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There will always be a 2:1 ratio between numbers not divisible by 3 and numbers divisible by 3. For example: 100, 101, 102... 103, 104, 105... 997, 998, 999

We can infer that the 900 integers in the desired range will also fulfil this 2:1 ratio. Thus, 600 integers are not divisible by 3 and 300 are divisible by 3.

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There will always be a 2:1 ratio between numbers not divisible by 3 and numbers divisible by 3. For example: 100, 101, 102... 103, 104, 105... 997, 998, 999

We can infer that the 900 integers in the desired range will also fulfil this 2:1 ratio. Thus, 600 integers are not divisible by 3 and 300 are divisible by 3.

Are you subtracting 102 from 999 because 102 is the first three-digit multiple of 3, and 999 is the last three-digit multiple of 3?

Yes this gives us the range of multiples of 3, and in this range every THIRD number will be div by 3.... so we divide the range by 3.. And add 1 as both the first and last integers are div by 3 BUT in our calculation we take ONLY one of them..
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We have an arithmetic sequence with 1st term 102 (smallest 3 digit multiple of 3) and 999 (last term), d=3. Using the formula, we have 999 = 102 + 3*(n1) <=> n = 300. Total number of possible 3 digit numbers is 999-100+1 = 900. Deduct those which are divisible by 3 from all possible numbers, 900-300 = 600 are NOT divisible by 3.