$(11,10!)=1$ so there will be a natural number $x, 11\cdot x \equiv 1 \pmod {10!}$ Let $P=(a+ 1\cdot 11)(a+2\cdot 11) \ldots (a+10 \cdot 11)$, then $x^{10} \cdot P \equiv (a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10) \equiv 0 \pmod{10!}$ because $\frac{(a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10)}{10!} = \binom{a \cdot x+10}{10}$. $(x,10!)=1$ so $P$ is divisible by $10!$

rms wrote: $(11,10!)=1$ so there will be a natural number $x, 11\cdot x \equiv 1 \pmod {10!}$ Let $P=(a+ 1\cdot 11)(a+2\cdot 11) \ldots (a+10 \cdot 11)$, then $x^{10} \cdot P \equiv (a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10) \equiv 0 \pmod{10!}$ because $\frac{(a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10)}{10!} = \binom{a \cdot x+10}{10}$. $(x,10!)=1$ so $P$ is divisible by $10!$ Why will $x, 11\cdot x \equiv 1 \pmod {10!}$?

also how did you get $x^{10} \cdot P \equiv (a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10) \equiv 0 \pmod{10!}$?

MetaphysicalWukong wrote: rms wrote: $(11,10!)=1$ so there will be a natural number $x, 11\cdot x \equiv 1 \pmod {10!}$ Let $P=(a+ 1\cdot 11)(a+2\cdot 11) \ldots (a+10 \cdot 11)$, then $x^{10} \cdot P \equiv (a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10) \equiv 0 \pmod{10!}$ because $\frac{(a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10)}{10!} = \binom{a \cdot x+10}{10}$. $(x,10!)=1$ so $P$ is divisible by $10!$ Why will $x, 11\cdot x \equiv 1 \pmod {10!}$?

1) Use Bézout's identity: Click to reveal hidden text

for $d=1$. 2) Possibly, if $x<0$ you can replace $x$ with $k \cdot 10! + x>0$ for some $k \cdot 10!$ multiple of $10!$, to get ”will be a natural number $x, 11\cdot x \equiv 1 \pmod {10!}$”.

MetaphysicalWukong wrote: also how did you get $x^{10} \cdot P \equiv (a \cdot x+1)(a \cdot x +2) \ldots (a \cdot x+10) \equiv 0 \pmod{10!}$? For the first congruence start with $x \cdot (a+1 \cdot 11) \equiv a \cdot x+1\pmod{10!}, x \cdot (a+2 \cdot 11) \equiv a \cdot x+2\pmod{10!}$ and the others, then multiply them.

For the second congruence see binomial coefficient $\binom{n}{k}$ here Click to reveal hidden text

these numbers are natural: ”Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set. This number can be seen as equal to the one of the first definition, independently of any of the formulas below to compute it: if in each of the n factors of the power $(1 + X)^n$ one temporarily labels the term X with an index i (running from 1 to n), then each subset of k indices gives after expansion a contribution $X^k$, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that is a natural number for any natural numbers n and k.”