To find the sum of the numbers, we must first know the value of n. This was not included in your question.
n/3 + 27
If they are equal numbers then n = n Hence n + n = 54 2n = 54 n = 54/2 n = 27 So 27 + 27 = 54
let the number be n, then the sum of the number and 2 is n+2
This is simple algebra. Let the unknown number be n. Then the sum of this number and 8 = n + 8.
Let 'N' be the number. The sum of the number and 13 is (N+13).The square of the sum of the number and 13 is (N+13)2N2 + 26 N + 169.It's probably easier to add 13 to the number and then square that answer ... less arithmetic.
27+n=n+27
n/3 + 27
n + (n + 1) + (n + 2) = 27 3n + 3 = 27 n = 8 Thus, the numbers are 8, 9, and 10, and the least of these is 8.
12
If they are equal numbers then n = n Hence n + n = 54 2n = 54 n = 54/2 n = 27 So 27 + 27 = 54
let the number be n, then the sum of the number and 2 is n+2
If the number is n, then twice the number is 2n and the sum of the two is n + 2n = 3n.
The sum of the interior angles of a polygon with n sides is (n-2)*180 degrees. Here, n = 27, so the sum of the interior angles is (27-2)*180 = 25*180 = 4500 degrees.
n = number n + 6 ======
This is simple algebra. Let the unknown number be n. Then the sum of this number and 8 = n + 8.
void main () { int no1, sum, n; clrscr() sum = 0; n = 1; printf("\n enter the number to which sum is to be generated"); scanf("%d",&no1); while(n<=no1) { sum=sum+n; n=n+1 } printf("\n the sum of %d = %d, no1, sum"); getch (); }
class Sum_Of_Digits { public static void printSumandnoofdigits(int n) { int temp = n; int count = 0; int sum = 0; while ( n > 0 ) { sum = sum + n % 10; n = n / 10; count ++; } System.out.println("The number is..." + temp ); System.out.println("The sum of digits is..." + sum); System.out.println("The number of digits is..." + count); } }