(In reply to
Solution by Dej Mar)
As 12222...2221 has 2003 digits, dividing by 11 gives a quotient of 111...111 with 2002 digits, not 2001. If it had 2001 digits, the odd and even wouldn't balance, and you'd be left with an excess equal to 1, which is not divisible by 11.
When that 2002-digit repunit is divided by a second 11, you'd get 1010...0101 with 2001 digits (not 1001). Of these, 1001 are 1's, all with the same parity position, so the sum is 1001, which is divisible by 11. That's the third time it's divisible by 11, so the original number in question is divisible by at least 11^3.
Then the question becomes whether the quotient of 1010...0101 divided by 11 is itself divisible by 11. That number has 1999 digits, but it could best be considered with three leading zeros, as the cycle length of its repeated digits is 22:
0009182736455463728191...
With the three leading zeros the length is 2002, and so consists of 2002/22 = 91 times through this cycle of 22 digits. The alternating sums through one cycle are 45 and 46, for a difference of 1. As there are 91 times through the cycle, the same difference in alternate places is 91 for the whole number, which is not divisible by 11, and so we are left with 3 as the highest power of 11 by which the number is divisible.
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Posted by Charlie
on 2009-08-01 16:25:37 |
re: Solution
