Q.2 . Find the product of 101*10001*100000001*...*(1000...01) where the last factor has  zeroes between the ones. Find the number of ones in the product.

 

Solution

 

  We know,

                 101 x 10001 = 1010101

                 10001 has 2^2 -1 zeroes between the ones.

                  The product has 2^2-1 zeroes alternately placed between 2^2 ones. It starts with one and ends with one.

     

                  1010101 x 100000001 = 101010101010101

                  100000001 has 2^3 -1 zeroes between the ones.

                  The product has 2^3-1 zeroes alternately placed between 2^3 ones. It starts with one and ends with one.

 

   

                  This multiplication pattern shows that, for such a multiplication series till 1000...001 with 2^n-1 zeroes, the product will have 2^n-1 zeroes alternately        

                 placed between 2^n ones.(As above)

 

       

                     Therefore the product of the series till 1000...001 having 2^7-1 zeroes would be,

                      10101010........01 with 2^7-1 zeroes alternately placed between 2^7 ones. 

 

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