There is a simple formula to solve these type of equations. If the smallest digit being counted is 1, then the formula is (n x (n-1))/2. (x means multiply) Therefore, (1000 x 999)/2 = 499500
The sum of the integers from 1 to 100 inclusive is 5,050.
The sum of integers from 1 to 2008 = 2008*2009/2 = 2017063
1 + 1 = 2 The sum of the digits is therefore 2.
55
To calculate the sum of the numbers 1 to n, the formula is: sum = n(1 + n) / 2 So, an equation to find the sum of the integers 1 to 2010 is: sum = 2010 x (1 + 2010) / 2
There is only 1, the number 54.
The sum of all the the integers between 1 and 2008 (2 through 2,007) is 2,017,036.
The answer is 28 054
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.
Using Gauss's method, 1+2+3...1000= 500x1001=500500 Answer:500500
The sum of the integers from 1 to 100 inclusive is 5,050.
The sum of integers from 1 to 2008 = 2008*2009/2 = 2017063
The sum of all integers from 1 to 20 inclusive is 210.
1, 2, 4, 5, 8, and 10, so only 6 digits.
The sum of the integers from 1 through 300 is 44,850.
No.Let the four numbers be (n-1), n, (n+1), (n+2).Their sum is 4n + 2 = 2(2n + 1)If 2000 is the sum of four consecutive integers, then:2(2n+1) = 2000⇒ 2n + 1 = 1000⇒ 2n = 999but 999 is odd, not even and so n cannot be an integer; therefore 2000 is not the sum of four consecutive integers.