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If p is an integer between 1000 and 1030 and if the sum of the digits of p is odd then p must be odd?
If ( p ) is an integer between 1000 and 1030, it can be expressed as ( p = 1000 + n ), where ( n ) ranges from 0 to 30. The sum of the digits of ( p ) is given by ( 1 + \text{(sum of the digits of } n) ). Since 1 is odd, for the total sum of the digits to be odd, the sum of the digits of ( n ) must be even. As a result, if ( p ) is odd, ( n ) must be odd (e.g., 1, 3, 5, etc.), confirming that ( p ) is indeed odd. Thus, the statement is true: if the sum of the digits of ( p ) is odd, then ( p ) must be odd.
How do you translate The sum of 6 and a number is equal to twice the number increased by 5?
6+n = 2n+5
How do you write a equation when the sum of a positive number and several negative numbers will be positive?
if p represents your positive number, and n represents all of your negative numbers, then: |∑n| < p
What is the sum expression for the sum of n and 5?
The sum of n and 5 is n+ 5
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 (); }
