Find all positive integers $n$ such that the number $n^5+79$ has all the same digits when it is written in decimal represantation.
To solve the problem, we need to find all positive integers \( n \) such that \( n^5 + 79 \) has all the same digits when written in decimal representation. Let's denote this number by \( d \cdot \frac{10^t - 1}{9} \) for some \( d \in \{1, 2, \ldots, 9\} \) and \( t \in \mathbb{N} \).

1. **Expression for \( n^5 + 79 \)**:
   \[
   n^5 + 79 = d \cdot \frac{10^t - 1}{9}
   \]
   This implies:
   \[
   9(n^5 + 79) = d(10^t - 1)
   \]

2. **Considering \( t \) even**:
   If \( t \) is even, then \( 10^t - 1 \) is divisible by 11. Therefore, \( 11 \mid n^5 + 79 \). This implies:
   \[
   n^5 + 79 \equiv 0 \pmod{11} \implies n^5 \equiv -79 \equiv -2 \pmod{11}
   \]
   By Fermat's Little Theorem, \( n^{10} \equiv 1 \pmod{11} \). However, \( n^5 \equiv -2 \pmod{11} \) does not satisfy this condition, leading to a contradiction. Thus, \( t \) cannot be even.

3. **Considering \( t \) odd**:
   If \( t \) is odd, then:
   \[
   d \cdot \frac{10^t - 1}{9} \equiv d \pmod{11}
   \]
   Since \( n^5 + 79 \equiv d \pmod{11} \), we have:
   \[
   n^5 + 79 \in \{1, 2, 3\} \pmod{11}
   \]
   This leaves us with \( d = 1, 2, 3 \).

4. **Case \( d = 1 \)**:
   \[
   n^5 + 79 = 111 \implies n^5 = 32
   \]
   Since \( n \) must be an integer, \( n = 2 \) works:
   \[
   2^5 + 79 = 32 + 79 = 111
   \]
   Assume the number has more than three digits. Then:
   \[
   1111 \equiv 7 \pmod{16} \implies n^5 \equiv -8 \pmod{16}
   \]
   This is not possible since \( n \) is even. Therefore, \( n = 2 \) is a valid solution.

5. **Case \( d = 2 \)**:
   This case is more challenging. We get:
   \[
   2 \cdot 10^t - 23 \cdot 31 = 9n^5 \implies n^5 \equiv 14 \cdot 10^t \pmod{31}
   \]
   Hence:
   \[
   n^{30} \equiv 14^6 \cdot 10^{6t} \pmod{31}
   \]
   Therefore:
   \[
   10^{6t} \equiv \frac{1}{8} \equiv 4 \pmod{31}
   \]
   Inspecting, we find:
   \[
   6t \equiv 12 \pmod{15} \implies 3t \equiv 6 \pmod{15} \implies t \equiv 2 \pmod{5}
   \]
   Now, put \( t = 5i + 2 \). Then:
   \[
   2 \cdot 10^t - 23 \cdot 31 \equiv 200 - 23 \cdot 31 \equiv 20 \pmod{41}
   \]
   This implies:
   \[
   n^5 \equiv 20/9 \equiv -180 \equiv -16 \pmod{41}
   \]
   Hence:
   \[
   n^5 \equiv -2^4 \implies n^{40} \equiv 2^{32} \equiv 1 \pmod{41}
   \]
   By Fermat's Little Theorem, \( 2^{40} \equiv 1 \pmod{41} \). These together imply \( 41 \mid 2^8 - 1 = 255 \), which is a contradiction.

Therefore, the only solution is \( n = 2 \).

The final answer is \( \boxed{ n = 2 } \).