# If you add all the consecutive numbers is one of the totals 1000?

No. There is no number n, such that the sum of the first n consecutive integers is 1,000. Here's why: Let S = S(n) = 1 + 2 + 3 + . . . + n. Then, S = n(n + 1) / 2. There are several ways you can derive this formula; but you might find it helpful to refer to this: http://mathworld.wolfram.com/ArithmeticSeries.html or any of a large number of similar sources. (I tried Wiki, but down-loading took forever!) If you know S(n) (in this case, 1,000) and wish to find n, you have only to solve: n² + n = 2S = 2000. This is not readily factorised; but we can try completing the square: 2S = n² + n = 2,000; whence, n² + n + ¼ = (n + ½)² = 2,000¼, and n = √2,000.25 - 0.5. To the nearest integer, n = 44, approximately; but this is inexact, of course. Let's see what happens when we try this result. We get 2S(44) = n² + n = 44 * 45 = 1,980; whence, S(45) = 990, which is just a little too small; and any n smaller than 44 would give an S(n) that is even smaller. You can easily check this result, if you like: 1 + 2 + 3 + . . . + 22 + 23 + . . . + 42 + 43 + 44 = (1 + 44) + (2 + 43) + (3 + 42) + . . . + (22 + 23) {this makes 22 pairs, each adding up to 45.} = 45 * 22 = 990. On the other hand, S(45) = S(44) + 45 = 990 + 45 = 1,035, which is just a little too large; and any n larger than 45 would give an S(n) that is even larger. We may conclude, then, that no integer n satisfies the requirement that we have placed on it.