Problem 4

(a) For which integers

n>2

does there exist a set of

n

consecutive positive integers such that the largest number in the set divides the least common multiple of the remaining

n-1

numbers? (b) For which integers

n>2

is there exactly one such set?

Step 1 of 6: Bound the set size by the largest prime power in

k

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In plain words

The 'hardest' prime-power factor of $k$ dictates how many consecutive numbers you need before another multiple of that same prime power shows up.

k=\prod_i p_i^{e_i} \implies \text{need } n > \max_i p_i^{e_i}

Detailed analysis

If $k$ (the largest number) is divisible by a prime power $p_i^{e_i}$ , then $k$ divides the lcm of the other numbers only if that lcm is also divisible by $p_i^{e_i}$ ; the closest multiple of $p_i^{e_i}$ below $k$ is $k-p_i^{e_i}$ , so the set must reach down at least that far, i.e. it must have at least $p_i^{e_i}$ elements. Taking the maximum over all prime powers dividing $k$ gives the stated bound.