99, 100, and 101
The integers are 99, 100 and 101. There is also a set of consecutive even integers whose sum is 300. That set is 98, 100 and 102.
The sum of any two consecutive integers must be an odd number. So this question is incorrect.
-3,-4,-5
The integers are 11, 13 and 15.
The sum of the integers from 1 to 100 inclusive is 5,050.
They are 2n+2
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11
99, 100, and 101
101
To find the sum of the first 100 integers, you first add 1 plus 100 (the first and last numbers of the set) and get 101. Do the same with the next two integers, 2 and 99 and you'll get 101. Since you are adding two integers at a time and there are 100 integers between 1 and 100, you'll get 101, fifty times. Therefore, a shortcut would be to simply multiply 101 times 50 and get 5,050.
The integers are 99, 100 and 101. There is also a set of consecutive even integers whose sum is 300. That set is 98, 100 and 102.
It is 2500.
The sum of all the digits of all the positive integers that are less than 100 is 4,950.
The sum of the squares of two consecutive positive even integers is 340. Find the integers.
It is 100*(100+1)/2 = 50500.
The following will sum all integers from 1-100. Adjust according to your needs.int sum = 0;for (int i = 1; i