Let $n$ be a natural number. Prove that
$$ \frac{25n^{61} + 52n}{77}$$
is an integer.
Hint: You may find the result of Exercises 3.1, question 24 (d) helpful:
$ a \equiv b \pmod {m_k} \implies a \equiv b \pmod {m_1 \times m_2 \times \ldots \times m_n}$
where $k= 1, 2, \ldots , n$ and $\gcd (m_i, m_j) = 1$ for $i \neq j$.
We reformulate the fraction needing to be an integer as a congruence,
$$ 25n^{61} +52n \equiv 0 \pmod {77} $$
Taking the hint, we'll consider modulo 11 and modulo 7.
Using $n^{11} \equiv n \pmod {11}$, we have
$$ n^{61} \equiv (n^{11})^5 \times n^6 \equiv n^5 \times n^6 \equiv n^{11} \equiv n \pmod {11} $$
And so
$$ 25n^{61} +52n \equiv 25n + 52n \equiv 11 \times 7n \equiv 0 \pmod {11} $$
Using $n^7 \equiv n \pmod 7$ we have
$$ n^{61} \equiv (n^7)^8 \times n^5 \equiv n^8 \times n^5 \equiv n^7 \times n^6 \equiv n \times n^6 \equiv n^7 \equiv n \pmod 7 $$
And so
$$ 25n^{61} +52n \equiv 25n + 52n \equiv 7 \times 11n \equiv 0 \pmod {7} $$
Given
$$ 25n^{61} +52n \equiv 0 \pmod {11} $$
$$ 25n^{61} +52n \equiv 0 \pmod {7} $$
and $\gcd(11,7)=1$ we conclude
$$ 25n^{61} +52n \equiv 0 \pmod {77} $$
which is the same as stating the following fraction is an integer
$$ \frac{25n^{61} + 52n}{77}$$