Odd Or Even??

${ a }^{ 2 }+a=a\left( a+1 \right)$ which is even for all values of a.

so, $\left( { a }^{ 2 }+a+2011 \right)$ is odd for all values of a.

Now, $2b+1=b+b+1$ which is odd for all values of b.

$\therefore \quad \left( { a }^{ 2 }+a+2011 \right) \left( 2b+1 \right)$ is odd for all values of a and b.

$a^2$ and a both must be either odd or even and hence their sum is always even.

So $a^2$ +a+(2011, an odd number) is odd. (2b+1) is clearly odd. So multiplication of two odds is odd.

I tried substituting a/b =1 and 2, the result is odd. Since 1 and 2 represents already all odd and even number, then that would be enough.

First let's see ( a^2 + a + 2011) :

Case 1 : A is even :

a^2 will be even bcoz multiple of any even number will be even no matter how many times it is multiplied...

So a^2 + a + 2011 will be odd ...........( Since a^2 and a are even)

Case 2 : a is odd:

a^2 will be odd because multiple of every odd number will be odd if it is multiplied odd-number of times.....

So a^2 + a + 2011 will again odd.............( odd+odd+odd =odd)

From case 1 &2 : a^2+a +2011 is odd for all values of a........(1)

Now let's see (2b+1):

Case1: b is even:

2b will be EVEN.....(Since twice of any number is even)

So, 2b + 1 will be odd

Case 2: b is odd

2b will again be even ......(same reason as above)

Again 2b + 1 will be odd

From case 1 and 2: 2b +1 is odd for any value of b........(2)

Now (a^2 + a +2011) × (2b +1) can be written as :

Odd ×Odd ; which will be a odd number since multiple of any odd number will be odd if multiplied odd-numbers of time

So (a^2 + a+2011) × (2b+1) will be odd for any values of a and b . 😉

Just pick small value numbers and try them on. A=1 and B=2 for example. Replace them and solve. Solve them 3 times and you´ll know that the result is always odd no matter what.

Just pick two elements from odd integers for a&b and check the results. Same goes for even integers, and odd & even, even & odd, for a and b respectively. All results are elements of odd integers. :)