just don't do the question.
For any number n you could use * (n % 2 == 0), which would be true for an even number, false for odd For an integer i, a simpler method would be * (i & 1), which would be true for an odd number, false for even
29
2n + 1 where n is an integer.
25 apex
Every integer is either even (divisible by 2) or odd (not divisible by 2). Since an even number plus 1 is odd and an odd number plus one is even, because 1 does not divide 2. We know (n + 4) is odd. The next integer is (n + 4 + 1) = (n + 5), because an odd number plus 1 is even, (n + 5) is even. The integer after (n + 5) is (n + 6), since (n + 5) we know is even, (n + 6) must be odd. Since (n + 6) is the smallest integer that is greater than (n + 4) and is odd, so (n + 6) is the next odd integer.
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
printf ("%d is %s\n", n, n%2 ? "odd": "even");
There is no such number. Given any odd number, n, the number (n + 2) is a greater odd number. You can go on, for ever, finding larger odd numbers.
#include <stdio.h> int main() { printf("Program to find ODD or Even Number\n"); while(1) { int n = 0; printf("\nEnter a number(-1 for Exit): "); scanf("%d",&n); if( n 0) { printf("%d is a EVEN number.\n", n); } else { printf("%d is a ODD number.\n", n); } } return 0; }
There is a surprisingly easy formula for this. Sum of n odd numbers = n2 So the sum of the first 600 odd numbers (starting with 1 as the very first odd number) is 6002 = 360000.
All odd numbers are of the form 2n + 1, where n is an integer.So an odd number minus an odd number is (2n+1) - (2m+1) = 2n -2m = 2(n-m). Both n and m are integers, so while we don't know whether n-m is odd or even, we definitely know that it's an integer and that multiplying it by two cannot possibly give an odd number. So an odd number minus an odd number is an even number. For similar reasons, an odd number plus an odd number is also an even number.
For any number n you could use * (n % 2 == 0), which would be true for an even number, false for odd For an integer i, a simpler method would be * (i & 1), which would be true for an odd number, false for even
29
2n + 1 where n is an integer.
let n = first odd number; n + 2 = 2nd odd number ,etc. n + n+2 + n+4 + n +6 = 4n + 12 = -72 4n = -84 n = -21 -21,-19,-17 and -15
Sum = n/2[2Xa1+(n-1)d] where n is last number, a1 is the first number & d is the common difference between the numbers, here d=2 for the even /odd numbers. Sum = n/2 [2Xa1+(n-1)2]
Let's do a small experiment with a few numbers to see we can find out.3 - 2 = 1, Odd when the first number is odd and larger than the second number which is even and smaller.5 - 12 = -7, Odd when the first number is odd and smaller than the second number which is even and larger.Based on the Inverse commutative property, switching the first number for an even would yield the same number except it's inverse.(a - b) = c(b - a) = -c2 - 3 = -112 - 5 = 7So it looks that an odd number minus an even number results in an odd number.