Odd and Even

Let $m = 2j + 1$ for some integer $j$ and $n = 2k$ for some integer $k$ .

A.

$m^2 + n = (2j + 1)^2 + 2k = 4j^2 + 4j + 1 + 2k = 2{\color{#3D99F6}{(2j^2 + 2j + k)}} + 1 = 2{\color{#3D99F6}{p}} + 1 \implies$ A is odd .

B.

$m^2 + mn = (2j + 1)^2 + (2j + 1)(2k) = 4j^2 + 4j + 1 + 4jk + 2k = 2{\color{#3D99F6}{(2j^2 + 2j + 2jk + k)}} + 1 = 2{\color{#3D99F6}{p}} + 1 \implies$ B is odd .

C.

$m^2 + n^2 = (2j + 1)^2 + (2k)^2 = 4j^2 + 4j + 1 + 4k^2 = 2{\color{#3D99F6}{(2j^2 + 2j + 2k^2)}} + 1 = 2{\color{#3D99F6}{p}} + 1 \implies$ C is odd .

D.

$m + n^2 = 2j + 1 + (2k)^2 = 2j + 1 + 4k^2 = 2{\color{#3D99F6}{(j + 2k^2)}} + 1 = 2{\color{#3D99F6}{p}} + 1 \implies$ D is odd .

E.

$mn + n^2 = (2j + 1)(2k) + (2k)^2 = 4jk + 2k + 4k^2 = 2{\color{#3D99F6}{(2jk + k + 2k^2)}} = 2{\color{#3D99F6}{p}} \implies$ $\boxed{E}$ is even .

Odd x even =even

Odd x odd =odd

Even x odd =even

so, the correct answer will be=(mn+n^2)=odd x even +even^2

            so, it will be            =even+even^2   ...........(always even)

The way i like to think about if a number is even is: a number is even if and only if it has 2 in it's prime factors. If a number is not even then it's square is not even aswell since you just multiply by the same prime factors which don't contain 2. If a number is even then it's square is even since the factor of 2 is not lost. Even times odd is even since the factor of 2 is not lost. So by that logic we have:
- A. $m^2$ is odd, $n$ is even. odd + even = odd $\boxed{\times}$
- B. $m^2$ is odd, $mn$ is even. odd + even = odd $\boxed{\times}$
- C. $m^2$ is odd, $n^2$ is even. odd + even = odd $\boxed{\times}$
- D. $m$ is odd, $n^2$ is even. odd + even = $\boxed{\times}$
- E. $mn$ is even, $n^2$ is even. even + even = $\boxed{even}$
Therefore, only E is even

$mn + n^2 = n(m+n) = 2k(m+n)$ , $k \in \mathbb{N_0}$

∴ $mn + n^2$ is even.

Lets think for a minute, the question is asking us about even number. D efinition : A number which is divisible by 2. S o how can we get an even number?? T o get an even number we must add two even numbers (following the given question) . S o even + even = even. Now, we know that odd^2 is not even , but oddXeven is even therefore m*n will be even and n^2 is already even . So we have our answer i.e mn+n^2.

Much easier in mod-arithmetic.Let $m \equiv 1 \pmod 2$ and $n \equiv 0 \pmod 2$

$\begin{array}{lll} x & x \pmod 2& \textrm{parity} \\ \hline m^2+n & 1^2+0 = 1 & \textrm{odd} \\ m^2 + mn & 1^2+1\cdot 0 = 1 & \textrm{odd} \\ \vdots \end{array}$

To solve this problem we must know the following rules:
1- The square of an even number is even.
2- The square of an odd number is odd.
3- The product of an even and odd number is even.
4- The sum of an even and odd number is odd.
5- The sum of 2 odd numbers is even.
6- The sum of 2 even numbers is even.
So:
A is odd
B is odd
C is odd
D is odd
E is even

$mn+n^{2}=n(m+n)$

is always even if $n$ is even, no matter what is the parity of $m$ .

