11*101*10001*100000001*10000000000000001*100000000000000000000000000000001...
If factor number m has n zeros, then factor number m+1 will have 2n+1 zeros.
The product will always be a repunit with 2^m decimal digits, right?
+23252 I am not certain of the use of "repunit" in this problem.
The factors are (10^1 + 1)(10^2 + 1)(10^4 + 1)(10^8 + 1)(10^16 + 1)(10^32 + 1)...
The number of digits in the answer will be the sum of the powers of 10 in the product plus 1.
1 factor --> 2 digits --> 2^1
2 factors --> 4 digits --> 2^2
3 factors --> 8 digits --> 2^3
4 factors --> 16 digits --> 2^4
5 factors --> 32 digits --> 2^5
Yes, you are correct!
+23252 I am not certain of the use of "repunit" in this problem.
The factors are (10^1 + 1)(10^2 + 1)(10^4 + 1)(10^8 + 1)(10^16 + 1)(10^32 + 1)...
The number of digits in the answer will be the sum of the powers of 10 in the product plus 1.
1 factor --> 2 digits --> 2^1
2 factors --> 4 digits --> 2^2
3 factors --> 8 digits --> 2^3
4 factors --> 16 digits --> 2^4
5 factors --> 32 digits --> 2^5
Yes, you are correct!
