Question number 13. How many numbers are there which have a sum of all their digits in base 10 equal to 2010 and the product of these digits equal to 2? So let's try to write down an example number here. What we have is some placeholders for our digits. Let me just write some lines here and possibly several more here. And if we were to add them together these numbers, which we have to come up with later, their sum would be equal to 2010. So there is lots and lots of possibilities here. But we also know that their product, if we were to multiply them all together, each of these factors multiplied by the others would give us 2. And so the only possibilities here are that we use the number 1 as a factor and that we use the number 2 as a factor. We can use as many 1s as we like, but we may only use 1, 2. So let's write that down. There is only 1, 2 in these digits. And then what we see is that our number looks like this. We have 1s. We're not sure how many yet. So let's say dot, dot, dot, plus 1, plus 1. And then there is just that 1, 2. And the end result here is equal to 2010. So if we subtract the 2 from both sides, what we have is 1 plus 1, many 1s, plus 1 plus 1 is equal to 2008. And so here we have exactly 2008 terms. So our number has exactly 2,009 digits because we have the 2008 1s and then the 2. And the 2 that we know is there can be in any one of these spots. So the 2 can be placed in any one of the 2009 spots and the remaining 2008 digits are all 1s. So that's how many numbers are possible in our situation that have the sum of the digits equal 2010 and the product only equal to 2. And that is B, 2009.

digit sum

product condition

number positions

combinatorics

unique arrangements