Question number 28. How many three-digit numbers have the property that the middle digit is the arithmetic mean of the other two digits? We can make some notes on this. We can make an example number like that with three digits. Let's label them A, B, and C. And so we know that these are digits, so A, B, and C belong to the set of numbers containing 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0 also. And what we then also note is that A is not equal to zero. We know that B is not equal to zero because B is, in fact, equal to the average of A plus C. And if A is not equal to zero, even though C could be zero, the average will not come out as zero. And then we count. We know that with this relationship here, 2B is equal to A plus C, and then so then we count. If B is equal to 1, then what happens is there are two possibilities. We can have A is equal to 1 and C is equal to 1, or we can have A is equal to 2 and C is equal to 0. So then we have two possibilities here, and then we have to do basically the other cases counting in a similar way. If B is equal to 2, then the possible numbers would be 123, and then also 222, and reverse the first one for 321, and there is one more, 420. So we have four possibilities here, and we keep going like that. If B is equal to 3, then there are six possibilities, and we can just say 6, 3, 0. The average of 6 and 0 is 3. Then we decrease that to 5, 3, 1. The average of 5 and 1 is also 3. Then we have 1, 3, 5, 2, 3, 4, 4, 3, 2, and 3, 3, 3. So here we have exactly six possibilities. When B is equal to 4, there are eight possibilities, 8, 4, 0, 7, 4, 1, 1, 4, 7, 6, 4, 2, 2, 4, 6. Then we start reversing them, 2, 4, 6, 3, 4, 5, 5, 4, 3, and 4, 3, 4. Rather, 4, 4, 4. Excuse me, 4, 4, and 4. And so that's eight. When B is equal to 5, there are nine possibilities. I will not write them down. When B is equal to 6, we go through the possibilities, and there are seven. So seven numbers. When B is equal to 7, then we have five possibilities. When we get to 8, we have three possibilities. And finally, so let's write them down. When B is equal to 8, we have 9, 8, 7, 7, 8, 9, and then of course 8, 8, 8. And finally, so that's 3. And finally, when B is equal to 9, we only have one possibility, and that's 9, 9, and 9. So that's one possibility. And so here, for the ones I didn't write down, for lack of space, I can refer you to the suggested solutions. And that is problem 28 in the suggested solutions. All the possibilities are written out. And then if we add up these numbers, 2, 4, 6, 8, 1, 3, 9, 7, 5, their sum is 45. And that counts all the numbers whose arithmetic mean appears in the middle digit, all three-digit numbers with that property. So the answer here is E45.

three-digit numbers

arithmetic mean

digit conditions

valid numbers

number combinations