to solve this problem one needs to know what exactly the maximal power of a prime $p$ is for any of the binomial coeffs, $\binom{n}{k}\,.$suppose $a(n)=\text{lcm}(1,2,\ldots,n),b(n)=\text{lcm}(\binom{n}{1},\binom{n}{2},\ldots,\binom{n}{n})\,.$ writing $n,k$ in $p\,-\text{ary}$ notation, the power of $p$ dividing $\binom{n}{k}$ is simply the number of $1's$ that are "carried over" in $p-\text{ary}$ addition of $k$ with $n-k\,.$ now if $p^{\alpha}\leq n<p^{\alpha+1}$ then it is easy to see that the lcm of $1,2,\ldots,n$ will be exactly divisible by $p^{\alpha}$.Thus if the same power were to appear in $\binom{n}{k}$ (for some $k$) as well, then from the above remark it follows that (since $n$ in $p$-ary notation will necessarily have $\alpha+1$ digits) that there have to be precisely $\alpha$ 1's carried over (i.e. in every 'position' except the units place) in addition of $k$ and $n-k$, whatever $k$ may be. now if $n+1$ were not prime, say $p|n+1,p<n$, then we see that in $p$-ary notation $n$ has as its last digit, $p-1$. hence no matter what $k$ is, $n-k$ and $k$ will not carry over a $1$ onto the next place digit and so $b(n)<a(n)$. on the other hand, if $n+1$ were prime, then for every prime $p\leq n$, the last digit of $n$ in $p$-ary notation will be strictly smaller than $p-1$ and so carry overs are likely. in fact it is easy to 'construct' a $k$ such that $k$ and $n-k$ will have a "carryover" at each digit and that finishes the proof. i have written this latter part a bit vaguely because i don't have the patience to write out more details. but i hope that the idea is conveyed and in that case the formal proof is just a formality.

I knwo general case : Prove that: $\frac{1}{n+1}[1,2,\dots,n+1]=[\binom{n}{1},\binom{n}{2},\dots,\binom{n}{n}]$ So in this $n+1=p$ and problem will sove easily!

a_vakilian wrote: I knwo general case : Prove that: $ \frac {1}{n + 1}[1,2,\dots,n + 1] = [\binom{n}{1},\binom{n}{2},\dots,\binom{n}{n}]$ So in this $ n + 1 = p$ and problem will sove easily! Sorry,I don't understand this.Are there anything wrong?

This is a problem from American Mathematical Monthly 1979 KDS wrote: a_vakilian wrote: I knwo general case : Prove that: $ \frac {1}{n + 1}[1,2,\dots,n + 1] = [\binom{n}{1},\binom{n}{2},\dots,\binom{n}{n}]$ So in this $ n + 1 = p$ and problem will sove easily! Sorry,I don't understand this.Are there anything wrong?

Lemma : $ (n+1) lcm \left( \binom{n}{1}, \binom{n}{2} , \dots, \binom{n}{n} \right) = lcm \left( 1,2,\dots,n+1 \right) $ Proof : Let $p$ be a prime and $ p < n $. Let $ k \in \mathbb{N} $ such that $ p^k \leq n+1 < p^{k+1} $ Then $ v_p \left( lcm \left( 1,2,\dots,n+1 \right) \right) = k $ Now $ (n+1)\binom{n}{i} = (i+1)\binom{n+1}{i+1} $ Thus $ v_p \left( (n+1) lcm \left( \binom{n}{1}, \binom{n}{2} , \dots, \binom{n}{n} \right) \right) = \underset{ i \in \{ 1,2,\dots,n \}}{ max } v_p \left( (i+1) \binom{n+1}{i+1} \right) \geq v_p \left( p^k \binom{n+1}{p^k} \right) \geq k $ So it is sufficient to show that $ \underset{ i \in \{ 1,2,\dots,n \}}{ max } v_p \left( (n+1) lcm \left( \binom{n}{1}, \binom{n}{2} , \dots, \binom{n}{n} \right) \right) \leq k $ By Legendre's formula $ v_p \left( \binom{n+1}{i+1} \right) = \displaystyle{\sum_{r=1}^k \left( \left \lfloor \frac{n+1}{p^r} \right \rfloor -\left \lfloor \frac{n-i}{p^r} \right \rfloor -\left \lfloor \frac{ i+1}{p^r} \right \rfloor \right)} $ Now observe that if $ r \leq v_p( i+1) $ then $ \displaystyle{\left( \left \lfloor \frac{n+1}{p^r} \right \rfloor - \left \lfloor \frac{n-i}{p^r} \right \rfloor - \left \lfloor \frac{ i+1}{p^r} \right \rfloor \right)} = 0 $ Let $ i = p^r u - 1 $ then $\displaystyle{ \left \lfloor \frac{n+1}{p^r} \right \rfloor - \left \lfloor \frac{n-i}{p^r} \right \rfloor - \left \lfloor \frac{ i+1}{p^r} \right \rfloor = \left \lfloor \frac{n+1}{p^r}\right \rfloor - \left \lfloor \frac{n+1}{p^r} - u \right \rfloor - u }= 0 $ And $ \lfloor x+y \rfloor - \lfloor x \rfloor - \lfloor y \rfloor = \{0,1\} $ Hence $\displaystyle{ \sum_{r=1}^k \left( \left \lfloor \frac{n+1}{p^r} \right \rfloor - \left \lfloor \frac{n-i}{p^r} \right \rfloor -\left \lfloor \frac{ i+1}{p^r} \right \rfloor \right) \leq k - v_p( i+1) }$ $ \Rightarrow v_p \left( (i+1) \binom{n+1}{i+1} \right) \leq k $ Hence $ \underset{ i \in \{ 1,2,\dots,n \}}{ max } v_p \left( (n+1) lcm \left( \binom{n}{1}, \binom{n}{2} , \dots, \binom{n}{n} \right) \right) = k $ Thus our lemma holds true. So we have to show that $ lcm(1,2,\dots,n)= lcm \left( \binom{n}{1}, \binom{n}{2}, \dots , \binom{n}{n} \right) $ iff $ n+1 $ is a prime. Applying our lemma $ (n+1) lcm \left( \binom{n}{1}, \binom{n}{2} , \dots, \binom{n}{n} \right) = lcm \left( 1,2,\dots,n+1 \right) $ $ \Leftrightarrow (n+1) lcm( 1,2,\dots,n ) = lcm( 1,2,\dots,n+1) $ $ \Leftrightarrow gcd(n+1,i) = 1 \ \forall \ i \in \{ 1,2,\dots,n \} $ $ \Leftrightarrow n+1 $ is a prime. Hence proved.