Book IX, Proposition 35, proves that in a geometric series if the first term is subtracted from the second and last term in the sequence, then as the excess of the second is to the first—so is the excess of the last to all those before it. (This is a restatement of our formula for geometric series from above.) Applying this to the geometric progression 31, 62, 124, 248, 496 (which results from 1, 2, 4, 8, 16 by multiplying all terms by 31), we see that 62 minus 31 is to 31 as 496 minus 31 is to the sum of 31, 62, 124, 248. Therefore, the numbers 1, 2, 4, 8, 16, 31, 62, 124 and 248 add up to 496 and further these are all the numbers that divide 496. For suppose that divides 496 and it is not amongst these numbers. Assume is equal to , or 31 is to as is to 16. Now cannot divide 16 or it would be amongst the numbers 1, 2, 4, 8 or 16.