Self Ref  math puzzle
Find a number ABCDEFGHIJ such that A is the count of how many 0's are in the number, B is the number of 1's, and so on.
Explanation
For other numbers of digits
n=1: no sequence possible n=2: no sequence possible n=3: no sequence possible n=4: 1210, 2020 n=5: 21200 n=6: no sequence possible n=7: 3211000 n=8: 42101000 n=9: 521001000 n=10: 6210001000 n>10: (n4), 2, 1, 0 * (n7), 1, 0, 0, 0
No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and (2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits.
I'll first prove that x_0 > n3 if n>4. Assume not, then this implies that at least four of the x_i with i>0 are nonzero. But then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9, but it isn't possible in this case (51111100000 isn't valid).
Now I'll prove that x_0 < n1. x_0 clearly can't equal n; assume x_0 = n1 ==> x_{n1} = 1 by (2) if n>3. Now only one of the remaining x_i may be nonzero, and we must have that x_0 + ... + x_n = n+1, but since x_0 + x_{n1} = n ==> the remaining x_i = 1 ==> by (2) that x_2 = 1. But this can't be, since x_{n1} = 1 ==> x_1>0. Now assuming x_0 = n2 we conclude that x_{n2} = 1 by (2) if n>5 ==> x_1 + ... + x_{n3} + x_{n1} + x_n = 2 and 1*x_1 + ... + (n3)*x_{n3} + (n1)*x_{n1} + n*x_n = 3 ==> x_1=1 and x_2=1, contradiction.
Case n>5:
We have that x_0 = n3 and if n>=7 ==> x_{n3}=1 ==> x_1=2 and x_2=1 by (1) and (2). For the case n=6 we see that x_{n3}=2 leads to an easy contradiction, and we get the same result. The cases n=4,5 are easy enough to handle, and lead to the two solutions above.
n=1: no sequence possible n=2: no sequence possible n=3: no sequence possible n=4: 1210, 2020 n=5: 21200 n=6: no sequence possible n=7: 3211000 n=8: 42101000 n=9: 521001000 n=10: 6210001000 n>10: (n4), 2, 1, 0 * (n7), 1, 0, 0, 0
No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and (2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits.
I'll first prove that x_0 > n3 if n>4. Assume not, then this implies that at least four of the x_i with i>0 are nonzero. But then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9, but it isn't possible in this case (51111100000 isn't valid).
Now I'll prove that x_0 < n1. x_0 clearly can't equal n; assume x_0 = n1 ==> x_{n1} = 1 by (2) if n>3. Now only one of the remaining x_i may be nonzero, and we must have that x_0 + ... + x_n = n+1, but since x_0 + x_{n1} = n ==> the remaining x_i = 1 ==> by (2) that x_2 = 1. But this can't be, since x_{n1} = 1 ==> x_1>0. Now assuming x_0 = n2 we conclude that x_{n2} = 1 by (2) if n>5 ==> x_1 + ... + x_{n3} + x_{n1} + x_n = 2 and 1*x_1 + ... + (n3)*x_{n3} + (n1)*x_{n1} + n*x_n = 3 ==> x_1=1 and x_2=1, contradiction.
Case n>5:
We have that x_0 = n3 and if n>=7 ==> x_{n3}=1 ==> x_1=2 and x_2=1 by (1) and (2). For the case n=6 we see that x_{n3}=2 leads to an easy contradiction, and we get the same result. The cases n=4,5 are easy enough to handle, and lead to the two solutions above.
