sum of positive integers will be a positive integer
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.
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.
Yes the sum of two integers will always be an integer.
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.
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...
sum of positive integers will be a positive integer
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.
2+(-2) =0
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.
Yes the sum of two integers will always be an integer.