) = 43 - 2(a + b)\). Together we have \(11(a + b) + 20 + 43 - 2(a + b) = 99\), from which \(a + b = 4\). Only one option fits, \(a = 1\), \(b = 3\), which gives the second solution \(9 + 8 + 7 + 6 + 5 + 43 + 21 = 99\).

There are no more solutions, as the sum of three two-digit addends will be no less than \(65 + 43 + 21\), which exceeds 99.