Rand's found an answer, and the technique he used is very effective for this sort of puzzle, but perhaps it's worth seeing whether his is the only answer. So let's go at it from first principles.
The total number of digits is 16. Therefore the sum of all the digits is 16. Therefore the sum of (digit*position) is 16. Therefore the second half of the number contains at most one nonzero digit. If it contains none then we have 8 zeros and therefore at least one digit >= 8, contradiction, so the second half of the number contains exactly one nonzero digit and (equivalently) there is exactly one digit that occurs at least 8 times.
If we have seven or more nonzero digits then the sum of (digit*position) is at least 0+1+2+3+4+5+6=21>16, contradiction; so in fact there are at most six non0s and at least nine 0s (and so the first digit is at least 9). In fact, if we have six or more nonzero digits then, since one of them is at least 9, that sum is at least 0+1+2+3+4+9=19, contradiction. So there are at least eleven 0s and the first digit is at least B. Now, if we have five or more nonzero digits then since one of them is at least B the sum is at least 0+1+2+3+11=17, still impossible. So in fact there are at least twelve 0s and at most four nonzeros. (One of which is of course a C or more in position 0.)
At this point we've reduced the possibilities enough for a computer search to be effective, but it would be nice to finish it off by hand. Let's see. There are at most four nonzeros, so no nonzero digit can occur >4 times, so no position >4 (other than whichever one counts the zeros) is nonzero. If position 4 is nonzerothen there is at least one 4, so at most three of anything else nonzero, and clearly there cannot be four 4s, so nothing occurs 4 times, contradiction. So now we have at most positions 0,1,2,3 nonzero, together with whatever position counts the 0s.
There is exactly one of whichever digit counts the 0s, so position 1 is occupied. What by? Not a 1 because then we have exactly one 1 and exactly one whatever-counts-zeros, hence at least two 1s, contradiction; hence by a 2 or a 3. If by a 3 then we must have three of something, but whatever it is will make our total too high. So position 1 contains a 2, there are exactly two 1s, and now we have 1+1+2+[>=C]+[others]=16, so position 0 contains a C and other positions are full of zeros. And now we are done because we have all the counts.
Specifically, we have twelve 0s, two 1s, one 2, and one C, and our number is C210,0000,0000,1000. In other words, Rand's solution is the only one.