Question 333214
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*[tex \LARGE \ 911\ \ \ \ \ \ \ \ \ 43]


Double the first column, halve the second, always rounding down.  If the second column number is even, cross out the row.


*[tex \LARGE \ 1822\ \ \ \ \ \ \ \ \ 21]


*[tex \LARGE \ xxxxxxxxxxxxxxxxxxx3644\ \ \ \ \ \ \ \ \ 10] Cross it out.


*[tex \LARGE \ 7288\ \ \ \ \ \ \ \ \ 5]


*[tex \LARGE \ xxxxxxxxxxxxxxxxxxx14576\ \ \ \ \ \ \ \ \ 2] Cross it out.


*[tex \LARGE \ 29152\ \ \ \ \ \ \ \ \ 1] Stop because you reached 1 in the 2nd column

Add up the numbers in the 1st column that are not crossed out.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 29152\ +\ 7288\ +\ 1822\ +\ 911\ =\ 39173]


Actually, it doesn't matter which number you put into column 1, except that the larger the number in column 2, the more steps it takes to get halved down to 1, and therefore the more terms to add in the final step.  Use your smallest number in column 2 to do the least amount of work.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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