Add them up and divide the sum by 2.
Go through each of the numbers, and count how many of them are odd. If you count an odd number of them, then their sum is odd. If you count an even number of them, then their sum is even.
Any 2 negative numbers, whether even or odd, when multiplied are positive
If you multiply one even number by one odd number, the result is always even. In general, if you multiply several numbers, and at least one of the numbers is even, the product is always even. This is because "even" means "multiple of 2", and if one of the factors contains a 2 as a factor, so will the product.
Very easily: if the prime factorization includes 2, it's even. If not, it's odd.
You didn't state whether the two odd numbers to be added must be different odd numbers or whether they can be the same odd number. So I'll do both cases. Case 1: They can be the same odd number: They are all even numbers starting with 1+1=2 and ending up with 9+9=18. So you can get these 9 even numbers: 2,4,6,8,10,12,14,16,18. Case 2: They must be different odd numbers: If you require that the two single-digit odd numbers be DIFFERENT odd numbers, then you can't get 2 or 18, so you can only get these 7 even numbers 4,6,8,10,12,14,16. So the answer is either 9 or 7, depoending on whether the two odd numbers you are adding can or can't be the same odd number.
If at least one of the numbers is even, the result will be even. Otherwise all the numbers are odd and the result will be odd.
if there is an od quantity of odd numbers the sum will be odd....example 1 3 4 6 has two odds so it will be even =14 1 3 5 6 has three odds and it is = 15
It can be any number. Two numbers do not even determine whether the "sequence" is arithmetic, geometric or other.
Yes, all you have to do is to count the number of ODD numbers in the list. If it is odd, then the sum will be odd; if even, so will the sum. Knowing this can help you run a quick validity check when you sum up a list of numbers. (The method works because: a) the sum of two even numbers is even, and b) the sum pf two odd numbers is even, but c) the sum of an even number and an odd number is odd. Hence, if you only determine whether there are any unpaired odd numbers, you know the answer.)
Well, there is a clear definition, and at least in theory you can always determine whether a number is a primer number or not, so I would say, yes.
Go through each of the numbers, and count how many of them are odd. If you count an odd number of them, then their sum is odd. If you count an even number of them, then their sum is even.
Yes it is possible to determine if a product will be even or odd. To do this, we need to consider what an even number is. Even numbers are numbers with at least one factor of 2 (meaning they are divisible by 2). Thus, any product of numbers which contains at least one even number will result in an even product. If all of the numbers being multiplied together are odd, the product will be odd. If one or more of the numbers is even, the product will be even.
The parity rules are: Odd + Odd = Even Even + Even = Even Odd + Even = Odd Even + Odd = Odd So the parity where one number is even is that of the other number. This means that you can go through a list of numbers and ignore all the even numbers. Every PAIR of odd numbers has even parity and by the previous paragraph, even parity can be ignored. So you can pair off odd numbers and ignore them.
The set is well defined. Whether or not a given integer belongs to the set of prime numbers is clearly defined even if, for extremely large numbers, it may prove impossible to determine the status of that number.
None Of Them Equals 100 and 10 so 127 is 130
You have to look at the ones place digit for the quick way to this. If both ones digits are even then it will be even, if they are both odd then it will be even, if one is odd and one is even then it will be odd.
Even plus even equals even. Odd plus odd equals odd. Odd plus even equals odd. 13 + 45 (odd plus odd) will be even. Even plus 24 will be even. Even plus 17will be odd. 13 + 45 + 24 + 17 will be odd.