Question

The smallest prime number satisfying the inequality $\frac{2n-3}{3}\ge \frac{n-1}{6}+1$ is

• A. 2
• B. 3
• C. 5
• D. 7
• E. 11

Answer 1

Answer: (C)

$n=2,3$, fails
$n=5\Rightarrow \frac{2\times 5-3}{3}\ge \frac{4}{6}+1$
$\frac{1}{3}>\frac{1}{6}+1$
$\frac{7}{3}\ge \frac{2}{3}+1$
$\frac{7}{3}\ge \frac{4}{6}+1$
$1+\frac{4}{3}>\frac{2}{3}+1$
$1+\frac{4}{3}>\frac{2}{3}+1$