No.
This can be shown by the following proof.
For any even number it can be written as 2k where k is some integer (a countable number). e.g. 4 = 2k where k = 2.
Thus the sum of any even number can be represented as
2k + 2l = 2(k+l)
Now if we let m = (k + l) than m is also an integer. This gives us,
2k + 2l = 2m
And as I said earlier and even number can be written as 2k and the final sum is in that form therefore it must be even.
The same proofs can be constructed for even+odd, and odd+odd by using 2k+1 for the odd number respectively.
Comment: 2 is an odd number 2/2 2/1 so it is an odd number see where im going with this? 2+4=6 :D I came up with ths btw.
The sum of an odd and even number is always odd because there's always one left over. Think of any odd number as being any even number plus one. If you add two even numbers, the answer is always even. If you add an even and an odd (plus one), the answer is always odd.
The sum of an odd and even number is always odd because there's always one left over. Think of any odd number as being any even number plus one. If you add two even numbers, the answer is always even. If you add an even and an odd (plus one), the answer is always odd.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. Even plus odd is odd.
no, b/c you can think of an odd number as an even number plus one. So if you add two odd numbers, you would be adding 2 even numbers plus 2... Let # be even... (#+1) + (#+1) = (#+#) + 2 two even numbers added together are always even
It is not possible to add together 5 odd numbers to make an even number.
Always an even number. Think of an odd number as an even number plus a half. Add them up and the two halves make an even.
No it is not possible to get a even number by adding odd and even number. 1+2=3-odd number.3+4=7-odd number.
4 add 12 = 16 and that is an square number
Any odd number can be written as an even number plus one. Any even number plus any other even number makes an even number. Add one makes an odd number. So an odd number plus an even number is always an odd number. Algebraically, any odd number can be written as 2a+ 1 where a is a whole number. Any even number can be written as 2b, where b is a whole number. Add them together, the sum is 2(a + b) + 1, which is odd.
-- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number plus an extra ' 1 '. -- Add them. You get (an even number) + (an even number) + ( 2 ). That's an even number. ======================== Another mind-bending , brain-busting way to look at it : -- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number minus ' 1 '. -- Add them. You get (an even number) + (an even number) + ( [1 - 1] or zero ). That's an even number.
No because two plus 15 equals 17
The sum of an odd and even number is always odd because there's always one left over. Think of any odd number as being any even number plus one. If you add two even numbers, the answer is always even. If you add an even and an odd (plus one), the answer is always odd.
The sum of an odd and even number is always odd because there's always one left over. Think of any odd number as being any even number plus one. If you add two even numbers, the answer is always even. If you add an even and an odd (plus one), the answer is always odd.
The sum of an odd and even number is always odd because there's always one left over. Think of any odd number as being any even number plus one. If you add two even numbers, the answer is always even. If you add an even and an odd (plus one), the answer is always odd.
no...all even numbers that are added together are even.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. Even plus odd is odd.
no, b/c you can think of an odd number as an even number plus one. So if you add two odd numbers, you would be adding 2 even numbers plus 2... Let # be even... (#+1) + (#+1) = (#+#) + 2 two even numbers added together are always even