A cute number theory problem!
Number Theory
Level
pending
In the series 2006 ,2007, 2008 ………. 4015 find the summation of the maximum odd divisor of every number?
The answer is 4035074.
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As stated in the problem,we need the max odd divisor of each number,we can't have the powers of 2 as the factor of every number.An interesting thing to notice here that all the odd numbers in the series are the max odd divisor of themselves.So we have to work with the even number.We can write all the even number in the series taking the powers of '2' like this: 2 * 1003,2 * 1005,2 * 1007,..........,2 * 2007, 4 * 503,4 * 505,........,4 * 1003, 8 * 251,.......,8 * 501, 16 * 127,..........,16 * 249, 32 * 63,........,32 * 125, 64 * 33,........,64 * 61, 128 * 17,.........,128 * 31, 256 * 9,.......,256 * 15, 512 * 5,....,512 * 7, 1024 * 3, 2048 * 1, This step is really boring.But the bright day is ahead!Now we have to sum up all the odd number stated in the previous step.The sum of odd numbers as a factor with '2' is (1004^2-501^2),like this,the sum of the odd number as a factor with the power of '2' are (502^2 -251^2),(251^2-125^2),(125^2-63^2),(63^2-31^2),(31^2-16^2),(16^2-8^2),(8^2-4^2),(4^2-2^2),3,1(as the sum of the first nth odd number is n^2).Adding all these we get a sum of (1004^2-501^2+502^2)=1009019.And the sum of the odd numbers from 2007 to 4015 in the series is 2008^2-1003^2=3026055.So the final answer is 1009019+3026055=4035074.Done!