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⌊ 2020!/(1!+2!+3!+..+2019!)⌋


Well-known member
Apr 5, 2012
Finding value of $\displaystyle \bigg\lfloor \frac{2020!}{1!+2!+3!+\cdots +2019!}\bigg\rfloor$


Well-known member
Nov 26, 2013
My attempt:

For $n > 2$ the following is true:

\[F_n = \left \lfloor \frac{(n+1)!}{1!+2!+3!+...+n!} \right \rfloor = n-1\]

Proof by induction:
Base cases:
\[F_3 =\left \lfloor \frac{4!}{1!+2!+3!} \right \rfloor = \left \lfloor \frac{24}{9} \right \rfloor = 2. \\\\ F_4 = \left \lfloor \frac{5!}{1!+2!+3!+4!} \right \rfloor = \left \lfloor \frac{120}{33} \right \rfloor = 3.\]

Suppose the identity holds for some n = m > 4. We need to show, that the identity also holds for n = m+1.

We have the identity: \[ F_m =\left \lfloor \frac{(m+1)!}{1!+2!+3!+...+m!} \right \rfloor = m-1.\]

To ease the algebra, let \[\sigma = 1!+2!+3!+...+m!\]

Then, we can write:

\[F_{m+1}=\left \lfloor \frac{(m+2)!}{1!+2!+3!+...+(m+1)!} \right \rfloor \\\\ =\left \lfloor \frac{(m+1)!}{\sigma +(m+1)!}(m+2) \right \rfloor\\\\ =\left \lfloor \frac{\frac{(m+1)!}{\sigma }}{1+\frac{(m+1)!}{\sigma }}(m+2) \right \rfloor\]

We know, that \[\frac{(m+1)!}{\sigma } = m-1+\varepsilon\] for some $0< \varepsilon<1$.

In other words: \[F_{m+1} =\left \lfloor \frac{m-1+\varepsilon }{m + \varepsilon }(m+2) \right \rfloor =\left \lfloor \left ( 1-\frac{1}{m+\varepsilon } \right )(m+2) \right \rfloor\]

Now, the fraction $\frac{m+2}{m+\varepsilon}$ has the sharp limits: \[1 < \frac{m+2}{m+\varepsilon } <2\]

This follows from the inequalities: $\varepsilon < 2 < m +2\varepsilon$

Thus the fraction can be written as: $\frac{m+2}{m+\varepsilon} = 1+\delta$ for some $0 < \delta < 1$.
Finally, we get

\[F_{m+1} =\left \lfloor m+2- (1+\delta )\right \rfloor = \left \lfloor m \right \rfloor+\left \lfloor 1-\delta \right \rfloor = m.\] q.e.d.

- and we conclude, that $F_{2019}= 2018.$



Well-known member
Apr 5, 2012
Thanks Ifdahl for nice solution

Here is mine

Using $n!=n(n-1)! = [(n-1)+1](n-1)! = (n-1)(n-1)!+(n-1)(n-2)!$

So $(n-1)(n-1)!+(n-1)(n-2)!<(n-1)(n-1)!+(n-1)(n-2)!+(n-1)(n-3)!+\cdots (n-1)1!\;\forall n\geq 4$

So $n!<(n-1)\bigg[(n-1)!+(n-2)!+(n-3)!+\cdots +2!+1!\bigg]\cdots \cdots (1)$

And $n!=n(n-1)!=[(n-2)+2](n-1)!=(n-2)(n-1)!+2(n-1)!=(n-2)(n-1)!+2(n-1)(n-2)!$

So $n!=(n-2)(n-1)!+(n-2)(n-2)!+n(n-2)(n-3)!$

As $n(n-3)!>(n-3)!+(n-4)!+\cdots +2!+1!\forall n\geq 4$

So $n(n-2)(n-3)!>(n-2)\bigg[(n-3)!+(n-4)!+\cdots +2!+1!\bigg]$

So $n!>(n-2)\bigg[(n-1)!+(n-2)!+\cdots\cdots +2!+1!\bigg]\cdots \cdots (2)$

From $(1)$ and $(2),$ We have

$\displaystyle (n-2)<\frac{n!}{1!+2!+3!+\cdots \cdots +(n-1)!}<(n-1)$

So we get $\displaystyle \bigg\lfloor \frac{n!}{1!+2!+3!+\cdots \cdots +(n-1)!}\bigg \rfloor =(n-2)$

Now put $n=2010,$ We get $\displaystyle \bigg\lfloor \frac{2020!}{1!+2!+3!+\cdots \cdots +2019!}\bigg \rfloor =2018$