Floor + Factorial = ?

This can be solved using only fundamental knowledge of floor function.

For every real $n$ , $n-1<\left\lfloor n \right\rfloor \le n$ .

Applying this, we have:

$\frac { x }{ 1! } -1<\left\lfloor \frac { x }{ 1! } \right\rfloor \le \frac { x }{ 1! }$ $\frac { x }{ 2! } -1<\left\lfloor \frac { x }{ 2! } \right\rfloor \le \frac { x }{ 2! }$ $\frac { x }{ 3! } -1<\left\lfloor \frac { x }{ 3! } \right\rfloor \le \frac { x }{ 3! }$

Hence,

\frac { x }{ 1 } +\frac { x }{ 2 } +\frac { x }{ 6 } -3<224\le \frac { x }{ 1 } +\frac { x }{ 2 } +\frac { x }{ 6 }

By solving this inequality, we have $134.4 \le x < 136.2$ , and since $x$ is an integer, $x$ can only be $135$ or $136$ .

From this point, we can use trial and error to find the last solution, which is $\boxed{135}$

Let's try solving the following equation for x: $\frac{x}{1!} + \frac{x}{2!} + \frac{x}{3!} = 224$ $\frac{6x + 3x + x}{6} = 224$ $10x = 1344$ $x = 134.4$ Now we know that if we sum floor functions of positive numbers, and x must be positive becouse we have a som of x equal to a positive number (244), we get a result less or equal than the sum of the numbers without floor functions (for example $134.4 > \lfloor134.4\rfloor$ ). Hence, x must be greater than 134.4 to make true the equation above with floor functions. So let's try with some integer numbers greaters than 134.4. Let's begin trying with 135: $\lfloor\frac{135}{1!}\rfloor + \lfloor\frac{135}{2!}\rfloor + \lfloor{\frac{135}{3!}}\rfloor = 224$ $224 = 224$ We get that with x = 135 the equation is true, so the solution is $\boxed{135}$

Or easier: Another way to solve this problem is, after the first step (where we get x = 134.4), try the integer of 134.4 in the equation with the floor functions and see what we get: $\lfloor\frac{134}{1!}\rfloor + \lfloor\frac{134}{2!}\rfloor + \lfloor{\frac{134}{3!}}\rfloor = 224$

$134 + 67 + 22 = 224$

$223 = 224$

And that is not true just with a difference of 1, so if we think were we can put that 1 we see that the only place where it can survive is in $\lfloor\frac{x}{1!}\rfloor$ with the x = 135. So the solution is again $\boxed{135}$

1! = 1, 2! =2, and 3! = 6.

So if x = 6, 6/1 + 6/2 + 6/6 = 10.

if x = (6 * 22) = 132, 132/1 + 132/2 + 132/6 = 220.

If you increase x by 3, the first term goes up by 3, the second floor function goes up by 1, and the third stays the same. This gives a sum of 224.

therefore, x = 132 + 3 = 135

One tedious way can be by actually doing by brute force and finding the number.

Take x= 6n,6n+1...6n+5 and then solve the equation to get n=10 for x=6n+3

These floor signs make things a bit difficult, so me just remove them(for now): x/1 + x/2 + x/6 = 224 Now we multiply by 6 on both sides: 6x + 3x + 1x = 224 * 6 10x = 224 * 6 5x = 112 * 6 x = 672/5 Now, because we have to take the floor at the beginning, x must be rounded up now so if we round 672/5 up, we get 675/5 = 135

x + x/2 + x/6 = 224 6x + 3x + x = 1344 10x = 1344 x = 134.4 therefore the answer is 135

Since the factorial of 1,2 &3 are all integers so take them out of gif Then take lcm of 1,2&3 which is 6 Thus 6[x]+3[x]+[x] = 224 * 6 That is [x]= 1344/10= 134.4 And gif(greatest integer function) of 134.4 is 135

For intuition's sake, f(x) ~ x + (1/2)x + (1/6)x = 224 => (5/3)x = 224 x~ 134 & 2/5 (but has to be an integer)

Let f(x) be the given floor function above.

Examing the values given by x=1 through x =7, and then for verification analyzing from x=8 to x=14. Then, analyze the recursive pattern between f n+1 - f n e.g. f(1) = 0 then f(2) = 1 f(2)-f(1) = 1 , f(3)=3 f(3)-f(2) = 2, f(4)= 4 f(4)-f(3) = 2, f(5)=6 f(5)-f(4) = 2, f(6) = 7 f(6)-f(5) = 1, f(7)=10 f(7)-f(6) = 3, f(8) = 11 f(8)-f(7)= 1...

The difference between sequential terms, starting at n=1, has a pattern of 121213, that repeats itself forever, like 121213121213121213... forever. Because of this, you can calculate that each period has an overall increase of 1+2+1+2+1+3 = 10 . an increase of 10 per period constantly.

Since it follows a 6 element cycle with f(1) = 0 f(7)=10 f(13) = 20 f(19)=30 ... Y increases 10 every time x increases by 6, so you have a line with slope (5/3)

(5/3)x = 224

Since you know the answer is an integer and (5/3) does not divide evenly into 224, you divide it into 220 and get x=132

Since the cycle of the differences is 121213, and (5/3) divides evenly into 220, you know the next increase that happens will be of 1, then 2, then 1. At that point you have 220+1+2+1 = 224, which is the value expressed in the equation.

x=132 when y = 220 y= 220 + 1 =221 Now x = 133, y= 221 + 2 = 223 Now x = 134 y= 223 + 1 = 224 now x = 135

So, you reach y= 224 so x must equal 135.