The sum of any integer ( n ) and zero is ( n ).
When the positive integer is farther from 0 than the negative integer, example -4+5=1 why? the -4 is 4 places to the left of the 0 and the 5 is 5 places to the right.
sum of positive integers will be a positive integer
No. The sum of a positive integer and a negative integer has the same sign as the larger integer.
The sum of a positive integer and a negative integer is positive when the positive integer is greater. For example: 9 + (-5) = 4 In this case, the positive integer 9 is greater than the negative integer 5. Therefore, the sum is positive.
what is the sum of the first 10 positive integers? To me, if you include 0 as the first integer, then the tenth integer is 9 and the sum is 45. If you don't include 0, the tenth integer is 10, and the sum is 55.
The sum of any integer ( n ) and zero is ( n ).
Sum the digits in the odd positions in the integer. Sum = XSum the digits in the even positions in the integer. Sum = YIf X - Y is a multiple (including negative or 0), then the given integer is divisible by 11.
int i, sum; /* example using while */ sum = 0; i = 1; while (i <= 100) { sum += i; ++i; } /* example using for */ sum = 0; for (i = 1; i <= 100; ++i) sum += i;
The below method is written in the assumption that, the n numbers are sent as an ArrayList of Integer values. public void computeSumAndAvg(ArrayList lst){ Iterator itr = lst.iterator(); int sum = 0; float average = 0; int count = 0; while(itr.hasNext(){ Integer val = (Integer) itr.next(); sum = sum + val.intValue(); count++; } average = sum/count; System.out.println("Sum is: " + sum) ; System.out.println("Average is: " + average) ; }
The smallest 5 digit integer is -99999. The largest 5 digit integer is 99999. The sum is therefore 0.
#include<iostream> #include<sstream> using namespace std; unsigned sum_of_squares (const unsigned max) { if (max==0) return 0; if (max==1) return 1; return sum_of_squares (max-1) + (max*max); } int main () { unsigned num = 0; while (1) { cout << "Enter a positive integer (0 to exit): "; string s; cin >> s; if (s[0]=='0') break; stringstream ss; ss << s; if (ss >> num) { cout << "The sum of all squares from 1 to " << num << " is: " << sum_of_squares (num) << endl; continue; } cerr << "Invalid input: " << s << endl; } cout << "Quitting..." << endl; } Example output: Enter a positive integer (0 to exit): 1 The sum of all squares from 1 to 1 is: 1 Enter a positive integer (0 to exit): 2 The sum of all squares from 1 to 2 is: 5 Enter a positive integer (0 to exit): 3 The sum of all squares from 1 to 3 is: 14 Enter a positive integer (0 to exit): 4 The sum of all squares from 1 to 4 is: 30 Enter a positive integer (0 to exit): 5 The sum of all squares from 1 to 5 is: 55 Enter a positive integer (0 to exit): 6 The sum of all squares from 1 to 6 is: 91 Enter a positive integer (0 to exit): 7 The sum of all squares from 1 to 7 is: 140 Enter a positive integer (0 to exit): 0 Quitting...
When the positive integer is farther from 0 than the negative integer, example -4+5=1 why? the -4 is 4 places to the left of the 0 and the 5 is 5 places to the right.
sum of positive integers will be a positive integer
2+(-2) =0
Well, honey, any integer greater than 7 will do the trick. Just pick a number, add it to -7, and if the sum is greater than 0, you've hit the jackpot. Just remember, the bigger the integer, the better chance you have of getting a positive sum.
No. The sum of a positive integer and a negative integer has the same sign as the larger integer.