## Problem 1

Do there exist 14 consecutive positive integers each divisible by a prime less than 13 ? What about 21 consecutive positive integers each divisible by a prime less than 17 ?


## Solution

Answer: no, yes.

There are 7 odd numbers. At most one can be a multiple of 7 and one a multiple of 11. It is only possible to have two of them multiples of 5 if either the largest or smallest is a multiple of 5 . But it is only possible to have three of them multiples of 3 if both the largest and the smallest are multiples of 3 . So at most four numbers are multiples of 3 or 5 . That leaves one odd number unaccounted for.

A little juggling shows that we want the odd numbers to be divisible by: $3,7,5,3,11,13,3,5,7,3$ (or the 11 and 13 can be interchanged). We need the 6th odd to be 13 mod 143 to get the 11 and 13 correct, and hence the 1 st in the sequence to be 2 mod 143. But the 1st must be even, so it must be 2 mod 286 . The 2 nd must be divisible by 3 , so the 1 st must be $2 \bmod 858$. But we need the 4 th to be divisible by 7 , so the 1 st must be $3434 \bmod 6006$. Then the 6 th must be divisible by 5 , so the 1 st must be $9440 \bmod 30030$. Thus an example is $9440,9441, \ldots, 9460$.

