Consider six consecutive positive integers. Show that there is a prime number that divides exactly one of them.
(In reply to
re: n = 8... and more questions by JLo)
JLo:
I don't think you have said what you actually meant.
You have written:
"Consider n consecutive numbers. Then it is impossible that each prime number p<=n divides at least two of the n numbers!
- Can you prove or disprove the above, stronger statment?
- In case of disproval, what is a counter example with minimal n?
- For this minimal n, can you still prove the original puzzle?
"
The statement, as stated, is easily disproved:
N = 4
-------
3 4 5 6
The primes 2 and 3 are <= 4, and each divides two of the numbers. But the original problem is clearly true for n = 4.
N = 6
-------
5 6 7 8 9 10
The primes 2 and 3 and 5 are <= 6, and each divides at least two of
the numbers. But the original problem is for n = 6.
N = 10
--------
27 28 29 30 31 32 33 34 35 36
The primes 2 and 3 and 5 and 7 are <=10, and each divides at least
two of the numbers. I can disprove it with lower sequences,
but I chose this one because the beginning and ending numbers are not
divisible by 5 or 7.
re(2): n = 8... and more questions
