Let N(x) be the number 122....221 where the digit 2 occurs x times.
Twice in the
past we have determined the highest power of 11 that divides N(2001) is 11^3.
What is the smallest x for N(x) to be a multiple of 11^3? What about multiples of 11^4 and 11^5?
The numbers N(x) have an interesting property. If N(a) and N(b) both have f as a factor then so does N(a+b+1). This can be seen by the identity N(a)*10^(b+1) + N(b) = N(a+b+1). (
1) Since N(a) and N(b) both have f as a factor, f can be factored out of the left side of the identity which implies that N(a+b+1) also has f as a factor.
Let a=b and f=N(a). Then from (1) N(2a+1) is a multiple of N(a). Specifically N(2a+1) = N(a) * [10^(a+1) + 1]. (2)
Repeated application of (1) yields N(m*a+m-1) = N(a)*[10^((m-1)*(a+1)) + 10^((m-2)*(a+1)) + 10^(2*(a+1)) + 10^(a+1) + 1]. (3)
If a is even then the second term of the product of (2) is a multiple of 11. This then implies that N(2a+1) has at least one more factor of 11 than N(a). (4)
If a is odd and m is 11 then the second term of the product of (3) is a multiple of 11. This then implies that N(11a+10) has at least one more factor of 11 than N(a). (5)
Apply (4) to a=0. This shows N(1) has one more factor of 11 than N(0). Since N(0)=11 then N(1) is a multiple of 11^2
Apply (5) to a=1. This shows N(21) has one more factor of 11 than N(1). Since N(1) is a multiple of 11^2 then N(21) is a multiple of 11^3. (5) can be repeatedly applied to previously generated values to create a series where the nth term is a value such that N(n) is a multiple of 11^n. The series begins 0, 1, 21, 241, 2661, 29281, ...
Charlie's program proves by brute force the first five terms are the smallest for powers 1-5. An small modification proves N(29281) is the minimal term which is a power of 11^6. I suspect this sequence generates the minimal values for which N(x) is a specific power of 11.
Analytic Solution
