Given the equation:

$$
\(p^4+q^4=100\)
$$


We need to determine between which two consecutive integers the maximum possible value of $p$ lies. The options are:
- A. 0 and 1
-B. 1 and 2
- C. 2 and 3
- D. 3 and 4
- E. 4 and 5

Step 1: Understand the Constraint
The equation $\(p^4+q^4=100\)$ defines a relationship between $p$ and $q$. To find the maximum value of $p$, we need to consider the extreme case where $q$ is minimized.

Step 2: Maximize $p$ by Minimizing $q$
Since $\(q^4 \geq 0\)$, the smallest possible value of $q$ is $q=0$. Substituting $q=0$ into the equation:

$$
\(\begin{gathered}
p^4+0=100 \\
p^4=100 \\
p=\sqrt[4]{100}
\end{gathered}\)
$$


Step 3: Calculate $\(\sqrt[4]{100}\)$
We need to estimate $\(\sqrt[4]{100}\)$ :
1. Compute $\(3^4\)$ :

$$
\(3^4=81\)
$$

2. Compute 44:

$$
\(4^4=256\)
$$


Since $\(81<100<256, \sqrt[4]{100}\)$ lies between 3 and 4 .
Step 4: Narrow Down the Range
To refine the estimate:
1. $\(\operatorname{Try} p=3.1\)$ :

$$
\(3.1^4=\left(3.1^2\right)^2 \approx 9.61^2 \approx 92.35\)
$$

2. Try $\(p=3.2\)$ :

$$
\(3.2^4=\left(3.2^2\right)^2 \approx 10.24^2 \approx 104.86\)
$$


Since $\(92.35<100<104.86, \sqrt[4]{100}\)$ lies between 3.1 and 3.2.
Step 5: Conclusion
The maximum possible value of $p$ is $\(\sqrt[4]{100}\)$, which is between 3 and 4 .


D is the answer