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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?

 Sep 25, 2014

Best Answer 

 #1
avatar+17774 
+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!

 Sep 26, 2014
 #1
avatar+17774 
+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
 #2
avatar+101149 
0

Nice one, geno !!!

 

 Sep 26, 2014
 #3
avatar
0

Thx for your answer!

 

What I meant with "repunit" is a number where alle the digits are 1, such as 11, 1111 and 111111111111.

 Sep 26, 2014

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