One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer $ n$ such that \[133^5 \plus{} 110^5 \plus{} 84^5 \plus{} 27^5 \equal{} n^5.\] Find the value of $ n$.
1. **Units Digit Analysis:**
   - We start by analyzing the units digits of the numbers involved. The units digits of \(133\), \(110\), \(84\), and \(27\) are \(3\), \(0\), \(4\), and \(7\) respectively.
   - The units digit of any number \(k\) and \(k^5\) are the same. Therefore, the units digit of \(n\) must be the same as the units digit of \(133^5 + 110^5 + 84^5 + 27^5\).
   - Calculating the sum of the units digits: \(3 + 0 + 4 + 7 = 14\). The units digit of \(14\) is \(4\). Hence, the units digit of \(n\) must be \(4\).

2. **Modular Arithmetic Analysis:**
   - Next, we use modular arithmetic to further narrow down the possibilities for \(n\).
   - We calculate \(133^5 + 110^5 + 84^5 + 27^5 \mod 3\):
     \[
     133 \equiv 1 \pmod{3}, \quad 110 \equiv 2 \pmod{3}, \quad 84 \equiv 0 \pmod{3}, \quad 27 \equiv 0 \pmod{3}
     \]
     \[
     133^5 \equiv 1^5 \equiv 1 \pmod{3}, \quad 110^5 \equiv 2^5 \equiv 32 \equiv 2 \pmod{3}, \quad 84^5 \equiv 0^5 \equiv 0 \pmod{3}, \quad 27^5 \equiv 0^5 \equiv 0 \pmod{3}
     \]
     \[
     133^5 + 110^5 + 84^5 + 27^5 \equiv 1 + 2 + 0 + 0 \equiv 3 \equiv 0 \pmod{3}
     \]
   - This means \(n\) must be a multiple of \(3\).

3. **Finding \(n\):**
   - We know \(n\) must be greater than \(133\) and must end in \(4\) and be a multiple of \(3\).
   - The possible values for \(n\) that satisfy these conditions are \(144\), \(174\), \(204\), etc.
   - We need to check which of these values satisfies the equation \(133^5 + 110^5 + 84^5 + 27^5 = n^5\).

4. **Verification:**
   - We verify \(n = 144\):
     \[
     133^5 + 110^5 + 84^5 + 27^5 = 144^5
     \]
     - This is a known result from the disproof of Euler's sum of powers conjecture, and indeed \(n = 144\) satisfies the equation.

The final answer is \(\boxed{144}\).