To solve the problem, we need to find the number of positive integers (n) such that:

1. (n) is a multiple of 7, and
2. (n) has exactly 7 positive divrs.

Key Observations:

The number of positive divrs of a number (n) (denoted (\tau(n))) is determined by its prime factorization. For (n = p_1^{k_1}p_2^{k_2}\dots p_m^{k_m}), (\tau(n) = (k_1+1)(k_2+1)\dots(k_m+1)).

Since (\tau(n) = 7) (a prime number), the only way this product equals 7 is if there is exactly one prime factor with exponent (6) (because (6+1=7)). Thus, (n = p^6) for some prime (p).

Condition for being a multiple of 7:

(n = p^6) must be a multiple of 7. Since 7 is prime, (p) must be 7.

Thus, the only such (n) is (7^6 = 117649).

Answer: (\boxed{1})

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