Odd/Even Factorials | Number Properties

Today we will discuss how factorials affect the behaviour of odd and even integers. Since we are going to deal with factorials, positive integers will be our concern. Using a question, we will see how factorials are divided.

Question: If ${x}$ and ${y}$ are positive integers, is ${y}$ odd?

Statement 1: ${\frac{(y+2)!}{x!}}$ = odd

Statement 2: ${\frac{(y+2)!}{x!}}$ is greater than 2

Solution: The question stem doesn’t give us much information – just that $x$ and $y$ are positive integers

Question: Is $y$ odd?

Statement 1: $\frac{(y+2)!}{x!}$ = odd

Note that odd and even are identified only for integers. Since $\frac{(y+2)!}{x!}$ is odd, it must be a positive integer. This means that $x$ ! must be equal to or less than $(y+2)!$

Now think, how are $y$ and $y+2$ related? If $y+2$ is odd, $y+1$ is even and hence $y$ is odd. If $y+2$ is even, by the same logic, $y$ is even.

$(y+2)! = 1\times2\times3\times4\times…\times y\times (y+1) \times (y+2)$

$x! = 1\times2\times3\times4\times…\times x$

Note that $(y+2)!$ and $x!$ have common factors starting from 1. Since $x!$ is less than or equal to $(y+2)!$ , $x$ will be less than or equal to $(y+2)$ . So all factors in the denominator, from 1 to $x$ will be there in the numerator too and will get cancelled leaving us with the last few factors of $(y+2)!$

To explain this, let us take a few examples:

Example 1: Say, $y+2 = 6$ , $x = 6$

$\frac{(y+2)!}{x!} = \frac{6!}{6!} = 1$

Example 2: Say, $y+2 = 7$ , $x = 6$

$\frac{(y+2)!}{x!} = \frac{7!}{6!} = \frac{(1\times2\times3\times4\times5\times6\times7)}{(1\times2\times3\times4\times5\times6)} = 7$ (only one leftover factor)

Example 3: Say, $y+2 = 6$ , $x = 4$

$\frac{(y+2)!}{x!} = \frac{6!}{4!} = \frac{(1\times2\times3\times4\times5\times6)}{(1\times2\times3\times4)} = 5\times6$ (two leftover factors)

If the division of two factorials is an integer, the factorial in the numerator must be larger than or equal to the factorial in the denominator.

So what does $\frac{(y+2)!}{x!}$ is odd imply? It means that the leftover factors must be all odd. But the leftover factors will be consecutive integers. So after one odd factor, there will be an even factor. If we want $\frac{(y+2)!}{x!}$ to be odd, we must have either no leftover factors (such that $\frac{(y+2)!}{x!} = 1$ ) or only one leftover factor and that too odd.

If we have no leftover factor, it doesn’t matter what $y+2$ is as long as it is equal to $x$ . It could be odd or even. If there is one leftover factor, then $y+2$ must be odd and hence $y$ must be odd. Hence $y$ could be odd or even. This statement alone is not sufficient.

Statement 2: $\frac{(y+2)!}{x!}$ is greater than 2

This tells us that $y+2$ is not equal to $x$ since $\frac{(y+2)!}{x!}$ is not 1. But all we know is that it is greater than 2. It could be anything as seen in examples 2 and 3 above. This statement alone is not sufficient.

Both statements together tell us that $y+2$ is greater than $x$ such that $\frac{(y+2)!}{x!}$ is odd. So there must be only one leftover factor and it must be odd. The leftover factor will be the last factor i.e. $y+2$ . This tells us that $y+2$ must be odd. Hence y must be odd too.

Answer (C).

Takeaways: Assuming $a$ and $b$ are positive integers,