It is equavalent to find largest m, suth that $\sum_{p-prime<m} [\frac{ln(m-1)}{ln(p)}]ln(p) <3ln(m).$ We have $2^2*3*5*7=420<8^3$ and $2^3*3*5*7=840>9^3,2^3*3^2*5*7=2520>10^3,\dots$. Therefore n=420.

When have cube root of n less than or equal to 8, n is divisible by 420 (the lcm of 1,2,3,4,5,6,7). then largest n is 420, because 8^3 = 512. Now let's see what happens when the cube root is greater than or equal to 8.Have: n is divisible by lcm(1 through 8)=840. So, n =< 9^3 = 729 doesn't work. Then n is divisible by lcm(1-9) = 7560. n >= 19 ^ 3. but then, n is divisible by 11,13,17 so lcm(1-cube root of n) gets huge, nad from there it's pretty easy to prove that n>= 19 doesn't work. So, it's 420

rem wrote: lcm(1-9) = 7560. It's 2520, not 7560. rem wrote: nad from there it's pretty easy to prove that n>= 19 doesn't work This is the only "complicated" part of the problem. Anyone can notice that lcm gets huge. The problem is actually asking you to prove that.

For any integer n, select k s.t. $ k^3 \leq n < (k + 1)^3$ We have $ lcm(1,2,3,.....,k) \geq k(k - 1)(k - 2)(k - 3)/6 > (k + 1)^3 > n$ for $ k > 12$ Hence, $ n < 13^3 = 2197$ Then we easily see that the answer is 420 Im unable to understand your solution Rust. Could you please explain it ?

First off, remark that for the region $7^3<n<8^3$, we have $\lfloor \sqrt[3]{n} \rfloor=7$ therefore we must have $\text{lcm}(1,2, 3, 4, 5, 6, 7)|n$ giving us $420|n$ or therefore since $343<n<512$ we must have $n=420$. However, for the region $8^3<n<9^3$, we have $\lfloor \sqrt[3]{n} \rfloor=8$ therefore we must have $\text{lcm}(1,2, 3, 4, 5, 6, 7, 8)|n$ giving us $840|n$ however, $512<n<729$ giving us an obvious contradiction. From here, I am going to prove that $\text{lcm}(1,2, 3, \cdots n)>(n+1)^3$ for all $n\ge 9$ which in turn proves that looking in the region $n^3, (n+1)^3$, there are no solutions for all $n\ge 8$ (since we have already proven $8$ above) giving $\boxed{420}$ as our maximum. First off, remark that the maximum power of $2$ less than $n$ is at least $\frac{n}{2}$, for $3$ it is$\frac{n}{3}$, for $5$ it is $\frac{n}{5}$ and for $7$ it is at least $\frac{n}{7}$. From here, I will prove that the given equation works for $n=9$ through $13$ then explain why this completes our proof. For $n=9$, we get $3*840>10^3$ obviously true. For $n=10$, we get $3*840>11^3$ or $2520>1331$ again true. For $n=11$, we get $2520*11>12^3$ or $27720>12^3$ which is obviously true along with $27720>13^3$ and $27720>14^3$ since $27720>20^3=8000$. From here, remark that the maximum power of $11$ less than $11$ is at least $\frac{n}{11}$ and for $13$ it is $\frac{n}{13}$. Therefore, since we are going to take the product of all of these different factors in the lcm, we are going to assume the minimal case and get $\frac{n^6}{2*3*5*7*11*13}$ which we desire to have being greater than $(n+1)^3$. Taking the cube root of both sides, we get an equation approximately the same as $\frac{n^2}{32}>(n+1)$ or therefore $n\ge 34$. Therefore we need to prove that for $n=14\to 33$, we are going to have Since we have at least $\text{lcm}(1,2, 3, \cdots, n)=\text{lcm}(1,2,3,4,\cdots, 13)$ which we desire to prove is greater than $(n+1)^3$ for $n=33$ proving our maximum case and thus proving the cases for $n=14\to 33$ as well since we are only using $n=13$ in our $\text{lcm}$. Note that $\text{lcm}(1,2, 3, 4, \cdots, 13)\approx 71^3$ therefore it is obviously greater than $34^3$ and our proof is complete $\blacksquare$.

