Let $p(x)$ be a polynomial of degree 4 with leading coefficients 1. Suppose $p(1)=1$, $p(2)=2$, $p(3)=3$, $p(4)=4$. Then $p(5)=$
[list=1]
[*] 5
[*] $\frac{25}{6}$
[*] 29
[*] 35
[/list]
1. Let \( p(x) \) be a polynomial of degree 4 with leading coefficient 1. This means we can write \( p(x) \) as:
   \[
   p(x) = x^4 + ax^3 + bx^2 + cx + d
   \]
2. Given the conditions \( p(1) = 1 \), \( p(2) = 2 \), \( p(3) = 3 \), and \( p(4) = 4 \), we can define a new polynomial \( g(x) \) such that:
   \[
   g(x) = p(x) - x
   \]
   This implies:
   \[
   g(1) = p(1) - 1 = 0, \quad g(2) = p(2) - 2 = 0, \quad g(3) = p(3) - 3 = 0, \quad g(4) = p(4) - 4 = 0
   \]
3. Since \( g(x) \) has roots at \( x = 1, 2, 3, 4 \), we can express \( g(x) \) as:
   \[
   g(x) = c(x-1)(x-2)(x-3)(x-4)
   \]
   where \( c \) is a constant.
4. The leading coefficient of \( p(x) \) is 1, and since \( p(x) = g(x) + x \), the leading coefficient of \( g(x) \) must also be 1. Therefore, \( c = 1 \).
   \[
   g(x) = (x-1)(x-2)(x-3)(x-4)
   \]
5. To find \( p(5) \), we calculate \( g(5) \):
   \[
   g(5) = (5-1)(5-2)(5-3)(5-4) = 4 \cdot 3 \cdot 2 \cdot 1 = 24
   \]
6. Therefore, \( p(5) \) is:
   \[
   p(5) = g(5) + 5 = 24 + 5 = 29
   \]

The final answer is \( \boxed{29} \)