Floor and Factorial

Number Theory Level 4

$\large\left\lfloor \frac { x }{ 1! } \right\rfloor +\left\lfloor \frac { x }{ 2! } \right\rfloor +\left\lfloor \frac { x }{ 3! } \right\rfloor =n$

Let $x$ be integer. The integer $n$ is achievable if there exists $x$ satisfying the equation, otherwise it is NOT achievable .

How many positive integers $n$ , where $1\le n \le 2016$ , is achievable?

Note: $n=224$ is achievable while $n=222$ is NOT achievable.

Inspiration

The answer is 1210.

1 solution

Tran Hieu
Jan 22, 2016

Let's $f(x)=\left\lfloor \frac { x }{ 1! } \right\rfloor +\left\lfloor \frac { x }{ 2! } \right\rfloor +\left\lfloor \frac { x }{ 3! } \right\rfloor$ .

We have $f(x+1)-f(x)=\left\lfloor \frac { x+1 }{ 1! } \right\rfloor +\left\lfloor \frac { x+1 }{ 2! } \right\rfloor +\left\lfloor \frac { x+1 }{ 3! } \right\rfloor -\left\lfloor \frac { x }{ 1! } \right\rfloor -\left\lfloor \frac { x }{ 2! } \right\rfloor -\left\lfloor \frac { x }{ 3! } \right\rfloor$

$=(\left\lfloor \frac { x+1 }{ 1 } \right\rfloor - \left\lfloor \frac { x }{ 1 }\right\rfloor)+(\left\lfloor \frac { x+1 }{ 2} \right\rfloor-\left\lfloor \frac { x }{ 2 } \right\rfloor) +(\left\lfloor \frac { x+1 }{ 6 } \right\rfloor -\left\lfloor \frac { x }{ 6 } \right\rfloor)$

$\geq 1+0+0 = 1$

So we can conclude that $f(a)\neq f(b) \forall a\neq b$ . Also with a little calculation we have $f(1210)=2016$ , that means from 1 to 2016 there are $\boxed{1210}$ value of $f(x)$

Floor and Factorial

The answer is 1210.

1 solution

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