# a few examples of solutions to Znám’s problem

For $k=5$ in Znám’s problem, two sets of integers can be constructed. To check each set involves calculating $k$ different deleted products  .

To check that 2, 3, 7, 47, 395 is a solution, we verify that

$3\cdot 7\cdot 47\cdot 395+1=389866$, and $2|389866$

$2\cdot 7\cdot 47\cdot 395+1=259911$, and $3|259911$

$2\cdot 3\cdot 47\cdot 395+1=111391$, and $7|111391$

$2\cdot 3\cdot 7\cdot 395+1=16591$, and $47|16591$

$2\cdot 3\cdot 7\cdot 47+1=1975$, and $395|1975$.

To check that 2, 3, 11, 23, 31 is a solution, we verify that

$3\cdot 11\cdot 23\cdot 31+1=23530$, and $2|23530$

$2\cdot 11\cdot 23\cdot 31+1=15687$, and $3|15687$

$2\cdot 3\cdot 23\cdot 31+1=4279$, and $11|4279$

$2\cdot 3\cdot 11\cdot 31+1=2047$, and $23|2047$

$2\cdot 3\cdot 11\cdot 23+1=1519$, and $31|1519$.

$3\cdot 7\cdot 43+1=904$, and $2|904$

$2\cdot 7\cdot 43+1=603$, and $3|603$

$2\cdot 3\cdot 43+1=259$, and $7|259$

$2\cdot 3\cdot 7+1=43$, and obviously $43|43$ because $43=43$.

The vast majority of the sets in the known solutions include 2. A set is known for $k=13$ with its smallest two elements being 3 and 4. It’s not known if there are any sets consisting entirely of odd numbers   . But since 2 is a prime number  , sets consisting entirely of prime numbers are possible, such as the second one given above, and for $k=7$ there is 2, 3, 11, 17, 101, 149, 3109.