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Q: Is 144 a n odd or even number?

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Will always be odd. Suppose that a number n is an odd number. Any n times 2 must equal an even number. 2n - n = n

There are two cases here: one where n is even and one where n is odd. Let's consider the case where n is even: If n is even then n2 has to be even (since multiple of an even number must always be even.) In this case, we are subtracting an even number from and even number, the result must be even. This proves than n2 - n is even when n is even. Now let's consider the case where n is odd: If n is odd, then n2 must be odd. This is because an odd number times an odd number is always odd. (You could think if this as an odd number times and even number and then adding an odd number. For example, say that n is odd. n-1 is then even, and n2 = n(n-1) + n. n(n-1) must be even, since it is a multiple of an even number. And even number plus and odd number then has to be odd.) Now we know we have and odd number minus and odd number, which has to be even. So this proves that n2 - n is even when n is odd. Since we have proved that n2 - n is even for both when n is even and when n is odd, and there are no other cases, n2 - n must be even for any natural number n. or Let n be a natural number. Then n can be even or odd. We want to show that n2 - n = 2m where m is any positive nteger (by the def. of even). Case 1: Let n be even. Then n = 2k (def. of even), where k is any positive integer. Then, n2 - n = (2k)2 - (2k) = 2(2k2 - k); let 2k2 - k = m = 2m Therefore, n2 - n is even. Case 2: Let n be odd. Then n = 2k +1 (def. of odd), where k is any positive integer. Then, n2 - n = (2k - 1)2 - (2k - 1) = 4k2 - 4k + 1 - 2k + 1 = 4k2 - 6k + 2 = 2(2k2 - 3k + 1); let 2k2 -3k + 1 = m = 2m Therefore, n2 - n is even. Therefore, for any natural number n, n2 - n is even.

It will always be odd. A proof: Call the even number m, call the odd number p. 'p' is equal to an even number 'n' + 1. Adding m and p is equal to m + n + 1; since m and n are both even, their sum is also even (see the related questions). m + n + 1 is one more than an even number, so it is odd, and therefore an odd number plus an even number is also always odd. QED

if (number & 1) printf("Number is odd\n"); else printf("Number is even\n");

Here is a JavaScript option for determining if a number is odd or even. It even lets you know if the number is zero (if you want zero to be neither odd nor even). var n = prompt("Enter a number to identify as odd or even", "Type your number here"); n = parseInt(n); if (isNaN(n)) { alert("Please Enter a Number"); } else if (n == 0) { alert("The number is zero"); } else if (n%2) { alert("The number is odd"); } else { alert("The number is even"); }

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.

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.

# includevoid main(){int a;printf("\n Enter a number to find if its odd or even");scanf("%d",&a);if(a%2==0){printf("\n The number entered is even");}else{printf("\n The number entered is odd");}getch();}

Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.

read a number n if(n%2)=0 /*% indicates modulo division(remainder after division)*/ then n is even else n is odd

The sum of two even numbers is always an even number. The sum of two odd numbers is always an even number.The sum of an even number and an odd number is always an odd number.Because integers alternate between even and odd, adding two consecutive integers will always result the sum of an odd number and an even number, which, as stated above, will always be an odd number.Another way to look at it:An even number is divisible by 2, therefore, 2*n (where n is an integer) is always an even number.2*n+1 would be the next number after 2*n (so 2*n & 2*n+1 are consecutive integers).If we add these numbers together, we get 4*n+1 = 2(2*n)+1. Since 2(2*n) is an even number, adding 1 makes it an odd number._________________________________________________________________________I'm not going to but in or anything, seriously that was a good answer

#include <stdio.h> main() { int n, odd=0, even=0; while (scanf("%d",&n)&&(n!=0)) (n%2)?++odd:++even; printf("odd: %d even: %d\n",odd,even); }

3 if n is odd 2 if n is even where n is the number of vertices.

printf ("%d is %s\n", n, n%2 ? "odd": "even");

odd x even = even (2n + 1)(2n) = 2[n(2n + 1)] let n(2n + 1) = t = 2t (even)

echo -n "Enter numnber : " read n rem=$(( $n % 2 )) if [ $rem -eq 0 ] then echo "$n is even number" else echo "$n is odd number" fi

#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; }

The sum of two even numbers is always EVENYes.Proof:We can consider an even number as 2*n, and an odd number as 2*n+1.The sum of two even numbers would be 2*n+2*m=2*(n+m), an even number.

For any even number n, the following integer is odd (n+1), so for any odd number other than 1, you are adding the preceding even number, plus 1. Because all even numbers are multiples of 2, their sum is even, and 1 more will be an odd number.

Yes. If n is odd, then n + c where c is an even constant will be odd. n + d where d is an odd constant will be even.

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

/*C program to find the entered number is even of odd*/ #include #include void main() { int n; clrscr(); printf("Enter the number\n"); scanf("%d",&n); if((n%2)==0) printf("%d is even number",n); else printf("%d is odd number",n); getch(); }

Sum of two odd numbers is always an even number.e.g. 1 + 7 = 8Explanation:The successor of an odd number is an even number and the successor is obtained by adding 1.Let us assume any odd number, say n.On adding 1(i.e. an odd number) to n we get an even number and again on addition of 1 or addition of 2(an even number) to n we get an odd number. If we continue to add like this we get to know that addition of two odd numbers is always an even number.

It is odd. The definition of an odd number is "a number a is odd if there exists an integer n such that a = 2n + 1." Let a = 1 and let n = 0, so 1 = (2x0) + 1.

The sum of an odd and an even number is odd. Any odd number can be expressed as 2n + 1 (for some integer "n"). Any even number can be expressed as 2m (for another integer, "m"). The sum of the two is 2(m+n) + 1. Since the expression in parentheses is an integer, multiplying it by 2 gives you an even number. Adding 1 makes the entire expression odd.