solution: One gear turns $33\frac{1}{3}=100/3$ times in 60 seconds, so it turns 5/9 times in one second, or 5 times in 9 seconds. The other gear turns 45 times in 60 seconds, so it turns 3/4 times in one second, or 3 times in 4 seconds. To find out after how many seconds the two gears next have both their marks pointing due north, we have to find the least common multiple of $4=2^2$ and $9=3^2$, which is $2^2\cdot3^2=36$. Therefore, the two gears next have both their marks pointing due north after $\boxed{36}$ seconds. (One gear turns exactly $5\times4=20$ times, and the other gear turns exactly $3\times9=27$ times.)
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  One gear turns $33\frac{1}{3}$ times in a minute.  Another gear turns   45 times in a minute.  Initially, a mark on each gear is pointing  due north.  After how many seconds will the two gears next  have both their marks pointing due north?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 36
extra_info.question: One gear turns $33\frac{1}{3}$ times in a minute.  Another gear turns   45 times in a minute.  Initially, a mark on each gear is pointing  due north.  After how many seconds will the two gears next  have both their marks pointing due north?
extra_info.split: test