Recreational Mathematics?

A. Yep, your formula is right. It comes from the fact that, in an evenly spaced set of numbers, the sum of the 1st number and the last will equal the sum of the second number and the penultimate one, will equal the sum of the third and the antepenultimate, which will equal the sum of the fourth and the preantepenultimate, and so forth.

So 1000 + 1 = 999 + 2 = 998 + 3 = . . .

Each of those sums, in the case of a set of consecutive numbers starting with one, is L + 1, so we take L + 1 and multiply it by the number of such sums, which is half of L.

There is a story, and it could be apocryphal, that Carl Friedrich Gauss, as a young boy, was assigned to add up the numbers from one to one hundred as a punishment for misbehaving in class. He thought for a moment and then answered "5050." His teacher was sure Gauss was pulling his leg, until Gauss explained how he got that answer. (When I heard the story, I wasn't told whether that got Gauss of the hook or if his teacher beat him for being a smart ***. *grin*)

B. A pitcher could theoretically only throw 27 pitches, because he could get each player to pop up.

But I'm going to go one better than that: I'd say a pitcher could pitch a complete 9 inning game could do so while only throwing 25 pitches, and here's how: he could be the losing pitcher. He could get 24 batters to each pop out with one pitch, and, somewhere along the way, give up a single first pitch home run. If his opponent throws a shut-out, and if he's the visiting pitcher, he won't have to pitch the bottom of the ninth, because he will have already lost. It's still a 9 inning game, though, even if you don't play the bottom of the ninth. After all, the outcome is already decided by that point.

Or heck, he could give up that single pitch home run in the bottom of the ninth in a zero zero game, and then he would have pitched in all nine innings, and that home run would be a walk-off.

:)

B sounds right, but i don't Know the rules of baseball.

it could be 27 pitch and hit and caughtout

Can you add all the numbers from 1 to 1000 in one multiplication?

i came across a formula but I'm not sure if this is right

S = L (L+1) / 2

S= 1000(1000+1)/ 2 = 500500 ???

this is correct

you can see this by spliting the 1000 numbers into 500 pairs.

1 + 1000 = 1001

2+ 999 = 1001

etc

500+ 501 = 1001

500 * 1001 = 500500