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 the two equations:

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

Using the binomial expansion $(x+y)^n + (x-y)^n = 2 [^{n}C_{0} x^n + ^{n}C_{2} x^{n-2} y^2 + \dots]$:

$u + f + v = 2 [^{10}C_{0}(\sqrt{2})^{10} + ^{10}C_{2}(\sqrt{2})^8 + ^{10}C_{4}(\sqrt{2})^6 + ^{10}C_{6}(\sqrt{2})^4 + ^{10}C_{8}(\sqrt{2})^2 + ^{10}C_{10}]$

$u + f + v = 2 [1(32) + 45(16) + 210(8) + 210(4) + 45(2) + 1(1)]$

$u + f + v = 2 [32 + 720 + 1680 + 840 + 90 + 1]$

$u + f + v = 2 [3363] = 6726$

Since $0 < \sqrt{2}-1 < 1$, we have $0 < (\sqrt{2}-1)^{10} < 1$, which means $0 < v < 1$.

We are given $0 < f < 1$. Adding these inequalities gives $0 < f + v < 2$.

Since $u$ and $6726$ are integers, $f + v$ must be an integer. The only integer strictly between $0$ and $2$ is $1$.

Therefore, $f + v = 1$.

Substituting this back into the equation:

$u + 1 = 6726$

$u = 6725$

Answer: $6725$