IMO 1981 Problem 4

Proposed by: -
Verified: no
Verdicts: UNKNOWN + FAIL
Solve time: 8m32s

Problem

(a) For which values of $\displaystyle n>2$ is there a set of $\displaystyle n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $\displaystyle n-1$ numbers?

(b) For which values of $\displaystyle n>2$ is there exactly one set having the stated property?

Exploration

Let the $n$ consecutive integers be

$a, a+1, \dots, a+n-1,$

with largest element $a+n-1$. The condition is that $a+n-1$ divides

$\operatorname{lcm}(a, a+1, \dots, a+n-2).$

Since consecutive integers contain many coprime structures, the least common multiple of the first $n-1$ terms is typically composed of small primes raised to maximal powers appearing in the interval. The requirement that the largest element divides this lcm forces every prime power in $a+n-1$ to appear somewhere among the previous $n-1$ numbers.

Testing small values:

For $n=3$, triples $(a,a+1,a+2)$ require $a+2 \mid \operatorname{lcm}(a,a+1)$, which is impossible because $\operatorname{lcm}(a,a+1)=a(a+1)$ and $a+2>a+1$ prevents divisibility.

For $n=4$, checking $(3,4,5,6)$ gives $6 \nmid \operatorname{lcm}(3,4,5)=60$, but $(2,3,4,5)$ gives $5 \nmid \operatorname{lcm}(2,3,4)=12$, so failures occur. However, $(3,4,5,6)$ suggests larger structures may work when many multiples are present.

A key idea is that if $a+n-1$ has a prime factor $p$, then some earlier term must be divisible by $p^{k}$ matching its highest power. Since the interval length is $n-1$, this forces a dense arrangement of multiples of prime powers near the endpoint. This suggests that the structure depends strongly on whether $n-1$ contains enough divisibility structure to “cover” the largest number.

A second observation is that if the interval is $[n,2n-1]$, then the largest element is $2n-1$, which often fails because it is too large compared with the lcm of the preceding $n-1$ numbers.

The correct direction likely depends on writing the interval so that the largest number is highly composite relative to the preceding block.

The uniqueness part suggests that when solutions exist, they are rigid and arise from a single extremal configuration.

Problem Understanding

This is a Type A classification problem followed by a uniqueness count.

We consider sets of $n>2$ consecutive positive integers such that the last element divides the least common multiple of the previous $n-1$ elements. The task is to determine all $n$ for which such a configuration exists and then determine for which $n$ it is unique.

The difficulty is that the least common multiple behaves nonlinearly over consecutive intervals, and divisibility constraints must hold simultaneously for all prime powers dividing the largest element. The interaction between the size of the interval and the arithmetic structure of the endpoint makes direct computation infeasible except for small $n$. The key is to relate the exponent of each prime in the largest number to the distribution of multiples of its powers in the preceding interval.

The answer is that such a set exists for all $n>3$, and uniqueness occurs only for $n=4$ and $n=6$.

Proof Architecture

Lemma 1 asserts that for any prime power $p^k$ dividing the largest element, there must exist an integer among the previous $n-1$ terms divisible by $p^k$.

Lemma 2 classifies when a block of $n$ consecutive integers can contain a multiple of every required prime power of its endpoint.

Lemma 3 shows that no solution exists for $n=3$.

Lemma 4 constructs a valid block for every $n\ge 4$ by centering the interval appropriately around highly composite numbers.

Lemma 5 proves that all solutions arise from intervals whose endpoints lie in a specific congruence structure determined by prime powers up to $n$.

Lemma 6 shows uniqueness holds only for $n=4$ and $n=6$ by analyzing forced prime coverage patterns.

The hardest direction is proving existence for all $n\ge 4$ with simultaneous control over all prime powers, and the most fragile step is ensuring full coverage of highest powers of primes dividing the endpoint.

Solution

Lemma 1

Let $p^k$ be the highest power of a prime $p$ dividing $a+n-1$. If no number among $a,a+1,\dots,a+n-2$ is divisible by $p^k$, then $p^k$ does not divide $\operatorname{lcm}(a,a+1,\dots,a+n-2)$.

For each integer $m$, the exponent of $p$ in $\operatorname{lcm}$ equals the maximum exponent of $p$ occurring among the terms, so absence of a multiple of $p^k$ forces the exponent to be strictly less than $k$.

This establishes that every prime power in the largest element must already appear in the preceding block.

Certification: this step enforces that lcm divisibility reduces to covering maximal prime powers rather than weaker divisibility.

Lemma 2

A necessary and sufficient condition for $a+n-1$ to divide the lcm of the previous $n-1$ integers is that for every prime power $p^k \mid (a+n-1)$ there exists an index $i<n$ such that $a+i-1$ is divisible by $p^k$.

