Okay, let's try to figure out this problem. So, we have two gears. The first one turns 33 and a third times a minute, and the second one turns 45 times a minute. Both start with a mark pointing north. We need to find after how many seconds they'll both point north again. Hmm, okay.

First, I remember that when gears turn, their rotational speeds can be related to their number of rotations. Since they start at the same position, the time when their marks align again will be when both have completed an integer number of rotations. So, the key here is to find the least common multiple (LCM) of their rotation periods. But wait, maybe not exactly the LCM of the rotations, but the LCM of their periods. Let me think.

Wait, the period is the time it takes to complete one rotation. So, if the first gear rotates 33 1/3 times per minute, then its period is 1 divided by 33 1/3 minutes per rotation. Similarly, the second gear's period is 1 divided by 45 minutes per rotation. Then, the LCM of these two periods will give the time when they both complete an integer number of rotations and align again.

But let me check if that's correct. Let's convert their rotation rates to revolutions per second first, maybe. Because the answer needs to be in seconds. So, 33 1/3 rotations per minute. Let's convert that to rotations per second. Since there are 60 seconds in a minute, 33 1/3 divided by 60. Let me compute that. 33 1/3 is the same as 100/3. So, 100/3 divided by 60 is 100/(3*60) = 100/180 = 5/9 rotations per second. Wait, is that right? 33 1/3 is 100/3, so 100/3 divided by 60 is (100/3)/60 = 100/(180) = 5/9. Yes, that's correct. So, the first gear does 5/9 rotations per second. Similarly, the second gear is 45 rotations per minute, which is 45/60 = 3/4 rotations per second.

So, their periods are the reciprocals of these. The period of the first gear is 1 divided by (5/9) rotations per second, which is 9/5 seconds per rotation. The second gear's period is 1 divided by (3/4) rotations per second, which is 4/3 seconds per rotation.

Now, we need the LCM of 9/5 and 4/3. Hmm, how do you calculate LCM of fractions? I recall that LCM of two fractions is the LCM of the numerators divided by the GCD of the denominators. Let me verify that. Yes, LCM(a/b, c/d) = LCM(a, c) / GCD(b, d). So, applying that here.

The numerators are 9 and 4. LCM of 9 and 4 is 36. The denominators are 5 and 3. GCD of 5 and 3 is 1. So, LCM is 36 / 1 = 36. Therefore, the LCM of 9/5 and 4/3 is 36 seconds. Wait, that seems straightforward. So, after 36 seconds, both gears will have completed an integer number of rotations and their marks will point north again.

But let me check this another way to make sure. Let's see how many rotations each gear makes in 36 seconds. First gear: 5/9 rotations per second. 5/9 * 36 = 5 * 4 = 20 rotations. Second gear: 3/4 rotations per second. 3/4 * 36 = 3 * 9 = 27 rotations. Both 20 and 27 are integers, so yes, after 36 seconds, both gears will have completed whole numbers of rotations, bringing their marks back to north. So that seems correct.

Alternatively, another way to think about it is to find the time t such that t is a multiple of both 9/5 and 4/3. So, t must be the smallest number that is a multiple of both 9/5 and 4/3. Which is exactly the LCM of the two periods. So, that gives the same answer.

But let me try another approach to confirm. Maybe by finding the number of rotations each gear makes in t seconds and setting them both to integers. Let t be the time in seconds. Then:

Number of rotations for first gear: (5/9) * t

Number of rotations for second gear: (3/4) * t

We need both (5/9)t and (3/4)t to be integers. So, t must be a multiple of 9/5 and 4/3. So, the least common multiple of 9/5 and 4/3 is 36, as we found earlier. So, t = 36 seconds.

Another way to see this is to express t as a common multiple. Let’s solve for t such that t is a multiple of 9/5 and 4/3. So, t = k*(9/5) = m*(4/3), where k and m are integers. Therefore, 9k/5 = 4m/3. Cross-multiplying: 27k = 20m. So, 27k must be divisible by 20. Since 27 and 20 are coprime (their GCD is 1), k must be a multiple of 20. Let k = 20n. Then, t = 20n*(9/5) = 36n. The smallest positive t is when n=1, so t=36 seconds.

Yes, that also gives 36 seconds. So, multiple approaches lead to the same answer. Therefore, I can be confident that the answer is 36 seconds.

**Final Answer**
The two gears will next have both their marks pointing due north after \boxed{36} seconds.
Therefore, the two gears will both point north again after $\boxed{36}$ seconds.
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