Odd $\times$ even = $\boxed { even }$

Odd $\times$ odd = $\boxed { Odd }$

Even $\times$ odd = $\boxed { Even }$

$m^2 \Rightarrow \boxed { Odd }$

$n^2 \Rightarrow \boxed { Even }$

From the selection of available answers, the correct answer is E because $mn \Rightarrow \boxed { Even }$ and $n^2 \Rightarrow \boxed { Even }$ . So, $\boxed { Even } + \boxed { Even } = \boxed { Even }$ .

If the letter m is odd, and the number n is even, and we consider the following properties:

• odd + odd = even
• odd + even = odd
• even + even = even
• odd * odd = odd
• odd * even = even
• even * even = even

then we can easily figure out which of the expressions will always be even.

Expression 1 can be rewritten as (odd * odd) + even, which can then be simplified as odd + even, which means that the solution will be odd.

Expression 2 can be rewritten as (odd * odd) + (odd * even), which can be simplified as odd + even, which also makes it odd.

Expression 3 can be rewritten as (odd * odd) + (even * even), which automatically means that the expression will be equivalent to odd + even.

Expression 4 can be rewritten as odd + (even * even), which can be simplified as odd + even.

Expression 5 can be rewritten as (odd * even) + (even * even), which can be simplified as even + even, which makes expression 5 always equal an even number no matter what values you assign m or n.

Therefore, Statement $\boxed{5}$ will always be even.

In short: Even x Anything = Even. While Odd+Even and Odd*Odd = Odd.

Every number consists of two parts (before and after the addition sign). Notice that they all have an even "after part" because all of them is $n$ multiplied by something ( $n$ is even). So it all depends on the "before part". We know that even + even = even & even + odd = odd. E is the only number that has an even "before part". << because odd $\times$ odd = odd & even $\times$ odd = even.

The only one that satisfies our conditions is $E$ , the product of ODD and EVEN is EVEN.

The square of an EVEN is EVEN, implies the sum of an EVEN and EVEN is EVEN.

let m=3 and n=2 as we have given m is odd and n is even. then put 3 as m and 2 as n in each option,then u will get last option correct..

Because m is odd, n is even so m.n will sure be even And n2 is always even so We have m.n + n2 is even

Always:

• mn = even = n ( $3 \times 4$ = 12)

• $m^2$ = odd = m ( $5^2$ = $5 \times 5$ = 25)

• $n^2$ = even = n ( $6^2$ = $6 \times 6$ = 36)

• m + n = n + m = odd = m (7 + 2 = 2 + 7 = 9)

• m + m = even = n (7 + 3 + 10)

• n + n = even = n (4 + 12 = 16)

E. mn + $n^2$ = n + n = n

We can factor $n$ out of expresion $mn-n^2$ , so we have $E=mn-n^2=n\cdot (m-n)$ , and $n$ is an even number. So, definitely, E is the answer.

Apply the following trick Odd×odd=even Even×even=even Odd×even=odd

As n is even n^2 is even too. Now odd×even=even. And sum of two even is always even

Once we remember that 1 is an odd number, the solution can only be mn + n^2.

An even times any number is even. An even squared is even. An even plus an even is even.

PAC-MAN eats dots. His cousin 2PAC is similar except he insists on eating 2 dots at a time, or he becomes sad.

2PAC can happily eat any even number $n$ of dots, as it's a multiple of 2 (by definition).

He won't be happy with an odd number $m$ as there will be a dot left over. Nor with the sum $n+m$ , because he'll eat $n$ and then be left with an odd number.

What about $m^2$ ? That's $m$ rows of $m$ dots, in a square arrangement. He eats most of the first row but has a single leftover dot. No problem: he eats the second row, has another leftover, but now he has a pair of leftovers, so he can eat them. He continues clearing pairs of rows. But... there's an odd number of rows! So the last row will have one leftover and no subsequent row to generate another leftover dot. So $m^2$ ends up making 2PAC sad.

For $nm$ he's fine: he can just clear each row of n and he doesn't care that he has to do this an odd number of times. And in the same way $n^2$ is no problem at all.

So from the options, only $nm + n^2$ will keep 2PAC happy.