Solution
Consider the value \(N=7\) and check whether \(N+2p\) is prime
for all \(p\) smaller than \(N\):
- if \(p=2\), then \(7+2 \cdot 2=11\) (prime);
- if \(p=3\), then \(7+2 \cdot 3=13\) (prime);
- if \(p=5\), then \(7+2 \cdot 5=17\) (prime).
Let us justify that numbers larger than \(7\) do not work. If \(N>7\),
then one of the three numbers \(N+4, N+6, N+14\) is not a prime, because
- \(N+6\) is divisible by \(3\) if \(N=3k\),
- \(N+14\) is divisible by \(3\) if \(N = 3k+1\),
- \(N+14\) is divisible by \(3\) if \(N=3k+1\),
- \(N+4\) is divisible by \(3\) if \(N=3k+2\).
Atrisinājums
Aplūkojam vērtību \(N=7\) un pārbaudām, vai \(N+2p\) ir pirmskaitlis
visiem \(p\), kas mazāki nekā \(N\):
- ja \(p=2\), tad \(7+2 \cdot 2=11\) (pirmskaitlis);
- ja \(p=3\), tad \(7+2 \cdot 3=13\) (pirmskaitlis);
- ja \(p=5\), tad \(7+2 \cdot 5=17\) (pirmskaitlis).
Pamatosim, ka neder skaitḷi, kas lielāki nekā \(7\). Ja \(N>7\),
tad kāds no trīs skaitļiem: \(N+4, N+6, N+14\) nav pirmskaitlis, jo
- \(N+6\) dalās ar \(3\), ja \(N=3k\),
- \(N+14\) dalās ar \(3\), ja \(N = 3k+1\),
- \(N+14\) dalās ar \(3\), ja \(N=3k+1\),
- \(N+4\) dalās ar \(3\), ja \(N=3k+2\).