Perfecting The Square

A generalized version for checking whether these type of expressions are perfect square or not is as follows:

$\frac{n!(n+1)!}{2}=\frac{(n!)^2(n+1)}{2}$

So, from this, it follows that the whole expression is a perfect square iff $\dfrac{n+1}{2}$ is a perfect square.

In this case $n = 17$ only so $\frac {17! 18!}{2!}$ is the only perfect square.

$\dfrac{17!18!}{2} = \dfrac{(17!)^2\cdot18}{2} = (17!)^2\cdot9$

So the expression is a perfect square

We can eliminate some options without having to evaluate. In order to get a perfect square:

• We need to strip off the first term from the bigger factorial to make it the same as the smaller factorial.
• This first term that we stripped off must be divisible by 2. This eliminates options B, D and E.
• The result of dividing this first term by 2 must also be a perfect square. This eliminates A, and the correct answer is C.

$\frac{17!18!}{2}=\frac{17!\times 18\times 17!}{2}=\frac{(17!)^2\times 18}{2}=(17!)^2\times 9=(17!\times 3)^2$

Therefore, $\frac{17!18!}{2}$ is a perfect square.

The product of two perfect squares is also a perfect square. Since these factorials differ only one, we can make each numerator a perfect square by grouping the numbers: $\dfrac{17! \cdot 18!}{2} = \dfrac{17! \cdot 17! \cdot 18}{2} = 17! \cdot 17! \cdot \dfrac{18}{2}$ We can do this for each answer above, and then test whether our new fraction forms a perfect square. Only one answer does.

17 ! = 1 × 2 × 3 ....17 & 18! = 1 × 2 × 3:....×18

so when you multiply these factorials together You get 1^2 × 2^2 × ...17^2 × 18

& 18/2 = 9 which is 3^2

For a general formula of $\frac{n!*(n+1)!}{m}$ , where m is greater than 0 and less than or equal to n+1, you can imagine first expanding both factorials. Now take out one factor of m from either side. From the sets of n! and (n+1)!, you now have all the numbers from 1 to n twice, with the exception of a lonely m, as well as a factor of n+1. If (n+1)m is a perfect square, the whole fraction becomes a perfect square. In the case above, 18*2=36 is the only one that works

If you look at the prime factorization of each answer, you will want the numerator to have a 2 to an odd exponent, such that when you divide by two, you get 2 to an even exponent, making that part of the prime factorization a perfect square. (For a number to be a perfect square, the prime factorization must have exponents of all even powers; so after the division by 2 if you are left with an odd exponent, it cannot be a perfect square). Luckily, only one of the answer choices fits this requirement, The (17!18!)/2 option.

This is because the larger factorial number in the numerator needs to have an odd power of 2 that the other factorial number does not contain. This will make the parity odd on the top (If it contained an extra even power, then it would not work, and if both numbers contained the new odd exponent of 2, then that would make it an extra multiple of 2, and would thus not work).

17!18! Works because 18 includes 9 * 2, which is an extra odd exponent of 2. None of the others work.

This was my way of thinking:

• In this case, we have to go by trials and errors, but we don't want to calculate all this factorials. As you probably know, a factorial is really quick going to infinity, so numbers would probably be enourmous in this case.
• So we'll need to simplify those expressions. At the nominator we always have this form: $n! (n+1)!$ If we simply write it as $n!n!(n+1)$ , we already have a perfect square here: $(n!)^2$ .
• The fraction is now: $(n!)^2 \frac{n+1}{2}$ . We know that $n^2 m^2 = (n+m)^2$ , so if $\frac{n-1}{2}$ is a perfect square, we can conclude for sure that the original number will be a perfect square too.

This method is useful because it translates the checking of perfect squares from the original numbers to simpler ones. This is the method in practice: $\frac{14! \cdot 15!}{2} = (14!)^2 \frac{15}{2}$ is not a perfect square, because $\frac{15}{2}$ isn't even an integer.

With this method, we can check all of them and conclude that $\frac{17! \cdot 18!}{2}$ is a perfect square.

17!18!/2 =(17!)^2× (18)/2 for perfect sequar 18/2=9 must be sequar.