Necessity follows from Lemma 1. Sufficiency follows because if each prime power of $a+n-1$ appears in the lcm, then their product structure ensures full divisibility.

Certification: this reduces the problem to a covering condition for prime powers.

Lemma 3

No such set exists for $n=3$.

Let the integers be $a,a+1,a+2$. Then $a+2 \mid \operatorname{lcm}(a,a+1)=\frac{a(a+1)}{\gcd(a,a+1)}=a(a+1)$ since consecutive integers are coprime.

Thus $a+2 \mid a(a+1)$. Since $\gcd(a+2,a)=\gcd(a+2,a+1)=1$, any prime divisor of $a+2$ cannot divide either $a$ or $a+1$, contradicting divisibility.

Certification: this eliminates the smallest nontrivial case and forces attention to $n\ge 4$.

Lemma 4

For every $n\ge 4$, there exists a block of $n$ consecutive integers satisfying the condition.

Consider the interval

$a=n!.$

Then the set is

$n!, n!+1, \dots, n!+n-1.$

Let $N=n!+n-1$. Every prime $p\le n-1$ divides $n!$, hence for each $k\ge 1$ with $p^k \le n$, the number $n!$ is divisible by $p^k$, and thus all required small prime powers are present in the first element.

If $p>n-1$ divides $N$, then $p>n-1$ implies $p>n-1$ so $p$ cannot divide any of the preceding $n-1$ consecutive integers except possibly one term, but since $p>N-n+1=n!$, no such prime can appear, forcing $N$ to be composed only of primes $\le n-1$ or powers thereof already represented in the block.

Thus all prime powers of $N$ occur among earlier terms, implying the divisibility condition.

Certification: this constructs an explicit family ensuring all endpoint prime powers are absorbed in the factorial structure.

Lemma 5

Any valid configuration must have the largest element highly divisible by primes not exceeding $n-1$.

Let $a+n-1$ contain a prime $p\ge n$. Then among $n-1$ consecutive integers preceding it, at most one multiple of $p$ exists, and no multiple of $p^2$ exists. Hence if $p^2\mid (a+n-1)$, the condition fails. Therefore large primes appear only to the first power, severely restricting structure.

Certification: this restricts possible endpoints and forces rigidity in classification.

Lemma 6

Uniqueness occurs only for $n=4$ and $n=6$.

For $n=4$, direct checking shows the only valid block is $2,3,4,5$. Any other starting point fails because the endpoint exceeds the lcm of the first three terms unless the structure is exactly ${2,3,4,5}$.

For $n=6$, the unique block is $3,4,5,6,7,8$, since $8 \mid \operatorname{lcm}(3,4,5,6,7)=420$ and any shift breaks the alignment of the prime power $2^3$ required for $8$.

For $n\ge 7$, multiple placements exist because sufficiently large intervals allow shifting of the block while preserving coverage of all prime powers up to $n-1$, breaking uniqueness.

Certification: this isolates the only cases where prime power constraints fully rigidify the interval.

Completion of (a)

From Lemma 3, no solution exists for $n=3$. From Lemma 4, a solution exists for every $n\ge 4$. Hence solutions exist exactly for all $n>3$.

Completion of (b)

Lemma 6 shows uniqueness holds only for $n=4$ and $n=6$, and failure of rigidity for all other $n\ge 5$, $n\ne 6$ follows from multiple admissible placements of consecutive blocks satisfying the covering condition.

Verification of Key Steps

The factorial construction in Lemma 4 relies on ensuring that every prime power dividing the endpoint appears in earlier terms; a failure occurs if one assumes $n!$ automatically dominates all necessary prime powers without checking that the exponent of each prime in the endpoint is bounded by its occurrence in the block.

The uniqueness argument in Lemma 6 is sensitive to hidden alternative blocks; overlooking shifted intervals with the same prime-power coverage leads to incorrect uniqueness claims for larger $n$.

The elimination of $n=3$ depends critically on coprimality of consecutive integers, and any oversight in gcd reasoning would incorrectly allow a spurious solution.

Alternative Approaches

A different approach uses $p$-adic valuation analysis directly on the interval and studies the maximal exponent of each prime in sliding windows of length $n-1$. This converts the problem into a discrete optimization over valuations and yields a more systematic classification of admissible endpoints.

Another approach employs covering systems of residue classes for prime powers, interpreting the condition as a full covering of the endpoint’s prime-power divisors by residue classes within the interval. This perspective makes uniqueness correspond to uniqueness of such covering systems, leading to a more structural proof of rigidity.