The remainder in 1%3 refers to what remains of 1 (not 3) after you divide by 3. As you have already said, 3 goes into 1 zero times. So -- when you remove 0 multiples of 3 from 1, all of 1 remains. Thus 1 % 3 = 1.
The result of a modulo operation n % m is just that number r for which q * m + r = n (q may be anything). The only requirement we have is that 0 <= r < m.
So for instance:
7 % 5 --> 1 * 5 + 2 == 7 --> r = 2
1 % 3 --> 0 * 3 + 1 == 1 --> r = 1
Getting started. Springfield Mod 3? - Miscellaneous BEginners - Brian Enos's Forums... Maku mozo!
I have the solution to this problem, but I don't understand the logic behind the solution.
Any rootless One UI 3 mods out yet? (miss One UI 1 & 2 style :c )
Hex works for me fine. Even has custom fonts now
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Haven't CUP already done it?
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This is the problem, and this is the solution.
Already, on the first sentence I have a question. They just multiplied the end points of the range of integers the question asked us to consider by 3 and just said that would give us the same answer. I don't understand how those things are the same, nor do I understand why they multiplied by three specifically.
Then, they take the mod 3 of all the numbers the digits could be. They take 9 mod 3 and 3 mod 3. They get 0 for both, makes sense. 1 mod 3 and 7 mod 3 are 1, makes sense. But they take 5 mod 3 and get -1? What? Wouldn't it be 2? It is my understanding that mod is never negative.
Also, why take the mod 3 of the possible digits? How did that help? I assume it's somehow connected with the fact that a number divisible by 3 has the sum of its digits divisibly by three, but I don't see the connection.
Then, they use the modulus in a completely unclear way, at least to me. So, the first bullet point states that the amount of numbers that start with 3 followed by three digits that have a modulus of zero is the same as 2 * 2 * 2. I don't know how having the first digit be 3 followed by three numbers that divide evenly into 3 result in 23.
Similarly, the second bullet point states the amount of numbers that start with 3 and are followed by "one [0], one [1], and one [-1]" results in 3! * 2 * 2 * 1. I have no clue how they got that.
I have the same question for all the bullet points. How did the initial statement after the bullet point lead to the following statement after the dash.