IMO 1970 SL 4

Origin: CZS

Problem

For what natural numbers n can the product of some of the numbers n, n + 1, n + 2, n + 3, n + 4, n + 5 be equal to the product of the remaining ones?

Solution

Suppose that n is such a natural number. If a prime number p divides any of the numbers n, n+1, . . ., n+5, then it must divide another one of them, so the only possibilities are p = 2, 3, 5. Moreover, n + 1, n + 2, n + 3, n + 4 have no prime divisors other than 2 and 3 (if some prime number greater than 3 divides one of them, then none of the remaining numbers can have that divisor). Since two of these numbers are odd, they must be powers of

3 (greater than 1). However, there are no two powers of 3 whose difference is 2. Therefore there is no such natural number n. Second solution. Obviously, none of n, n + 1, . . . , n + 5 is divisible by 7; hence they form a reduced system of residues. We deduce that n(n +

1. \cdot \cdot \cdot (n+5) \equiv1\cdot2 \cdot \cdot \cdot6 \equiv−1 (mod 7). If {n, . . . , n+5} can be partitioned into two subsets with the same products, both congruent to, say, p modulo 7, then p2 \equiv−1 (mod 7), which is impossible. Remark. Erd˝os has proved that a set n, n + 1, . . . , n + m of consecutive natural numbers can never be partitioned into two subsets with equal products of elements.