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When the sum of a number pluse 3 is squated it is 11 more than the sum of the number plus 2 when squared?
Call the unknown number n. From the problem statement (n + 3)2 - (n + 2)2 = 11. Expanding both squares yields n2 + 6 n + 9 - (n2 + 4n + 4) = 11, or 2 n + 5 = 11, 2n = 6, n = 3.
The sum of a number and 6?
n = number n + 6 ======
C program to find sum of 'n' number using function?
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 (); }
Write a java program to find out the sum of a number of n digits?
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); } }
What is a number n where if the sum of the digits of the number is divisible by n, then the number itself is divisible by n?
(The assumes that "the number" in the question is not n, although if they are they same number, this is still true.) "If the sum of the digits of the number is divisible by n, then the number itself is divisible by n" is true if n is 3 or if n is 9.
