**Solution. **We show that two such primes exist: 3 and 7.

*p*=3 satisfies the conditions since 1+1^{.}2=3 is a prime.

*p*=5 does not satisfies the condition since 1+2^{.}4=9 is composite.

*p*=7 satisfies the conditions since 1+1^{.}6=7, 1+2^{.}6=13 and 1+3^{.}6=19 are all primes.

*p*=11 does not satisfies the condition since 1+2^{.}10=21 is composite.

*p*=13 does not satisfies the condition since 1+2^{.}12=25 is composite.

In the rest of the solution we assume *p*17.

Suppose that some prime does not divide (*p*-1). Then *q* and *p*-1 are coprime, so there is such a , for which *q*|1+*k*(*p*-1). By *q**p*-1<1+*k*(*p*-1), the number *q* is a proper divisor of (1+*k*(*p*-1)), contradicting the assumption that 1+*k*(*p*-1) is a prime.

Hence, *p*-1 is divisible by all prime numbers less than *p*/2. Let *r* be the greatest prime which is less than *p*/2. By Chebysev's theorem we have . The numbers 2,3,*r* are distinct prime divisors of *p*-1, so 6*r*|*p*-1. But this is not possible because *p*-14*r*+2<6*r*.