Okay, let's tackle this problem. The question is asking how many of the first one hundred positive integers are divisible by 3, 4, and 5. Hmm, so we need to find numbers between 1 and 100 that are divisible by all three numbers: 3, 4, and 5. 

First, I remember that if a number is divisible by multiple numbers, it's the least common multiple (LCM) of those numbers that we need to consider. So maybe I should start by finding the LCM of 3, 4, and 5. Let me recall how LCM works. The LCM of two or more numbers is the smallest number that is a multiple of each of them. 

Alright, let's compute the LCM step by step. For 3, 4, and 5. Let's break them down into prime factors:

- 3 is a prime number, so its prime factorization is 3.
- 4 is 2 squared, so 2^2.
- 5 is a prime number, so 5.

To find the LCM, we take the highest power of each prime number present in the factorizations. So for prime 2, the highest power is 2^2 (from 4). For prime 3, it's 3^1, and for prime 5, it's 5^1. Therefore, LCM = 2^2 * 3 * 5 = 4 * 3 * 5. Let's compute that: 4*3 is 12, then 12*5 is 60. So the LCM of 3, 4, and 5 is 60. 

That means every 60th number is divisible by all three numbers. So in the range from 1 to 100, how many multiples of 60 are there? Let's see. The multiples of 60 are 60, 120, 180, etc. But since we're only considering up to 100, the first multiple is 60, then the next would be 120, which is beyond 100. So only 60 is within the first 100 positive integers. 

Wait, but let me verify that. Let's check if 60 is divisible by 3, 4, and 5. 

60 divided by 3 is 20, which is an integer. 60 divided by 4 is 15, also an integer. 60 divided by 5 is 12, which is an integer. So yes, 60 is divisible by all three. Then the next multiple would be 60*2=120, which is 120, but that's beyond 100. Therefore, only one number in the first 100 positive integers is divisible by 3, 4, and 5. 

But wait, let me make sure I didn't miss anything. Is there a possibility that there's another number less than 100 divisible by all three? For example, maybe 30? Wait, 30 divided by 3 is 10, 30 divided by 4 is 7.5, which is not an integer. So 30 isn't divisible by 4. How about 20? 20 divided by 3 is about 6.666, not an integer. 15? 15 divided by 4 is 3.75, nope. 12? 12 divided by 5 is 2.4, still no. So yeah, seems like 60 is the only one. 

Alternatively, maybe I should list all numbers divisible by 3, 4, and 5 up to 100. Let's see. Numbers divisible by 3 up to 100: 3,6,9,...,99. Numbers divisible by 4 up to 100: 4,8,12,...,100. Numbers divisible by 5 up to 100: 5,10,15,...,100. The intersection of these three sets would be numbers divisible by 3, 4, and 5. 

But since 3,4,5 are co-prime in pairs except for 4 and 3 which are co-prime, and 4 and 5 are co-prime, the LCM is indeed 60. So the intersection is multiples of 60. Hence, only 60 and 120, but 120 is over 100. Therefore, only 60. 

Another way to think about it is using inclusion-exclusion, but since we need numbers divisible by all three, it's the intersection. So the formula for the count of numbers divisible by all three is floor(100 / LCM(3,4,5)) = floor(100 / 60) = 1. 

Therefore, the answer should be 1. 

But let me check once more. Let's compute LCM of 3,4,5 again. LCM(3,4) is 12. Then LCM(12,5). Since 12 and 5 are co-prime, LCM is 12*5=60. Yep. So the LCM is 60. So every 60th number is divisible by all three. So in the first 100 numbers, only 60 is present. 

Therefore, the answer is 1.
</think>