For the product to be zero, one of the numbers must be 0. So the question is to find the maximum sum for fifteen consecutive whole numbers, INCLUDING 0. This is clearly achived by the numbers 0 to 14 (inclusive), whose sum is 105.
The sum of the integers between 101 and 300 inclusive is equal to ((101+300) x 200) / 2 = 40100.
There are 8 numbers between 100 and 200 (inclusive) that have a digit sum of 12, namely: 129, 138, 147, 156, 165, 174, 183, 192.
The sum of the integers from 1 to 100 inclusive is 5,050.
2550
The sum of the cubes of the first 100 whole numbers is 25,502,500.
The sum of the first 10 odd whole numbers is 100.
It is 180,001.
The sum is 10,000.
For the product to be zero, one of the numbers must be 0. So the question is to find the maximum sum for fifteen consecutive whole numbers, INCLUDING 0. This is clearly achived by the numbers 0 to 14 (inclusive), whose sum is 105.
We have to assume that you're talking about whole numbers. The sum is 5,050 .
100
5050
The sum of the integers between 101 and 300 inclusive is equal to ((101+300) x 200) / 2 = 40100.
Well, honey, the sum of counting numbers from 1 to 20 is 210. You add up 1 + 2 + 3... all the way to 20 and you get your answer. So, go ahead and impress your friends with your newfound math knowledge.
There are 8 numbers between 100 and 200 (inclusive) that have a digit sum of 12, namely: 129, 138, 147, 156, 165, 174, 183, 192.
The sum of the integers from 1 to 100 inclusive is 5,050.