An angle measuring \( (468 n)^{\circ} \) is in standard position. For which value of \( n \) will the terminal side fall on the \( x \)-axis? \( n=4 \) \( n=5 \) \( n=6 \) \( n=7 \)

Let the given angle be \[ \theta = 468n^\circ. \] For the terminal side of the angle to lie on the \(x\)-axis, the angle must be coterminal with an angle of \(0^\circ\) or \(180^\circ\). This means we need \[ 468n \equiv 0 \pmod{180} \quad \text{or} \quad 468n \equiv 180 \pmod{360}. \] A simpler requirement is that the angle must be a multiple of \(180^\circ\), i.e., \[ 468n = 180k \quad \text{for some integer } k. \] Solve for \( n \): \[ n = \frac{180k}{468}. \] Simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Notice that: \[ 468 = 6 \times 78 \quad \text{and} \quad 180 = 6 \times 30. \] Thus, \[ n = \frac{6 \times 30 \, k}{6 \times 78} = \frac{30k}{78} = \frac{5k}{13}. \] For \( n \) to be an integer, \( \frac{5k}{13} \) must be an integer. Since 5 and 13 are relatively prime, \( k \) must be a multiple of 13; let \( k = 13m \) where \( m \) is an integer. Then, \[ n = \frac{5(13m)}{13} = 5m. \] Among the answer choices \( n = 4,\, 5,\, 6,\, 7 \), the only one of the form \(5m\) is \( n = 5 \) (with \( m=1 \)). We can verify: \[ 468 \times 5 = 2340^\circ. \] Now reduce \(2340^\circ\) modulo \(360^\circ\): \[ 2340^\circ \div 360^\circ = 6 \text{ remainder } 180^\circ, \] so \[ 2340^\circ \equiv 180^\circ \pmod{360}. \] Since an angle of \(180^\circ\) lies on the \(x\)-axis, the correct value is \[ n = 5. \]