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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?

Guest Sep 25, 2014

#1**+10 **

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!

geno3141 Sep 26, 2014

#1**+10 **

Best Answer

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!

geno3141 Sep 26, 2014