Estonian Math Competitions

Answer: (a) Yes; (b) No.

(a) Consider $n$ consecutive integers $k+1,k+2,\dots ,k+n$. As $i\le n$, there must be a multiple of $i$ among them; denote it $i^{\prime }$. As $j\le n$, there must also be a multiple $x$ of $j$ among them. If $x\ne i^{\prime }$ then one may choose $j^{\prime }=x$; then the product $i^{\prime }{j}^{\prime }$ is divisible by the product $ij$. Suppose in the rest that $x=i^{\prime }$; then $i^{\prime }$ as a common multiple of $i$ and $j$ is divisible by $\text{lcm}(i,j)$. Note that if $\text{gcd}(i,j)>\frac{n}{2}$ then $\text{gcd}(i,j)$ could not have two distinct multiples $i$ and $j$ among $1,2,\dots ,n$. Thus $\text{gcd}(i,j)\le \frac{n}{2}$, which in turn implies that $\text{gcd}(i,j)$ must have two distinct multiples among $k+1,k+2,\dots ,k+n$. At least one of them differs from $i^{\prime }$; let that be $j^{\prime }$. Then the product $i^{\prime }{j}^{\prime }$ is divisible by the product $\text{lcm}(i,j)\text{gcd}(i,j)$ which equals $ij$.

(b) Let $n=143$ and $i=77=7\cdot 11$, $j=91=7\cdot 13$, and $k=143=11\cdot 13$; then $ijk=7^2\cdot 11^2\cdot 13^2$. We show that among $143$ consecutive integers $p-71,p-70,\dots ,p+70,p+71$, where $p=6006=2\cdot 3\cdot 7\cdot 11\cdot 13$, no three distinct numbers can have a product divisible by $7^2\cdot 11^2\cdot 13^2$. For that, note that the largest number less than $p$ that is divisible by at least two numbers among $7,11$ and $13$ is $p-77$, and similarly, the least number greater than $p$ that is divisible by at least two numbers among $7,11$ and $13$ is $p+77$. Both lie outside the region under consideration. Consequently, at most one prime among $7,11$ and $13$ can belong to the canonical representation of any of the numbers $p-71,p-70,\dots ,p-1,p+1,\dots ,p+70,p+71$. Now choose any three numbers among $p-71,p-70,\dots ,p+70,p+71$. Let $p$ be among these three numbers. The primes $7,11$ and $13$ have exponent $1$ in its canonical representation. As shown above, only one of these three primes can occur in the canonical representation of either of the other two chosen numbers. Thus the product of the chosen three numbers cannot be divisible by $7^2\cdot 11^2\cdot 13^2$. Let all these three numbers differ from $p$. As shown above, each of these numbers can be divisible by at most one prime among $7,11$ and $13$. Hence, for their product to be divisible by $7^2\cdot 11^2\cdot 13^2$, one of the numbers should be divisible by $13^2$. But $\frac{p}{13}=6\cdot 77\equiv -6(\mathrm{mod}13)$ which implies $p\equiv -6\cdot 13(\mathrm{mod}13^2)$. Thus the nearest to $p$ numbers divisible by $13^2$ are $p-7\cdot 13$ and $p+6\cdot 13$ which lie outside the region under consideration. Consequently, the product of the chosen three number cannot be divisible by $7^2\cdot 11^2\cdot 13^2$.