## Trivia

Subject:
The single-digit solution
Response To:

Alex Y
This doesn't seem to be drawing any guesses or partial answers, so here is the solution and answer, repeating some from the previous hint.

the number of occurrences of 0 is _, of 1 is _, of 2 is _, of 3 is _, of 4 is _, of 5 is _, of 6 is _, of 7 is _, of 8 is _, and of 9 is _

1) the numbers in the blanks must add to 20.

2) the numbers in the blanks must all be at least 1

2a) Said another way, the numbers in the blanks can be stated as 1,1+a,1+b,1+c,1+d,1+e,1+f,1+g,1+h,1+i

2b) a+b+c+d+e+f+g+h+i = 10

3) a+1 = number of 1's in the sentence = 1 (from the original) + 1 (the number of 0's) + the number of zeros in a,b,c,d,e,f,g,h,i
or: a=1+number of zeros in a,b,c,d,e,f,g,h,i = 1+ 9-number of non-zeros in a,b,c,d,e,f,g,h,i
Since a is non-zero, a= 9-number of non-zeros in b,c,d,e,f,g,h,i
and a final rearrangement,

the number of non-zeros in b,c,d,e,f,g,h,i = 9-a

4) Using the relationships in (2) and (3), we can see what the other letters must equal for each possibility for a, and see which one works. As an example, if a=5, there are four non-zero numbers among b,c,d,e,f,g,h,i, whose sum is 5. The only way that can happen is for those numbers (in no particular order at this point) to be 2,1,1,1. If this is a solution, it means that 6,3,2,2,2 must be in the blanks, and the rest filled with 1's. But that means there will be 4 2's, but there is no 4 in this solution.

You can try other values for a and fail until you come up with a=6. 3 of b,c,d,e,f,g,h,i add to 4, meaning they are 2,1,1. In this case, the blanks in the original are filled with 7,3,2,2 and the rest 1's

the number of occurrences of 0 is 1, of 1 is 7, of 2 is 3, of 3 is 2, of 4 is 1, of 5 is 1, of 6 is 1, of 7 is 2, of 8 is 1, and of 9 is 1

That works!