X, Y and Z jump forward 4', 6' and 5', respectively. At 8 AM, they all land on mark 199'. How many times will they all land on the same mark (need not be at the same moment) between mark 195' and 1000', if all of them cross mark 1000' by 9 AM?

Correct Answer: (D) 14

All three jumpers land together on mark 199'. Since their jump lengths are 4', 6', and 5', any subsequent mark on which all three land must differ from 199' by a common multiple of their jump sizes. The least common multiple is $\text{LCM}(4,6,5)=60\text{ feet}$$\text{LCM}$(4, 6, 5) = 60$\text{ feet}$. Thus, common landing marks are given by $199 + 60k$ for integer $k$. In the range between mark 195' and 1000', $k$ can range from $k = 0$ ($M = 199'$) up to $k = 13$ ($M = 199 + 60$$\times$ 13 = 979'). Because $199 + 60(14) = 1039' > 1000'$, there are $13 - 0 + 1 = 14$ such landing marks.