Below are three questions on numbers on concepts dealing with odd and even and a new class called "even-odd" numbers.

If n is an even-odd number then which of the following must be false?

(A number is called "even-odd" if it is halfway between an even integer and an odd integer.)

a) n/2 is not an integer

b)(2n)

c)4n is an odd integer

d)none of these

Solution:

A number is called "even-odd" if it is halfway between an even integer and an odd integer. For example, consider an even integer 10 and an odd integer -5. Number halfway between them will be (10 - (-5)) / 2 = 7.5. Here 7.5 is an "even-odd" number.

i.e., an even-odd number will be in the form x + 1/2 = x.5 where x is any integer.

Let us see with each option:

Consider option a :

Since n is a fraction number then n/2 is also a fraction.

i.e., n/2 is not an integer.

Hence option a is true.

Consider option b :

(2n)

since n is an even-odd number then n^2 is not an integer.

Also n

When we multiply n

Hence option b is true

Consider option c:

We can express 4n as 2 x 2n.

Since n is X.5 then 2n must be an integer and any multiple of 2 gives an even integer.

But the statement says that 4n is an odd integer.

Hence option c is false.

If p is an odd number and q is an even number, which of the following must be an even?

I. 2p+3q

II. p

III.(pq)

IV. (1+pq)

Options:

a) I only b)II&IV only c)I&III only d)I,II&III only

Solution:

We know that any multiple of 2 is even and 3 times of an even number is also even.

Then 2p and 3q both are even.

Since the addition of two even numbers is again an even, then 2p+3q is even.

i.e., I is even.

Now, the square of any even number is even and the square of an odd is odd. Then p

And difference between even number and odd number is an odd number.

Then p

Hence II is odd.

The multiplication of an odd and even number will be an even number, then pq is even,

And the square of an even is even, so (pq)

Hence III is even.

since pq is even, then 1 + pq will be odd.

And the square of odd is odd. Therefore (1 + pq)

Hence IV is odd.

Therefore only I and III are even. Thus the answer is I & III only.

If n is an odd integer, then which of the options will be even.

a)(n

Solution :

Consider option a,

since n is an odd number then n

We can express (n

And n

Then ( n

Consider option b,

since n is odd, then n power anything is odd and n

The addition of 3 odd numbers is odd, then n

Consider option c,

Here n

Now, the square of an odd integer is odd, then (n

Hence the option c)(n

**Question 1**If n is an even-odd number then which of the following must be false?

(A number is called "even-odd" if it is halfway between an even integer and an odd integer.)

a) n/2 is not an integer

b)(2n)

^{2}is an integerc)4n is an odd integer

d)none of these

**Answer :**c)4n is an odd integerSolution:

A number is called "even-odd" if it is halfway between an even integer and an odd integer. For example, consider an even integer 10 and an odd integer -5. Number halfway between them will be (10 - (-5)) / 2 = 7.5. Here 7.5 is an "even-odd" number.

i.e., an even-odd number will be in the form x + 1/2 = x.5 where x is any integer.

Let us see with each option:

Consider option a :

Since n is a fraction number then n/2 is also a fraction.

i.e., n/2 is not an integer.

Hence option a is true.

Consider option b :

(2n)

^{2 }= 4n^{2}since n is an even-odd number then n^2 is not an integer.

Also n

^{2}will have 1/4 = .25 in its decimal place. (For example, consider n = 7/2. Then n^{2}will be 7/2 x 7/2 = 49/4. 49/4 can be written as 49 + 1/4. Here 1/4 will contribute to the decimal part.)When we multiply n

^{2}by 4 we have an integer, since .25 x 4 = 1.Hence option b is true

Consider option c:

We can express 4n as 2 x 2n.

Since n is X.5 then 2n must be an integer and any multiple of 2 gives an even integer.

But the statement says that 4n is an odd integer.

Hence option c is false.

**Question 2**If p is an odd number and q is an even number, which of the following must be an even?

I. 2p+3q

II. p

^{2}- q^{2}III.(pq)

^{2}IV. (1+pq)

^{2}Options:

a) I only b)II&IV only c)I&III only d)I,II&III only

**Answer :**c)I&III onlySolution:

We know that any multiple of 2 is even and 3 times of an even number is also even.

Then 2p and 3q both are even.

Since the addition of two even numbers is again an even, then 2p+3q is even.

i.e., I is even.

Now, the square of any even number is even and the square of an odd is odd. Then p

^{2}is odd and q^{2}is evenAnd difference between even number and odd number is an odd number.

Then p

^{2}- q^{2}is an odd.Hence II is odd.

The multiplication of an odd and even number will be an even number, then pq is even,

And the square of an even is even, so (pq)

^{2}is also even.Hence III is even.

since pq is even, then 1 + pq will be odd.

And the square of odd is odd. Therefore (1 + pq)

^{2}is odd.Hence IV is odd.

Therefore only I and III are even. Thus the answer is I & III only.

**Question 3**If n is an odd integer, then which of the options will be even.

a)(n

^{2})/2 + n b)n^{n}+ n + 1 c)(n^{2}+ 3)^{2}d)none of these**Answer :**c)(n^{2}+ 3)^{2}Solution :

Consider option a,

since n is an odd number then n

^{2}is also an oddWe can express (n

^{2})/2 + n as ( n^{2}+ 2n)/ 2.And n

^{2}is odd and 2n is even, then n^{2}+ 2n will be oddThen ( n

^{2}+ 2n)/ 2 is not an even integer.Consider option b,

since n is odd, then n power anything is odd and n

^{n}is odd.The addition of 3 odd numbers is odd, then n

^{n}+ n + 1 is an odd integer.Consider option c,

Here n

^{2}is odd and then n^{2}+ 3 is also odd.Now, the square of an odd integer is odd, then (n

^{2}+ 3)^{2}is an even integer.Hence the option c)(n

^{2}+ 3)^{2}is the answer.