Need help on this a lot. Thanks!

I've given some more thought to this problem....I may have hit upon a "solution" [but, I'm not sure]...anyway....here's what I discovered.....

First of all, any positive square root, a, can always be written in one of the following two forms....

floor (√a) * (√a) = a     or    floor (-√a) * (√a)  = a

Now, except for the special case noted below....any other positive integer, n, can be written in one of the following two forms.....

floor (n/√s₁) * (n/√s₁)   or  floor (-n/√s₂) * (-n/√s₂)    ....where s₁ and s₂ are the the smaller and larger perfect squares, respectively, on either side of n

Let's take the given question as an example .....note the the two perfect squares on either side of 27 are 25 and 36...so, let us guess that we might be able to  write 27 in terms of  s₁, i.e., 25.  So we have...

floor(27/ √25) * (27 /√25) = floor (27/5) * (27/5) = floor(5.4) * (27/5) = (5) * (27/5)  = 27 ....!!!

Now let's look at another number, say, 31.....and let's guess that we can write it in terms of s₁, also...so we have

floor(31/ √25) * (31 /√25) = floor (31/5) * (31/5) = floor(6.2) * (31/5) = (6) * (27/5)  = 32.4 ....nope, that doesn't work out....so, let's try s₂  - (i.e, 36).......and we have

floor(-31/ √36) * (-31 /√36) = floor (-31/6) * (/-31/6) = floor(-5.4) * (-31/6) = (-6) * (-31/6)  = 31....!!! .....just what we hoped for !!!

The small exception to the rule is any number that can be factored in terms of (p) (p+1)....let us consider the number 12, which can be written in terms of (3) * (4)

Now... the smaller and larger perfect squares on either side of 12 are 9 and 16....so...using our method, s₁, (i.e., 9) yields the following result

floor(12/ √9) * (12/√9) = floor (12/3) * (12/3) = floor(4) * (4) = (4) * (4)  = 16 ....!!!...that's no good

Let's try s₂ , (i.e., 16)...so we have

floor(12/ √16) * (12/√16) = floor (12/4) * (12/4)= floor(3) * (3) = (3) * (3)  = 9...!!!..and that's no good, either....so.. ..these types of numbers are "skipped" 

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I see that Melody has posted an answer, as well......I'll take a look at hers...maybe it's better than mine...!!

x * floor(x)  = n

Melody's solution will work for 27

But, consider 12

x  = 12 / floor (√12) 

x  = 12/3

x = 4

I don't think this works out in this case, Melody....if x = 4, then floor(x) = floor(4)  = 4

And 4 * floor(4) = 16, not 12

Also, consider 31

x  =   31 / floor (√31)

x   = 31 / 5 

So x = 31/5

And

(31/5) * floor (31/5)  =

 (6.2)* (6)=   37.2, not 31....

I'm not sure how to solve this algebraically, but WolframAlpha gives the answer as 27/5 = 5.4.....this males sense because (27/5)* floor(27/5)  = (27/5) *(5)  = 27

(Maybe another forum member can show a "mathematical" way to solve this....!!!)

You need a number such that  x  and the  integer value(x)  when multiplied will give you 27.

If the answer is a perfect square, then the square root will be your answer.

Since 27 is between the perfect squares 25 and 36, the answer will be between their square roots, 5 and 6.

The integer value of a number between 5 and 6 is 5, so now you are looking for a number x that satisfies the equation:  x · 5  =  27                      (because 5 will be the integer value of x)

Solving this, you get  27/5  =  5.4

Thanks, geno....I'll recommend this one for Melody's Daily Wrap...!!!!

Just playing around with this in WolframAlpha, I noticed that...

floor(x)*(x) = 91 returns the solution  -91/10

but

floor(x)*(x) = 101  returns the solution 101/10

and

floor(x)*(x) = 122  returns the solution  122/11

but

floor(x)*(x) = 143  returns the solution  -143/12

I wonder if there's some rationale for the fact that, depnding on the number on the RHS of the equation, the parity of the solutions seems to "oscillate" ???  .... anybody ????

Maybe only a coincidence, but when, in those four cases, the positive answer was a whole number, it gave you the negative answer. ???

General question

$$x \times \lfloor x \rfloor =n \qquad $Where n is a known constant n\ge 1$$

$$\\x\times \lfloor \sqrt{n}\rfloor=n\\\\ x=\frac{n}{\lfloor \sqrt{n}\rfloor}\\\\$$

The gaps in the following graph show that x.floor(x) = M has a solution for some values of M, but not others:

.