Corrcect odd primes are wanted
How many odd primes are there such that is a divisor of
The answer is 3.
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1 solution
Notation: [x] is the largest integer less than or equal to a real number x.
If p is a divisor of k,then
Otherwise ,by Fermat's little thorem
So we see that
since [2018/p] is exactly the number of multipliers of p between 1 and 2018. This means
......(※)
We see that p<2019,because p≥2019 means [2018/p] will be zero,and so p becomes a divisor of 2018,contradiction.
Then
Two cases to be discussed:
_ in this case we only need to check one by one,
as p∈{3,5,7,11,13,17,19,23,29,31,37,41,43} and must satisfy (※).The result turns out that p=5 or 31 .
Therefore,the answer is 3.( only p=5, 31, 2017 )