11) $n$ and $m$ are two positive integers for which: $\sqrt[3]{45+29 \sqrt{2}}=m+n \sqrt{2}$. What is the value of $m+n$?
(A) 3 , (B) 4 ,
(C) 5 ,
(D) 6 ,
(E) 7 .
11. The answer is $(\mathbf{B})$.

If we raise both sides of the equality to the third power, we get:

$$
45+29 \sqrt{2}=m^{3}+2 \sqrt{2} n^{3}+3 \sqrt{2} m^{2} n+6 m n^{2}=m\left(m^{2}+6 n^{2}\right)+n\left(3 m^{2}+2 n^{2}\right) \sqrt{2} \text {. }
$$

From this relation, knowing that $\sqrt{2}$ cannot be expressed as a ratio of two integers, it follows that

$$
m\left(m^{2}+6 n^{2}\right)=45, \quad n\left(3 m^{2}+2 n^{2}\right)=29 .
$$

29 is a prime number that can only be factored as 1.29. Therefore, since $n$ is less than $\left(3 m^{2}+2 n^{2}\right)$, it must be $n=1$ and $3 m^{2}+2=29$, from which we easily find $m=3$ and $m+n=4$.