Why????? First off, remark that the maximum power of 2 less than n is at least \frac{n}{2}, for 3 it is\frac{n}{3}, for 5 it is \frac{n}{5} and for 7 it is at least \frac{n}{7}

Since chek for n=7 the maximum power of 2 is 4 isnt it?

Also this one how do you get for all n>12 it holds? We have lcm(1,2,3,.....,k) \geq k(k - 1)(k - 2)(k - 3)/6 > (k + 1)^3 > n for k > 12 If you indut it tell how to induct that one

Why????? First off, remark that the maximum power of $2$ less than $n $ is at least $\frac{n}{2}$, for $3 $ it is $\frac{n}{3}$ , for $ 5 $ it is $ \frac{n}{5} $and for $ 7 $ it is at least $\frac{n}{7}$

nawaites wrote: Why????? First off, remark that the maximum power of 2 less than n is at least \frac{n}{2}, for 3 it is\frac{n}{3}, for 5 it is \frac{n}{5} and for 7 it is at least \frac{n}{7} nawaites wrote: Why????? First off, remark that the maximum power of $2$ less than $n $ is at least $\frac{n}{2}$, for $3 $ it is $\frac{n}{3}$ , for $ 5 $ it is $ \frac{n}{5} $and for $ 7 $ it is at least $\frac{n}{7}$ Huh! What's the need to double-post after just $\text{\LaTeX}\text{ifying}$ the previous post? You could have easily edited it there.

Sorry to revive. Could some mathlinkers explain Rust's solution? How can it be related with logrithm? Thanks in advance.

Of course, the general problem can also be shown to possess finitely many solutions: shobber wrote: Find the largest integer $n$ such that $n$ is divisible by all positive integers less than $\sqrt[k]{n}$ Let $\text{lcm}(1,2,\dots,m)=\mathcal{L}(m)$. Let the primes less than or equal to $m$ be $2,3,\dots,p$. We have $$\frac{\mathcal{L}(m)}{m^k}=\frac{2^{\left\lfloor \log_2(m)\right\rfloor}3^{\left\lfloor \log_3(m)\right\rfloor}\cdots p^{\left\lfloor \log_p(m)\right\rfloor}}{m^k}\geq\frac{2^{\log_2(m)-1}3^{\log_3(m)-1}\cdots p^{\log_p(m)-1}}{m^k}=\frac{m^{\pi(m)-k}}{2\cdot3\cdots p}$$Since the right is strictly increasing except when $n$ is prime (in which case it remains constant), after a certain point we must have $\tfrac{\mathcal{L}(m)}{m^k}>1\implies \mathcal{L}(m)\nmid m^k$ $\Box$

this qn is from SMO open 2007 q5 ans is 420. suppose $p_k$ is the largest prime divisor of $x$ and the $k$th prime we show $p_k\leq 7$ suppose $p_k\geq 11$ $p_{k+1}<2p_k<4p_{k-1}<8p_{k-2}$ by Bertrand's postulate $p_{k+1}^3 < 64 p_k p_{k-1} p_{k-2} \leq 2^3 3^2 p_{k-2} p_{k-1} p_k | lcm(1,...,y) |x$ where $y=\lfloor \sqrt[3]{x} \rfloor$ , the first ineq comes from bertrand's postulate and third relation (divides) comes from smallest two and largest three prime factors of lcm(1,...,y). so $p_{k+1} < \sqrt[3]{x}$ $p_{k+1}|x$ , contradicting the maximality of $p_k$ so $p_k\leq 7 \implies \sqrt[3]{x}<p_{k+1}\leq 11$ $x<11^3=1331$ and $420|x$ implies x=420,840,1260 but only 420 satisfies conditions i dont get @rust soln

The problems is casework

The main idea is to bound $n$ above; notice that for the condition to be true, we need $$\text{lcm}(1, 2, 3, \cdots, n) < (n+1)^3.$$We may lower bound $$\text{lcm}(1, 2, 3, \cdots, n) \geq \frac{n(n-1)(n-2)(n-3)}6$$because at most one factor of 2 and one factor of 3 are shared among these numbers. Thus, we must have $$\frac{n(n-1)(n-2)(n-3)}6 < (n+1)^3 \iff n \leq 12,$$so conducting a manual case check reveals $\boxed{420}$ to be the maximum when $n=7$.