a = 111...111 (a string of 2012 ones)
What is the 73rd digit from the end of a^{2}?
If you were to do out the long multiplication of a*a
the last digit would be 1
the second to last digit would be 1+1 = 2
the third to last digits would be 1+1+1 = 3
the nth to last digit would be the last digit of n plus any carries
the 73rd digit would be 3 plus a 7 carried from the 72nd digit. There would almost be another 1 carried from the 71st digit but the 72nd digit is just 2 plus 7. So
the solution is 0.The 74th digit would end as 7 plus 4
plus the extra 1 carried up from the 7 plus 3 at the 73rd place. This is the first extra carry so the few digits beginning with the 74th are
2
098765432...

Posted by Jer
on 20130129 12:22:18 