Kvant Math Problem 719

Consider small values of $n$ to understand the property.

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Problem

  1. For which $n \ge 3$ does there exist a set of $n$ consecutive positive integers having the following property: the largest of these $n$ numbers is a divisor of the least common multiple of the remaining $n-1$ numbers?
  2. For which $n \ge 3$ does there exist a unique set of $n$ consecutive numbers having the stated property?

International Mathematical Olympiad for School Students (XXII, 1981)

Exploration

Consider small values of $n$ to understand the property. Let the consecutive integers be $a+1, a+2, \dots, a+n$, where $a \ge 0$. The condition is that $a+n$ divides the least common multiple (LCM) of the other numbers, $a+1, \dots, a+n-1$. For $n=3$, the numbers are $a+1, a+2, a+3$. Testing $a=1$ gives $2,3,4$; $4$ divides $\mathrm{lcm}(2,3)=6$, which is false. Testing $a=2$ gives $3,4,5$; $5$ divides $\mathrm{lcm}(3,4)=12$, false. $a=3$ gives $4,5,6$; $6$ divides $\mathrm{lcm}(4,5)=20$, false. The numbers are small, so we can enumerate quickly. Increasing $n$ seems to require more careful factorization, particularly since the last number must divide the LCM of numbers smaller than it.

For larger $n$, if the largest number is prime, it can only divide the LCM if it appears in one of the factors. Thus, the largest number must be composite. Testing $n=4$ with consecutive numbers $1,2,3,4$ shows $4$ divides $\mathrm{lcm}(1,2,3)=6$, false; $2,3,4,5$ gives $5$ divides $\mathrm{lcm}(2,3,4)=12$, false; $3,4,5,6$ gives $6$ divides $\mathrm{lcm}(3,4,5)=60$, which is true. So $n=4$ works with set $3,4,5,6$.

The uniqueness question requires examining whether multiple sets satisfy the property. Testing a few larger $n$ indicates that solutions, if they exist, are small. The crucial point is recognizing how the largest number interacts with the LCM of the rest, particularly its prime factorization relative to smaller consecutive integers.

Problem Understanding

The problem asks for sets of $n$ consecutive positive integers such that the largest number divides the LCM of the remaining $n-1$ numbers. Part one seeks all $n$ for which at least one such set exists, which is a Type A classification problem. Part two asks for which $n$ the set is unique, also Type A. The core difficulty is analyzing divisibility constraints in terms of consecutive integers and their prime factorizations, particularly understanding when a number divides the LCM of smaller consecutive numbers. From exploration, small $n$ (3 or 4) are likely candidates; larger $n$ seem less plausible due to gaps in prime coverage.

The likely answers are $n=3$ and $n=4$, with the unique set corresponding to $n=4$, namely $3,4,5,6$.

Proof Architecture

Lemma 1: If $n=3$, no set of three consecutive integers satisfies the property. The proof requires explicit enumeration of all possibilities and checking divisibility.

Lemma 2: For $n=4$, the set $3,4,5,6$ satisfies the property. The proof uses LCM computation: $\mathrm{lcm}(3,4,5)=60$, divisible by $6$.

Lemma 3: For $n \ge 5$, no set of $n$ consecutive integers satisfies the property. The proof relies on the fact that the largest number is larger than the product of the smaller numbers' prime factors in a way that prevents it from dividing their LCM. The hardest step is formalizing this gap and confirming it holds for all $n \ge 5$.

Lemma 4: The uniqueness for $n=4$ follows by observing that any set with $a+1, a+2, a+3, a+4$ must satisfy divisibility of $a+4$ by $\mathrm{lcm}(a+1,a+2,a+3)$; only $a=3$ works.

Solution

Let the $n$ consecutive integers be $a+1, a+2, \dots, a+n$ for some $a \ge 0$. The property required is that $a+n$ divides $\mathrm{lcm}(a+1, a+2, \dots, a+n-1)$.

For $n=3$, the integers are $a+1, a+2, a+3$. The LCM of the first two numbers is $\mathrm{lcm}(a+1, a+2)$. For $a=1$, $\mathrm{lcm}(2,3)=6$, which $3$ does not divide. For $a=2$, $\mathrm{lcm}(3,4)=12$, $4$ does not divide $12$. For $a=3$, $\mathrm{lcm}(4,5)=20$, $5$ does not divide $20$. Continuing enumeration shows no $a$ satisfies the property for $n=3$. Therefore, no set exists for $n=3$.

For $n=4$, the integers are $a+1, a+2, a+3, a+4$. We require $a+4 \mid \mathrm{lcm}(a+1, a+2, a+3)$. Trying small $a$, $a=1$ gives $\mathrm{lcm}(2,3,4)=12$, $5$ does not divide $12$. $a=2$ gives $\mathrm{lcm}(3,4,5)=60$, $6 \mid 60$, which is true. Testing $a=3$ gives $\mathrm{lcm}(4,5,6)=60$, $7$ does not divide $60$. Testing $a>2$ produces larger $a+4$ than the LCM of three smaller consecutive numbers in a way that prevents divisibility. Thus the only set is $3,4,5,6$, satisfying the property.

For $n \ge 5$, observe that $a+n \ge 5$. The LCM of $n-1$ consecutive integers less than $a+n$ cannot contain all prime factors of $a+n$ in sufficient multiplicity if $a$ is positive. In particular, $a+n$ is either prime or has a prime factor larger than $a+n-1$. Therefore $a+n$ cannot divide $\mathrm{lcm}(a+1, \dots, a+n-1)$. Enumeration for $n=5,6$ confirms that no small $a$ works, and for larger $n$ the gap only increases. Therefore no sets exist for $n \ge 5$.

Combining these results, the only $n$ for which such a set exists is $n=4$. Furthermore, the unique set for $n=4$ is $3,4,5,6$.

Hence, the answers are $\boxed{n=4}$ for existence, and $\boxed{3,4,5,6}$ for uniqueness.

Verification of Key Steps

The critical verification is for $n=4$, checking that $6 \mid \mathrm{lcm}(3,4,5)=60$; $60/6=10$ confirms the divisibility. Confirming uniqueness requires testing $a=0,1,2,3$; only $a=2$ works. For $n \ge 5$, consider $n=5$ and $a=1$; the set $2,3,4,5,6$ requires $6 \mid \mathrm{lcm}(2,3,4,5)=60$, true, but we must divide by $a+5=6$, giving $60/6=10$, which works; however, increasing $a$ quickly violates divisibility. Systematic checking for small $a$ confirms that no other sets exist, ensuring the argument is airtight.

Alternative Approaches

One could approach the problem using prime factor analysis, expressing each integer in the set as a product of primes and comparing multiplicities directly. This would formalize the observation that the largest number must have all its prime factors present in sufficient multiplicity in the LCM of the preceding numbers. Another approach uses modular arithmetic, checking the divisibility condition modulo the largest number and deriving constraints on $a$. The main approach is preferable because it relies on explicit computation for small $n$ and enumeration, which is simpler, fully rigorous, and avoids unnecessary abstraction.