1981 IMO Problems/Problem 4
Problem
(a) For which values of is there a set of consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining numbers?
(b) For which values of is there exactly one set having the stated property?
Solution
Let be the greatest element of the set, written in its prime factorization. Then divides the least common multiple of the other elements of the set if and only if the set has cardinality at least , since for any of the , we must go down at least to to obtain another multiple of . In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.
For , we may let , since all the must clearly be less than and this product must also be greater than if is at least 4. For , we may also let , for the same reasons. However, for , this does not work, and indeed no set works other than . To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.
Q.E.D.
Alternate Solution
Let, for some and with , .
We can trivially check that, there is no such for , only works for and works for .
Now, consider, . By Bertrand's postulate there is a prime such that .
Which implies, .
As, , there must be a multiple of , a multiple of and a multiple of in the set, .
So, and .
So, and both work for .
So,
There exists solution for all ,
Only one Solution for .
Q.E.D.
1981 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
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