Logarithmic Inequality and Integer Solutions

Since base is $1/4 < 1$, reversing inequality gives: $\log_{1/4} (n^2 - 7n + 11) > 0 \iff 0 < n^2 - 7n + 11 < \left(\frac{1}{4}\right)^0 = 1$

1. $n^2 - 7n + 11 < 1 \implies n^2 - 7n + 10 < 0 \implies (n-2)(n-5) < 0 \implies 2 < n < 5$. Possible integer values: $n = 3, 4$.

2. Check $0 < n^2 - 7n + 11$ for these values: For $n = 3$: $3^2 - 7(3) + 11 = 9 - 21 + 11 = -1 \ngtr 0$. For $n = 4$: $4^2 - 7(4) + 11 = 16 - 28 + 11 = -1 \ngtr 0$.

Neither $n=3$ nor $n=4$ satisfies the condition. Thus, the number of distinct integers is $0$.