Passage: Let u be a positive integer and f be a real number lying...

Given $(\sqrt{2}+1)^{10} = u+f$ and $(\sqrt{2}-1)^{10} = v$.

Adding both equations:

$u+f+v = (\sqrt{2}+1)^{10} + (\sqrt{2}-1)^{10}$

Expanding using the binomial theorem:

$u+f+v = 2 [^{10}C_0 (\sqrt{2})^{10} + ^{10}C_2 (\sqrt{2})^8 + \dots + ^{10}C_{10}]$

The right hand side is an even integer. Let it be $2k$.

$u+f+v = 2k$

$f+v = 2k - u$

Since $2k$ and $u$ are integers, $f+v$ must be an integer.

We are given $0 < f < 1$.

Also, $0 < \sqrt{2}-1 < 1 \Rightarrow 0 < (\sqrt{2}-1)^{10} < 1 \Rightarrow 0 < v < 1$.

Adding the inequalities for $f$ and $v$:

$0 < f+v < 2$

Since $f+v$ is an integer strictly between $0$ and $2$, the only possible value is $1$.

Therefore, $v+f = 1$.

Answer: $1$