Oh, dude, adding numbers from 1 to 500? That's like... easy math, man. You just use the formula for the sum of an arithmetic series: n/2 * (first number + last number). So, for this case, it's 500/2 * (1 + 500) = 250 * 501 = 125250. Boom, math wizard!
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-499
The sum of the first 500 counting numbers (1-500) is 125,001.
There's a trick to this one. Let 1 + 2 + ... + 500 be the sum of all integers 1 to 500, called X Now, imagine we want to add the sum the integers from 1 to 500 and to the sum of the integers from 1 to 500. This would give us 2*X we can write it as 1 + 2 + ... + 500 + 1 + 2 + ... + 500 But that's not particularly useful. What if we look at it as: 1 + 2 + ... + 499 + 500 + 500 + 499 + ... + 2 + 1 adding the numbers that are directly on top of each other, we get 500 + 1 = 501, 499 + 2 = 501... 1 + 500 = 501 Thus every term is 501, and we have 500 terms. So, we have 501*500 = 2X Thus X, the sum from 1 to 500, is (500*501)/2 = 125,250
There are 232 numbers between 1 and 500 that are divisible by 3 or 5.
500.5 Because: 1+2+3+.......998+999+1000 Add 1 to 1000 equaling 1001. This makes a pair. There are 500 more of these so 500 pairs. So 1001x500= 500500 Finally divide it by 1000 because there are 1000 numbers. (1000-1=999+1=1000) Concluding the answer to be 500.5 :)
-499
500
Sum of first n numbers = n/2(n +1) = 500 x 1001 = 500500
Add them up.
The sum of the first 500 counting numbers (1-500) is 125,001.
There are 63 numbers 1 to 500 that are divisible by six but not by eight.
There's a trick to this one. Let 1 + 2 + ... + 500 be the sum of all integers 1 to 500, called X Now, imagine we want to add the sum the integers from 1 to 500 and to the sum of the integers from 1 to 500. This would give us 2*X we can write it as 1 + 2 + ... + 500 + 1 + 2 + ... + 500 But that's not particularly useful. What if we look at it as: 1 + 2 + ... + 499 + 500 + 500 + 499 + ... + 2 + 1 adding the numbers that are directly on top of each other, we get 500 + 1 = 501, 499 + 2 = 501... 1 + 500 = 501 Thus every term is 501, and we have 500 terms. So, we have 501*500 = 2X Thus X, the sum from 1 to 500, is (500*501)/2 = 125,250
1 000 500
if you add 1+1 it = 2 get it :D
There are 232 numbers between 1 and 500 that are divisible by 3 or 5.
500.5 Because: 1+2+3+.......998+999+1000 Add 1 to 1000 equaling 1001. This makes a pair. There are 500 more of these so 500 pairs. So 1001x500= 500500 Finally divide it by 1000 because there are 1000 numbers. (1000-1=999+1=1000) Concluding the answer to be 500.5 :)
There are 95 Prime #'s between 1 and 500