Prove that $\gcd{\left(\binom M1,\binom M2,\binom M3,\ldots,\binom Mn\right)}=1$ where $M=\mathrm{lcm}(1,2,3,\ldots,n)$

Question

Let $n$ be a positive integer and let $$M=\mathrm{lcm}(1,2,3,\ldots,n).$$ Show that $$\gcd{\left(\binom{M}{1},\binom{M}{2},\binom{M}{3},\ldots,\binom{M}{n}\right)}=1$$

Answer 1

HINT: Suppose that $p$ is a prime dividing $M$, and that $p^k$ is the highest power of $p$ dividing $M$.

• Show that $p^k\le n$.
• Show that if $M=ap^k$, and $1\le t\le p^k$, then $(a-1)p^k+t$ is divisible by the same maximum power of $p$ as $t$.
• Conclude that $p$ does not divide $\dbinom{M}{p^k}$.