Why are you being sarcastic?
It is a fact, add up the numbers of a 3 digit number (or any n digit number) and if they are divisible by 3 then the entire number is divisible by 3.
Streetwolf's way is simple,
add up 3 2s, 3 3s, or 3 4s, or 2+3+4 and they all are divisible by 3.
So how many ways can you arrange 222 or 333 or 444, only 1 way each.
as for 234, you can arrange it 3+2+1 ways
so total is 6+3= 9 ways and shouldn't take more than about half a min at most.
I didn't understand the first problem, sorry.
Actually it's 3*2*1 ways.
Theorem: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
Proof: I'm not proving this both ways as in a formal proof but this should convince you. Most of the time you add up the digits and if they add up to a number that is divisible by 3, then the entire number is divisible by 3. So I'm proving this in that direction: If the sum of a number's digits is divisible by 3, then the number itself is divisible by 3.
Say you have a number with any amount of digits in it (so 3125 has 4 digits, the number 8 has 1 digit, etc). Now consider each digit separately. Let the first digit be x1, the second x2, the third x3, and so on. The last digit is x
👎 where n = the number of digits in the number.
So the number is represented by x
👎 + 10*x(n-1) + 100*x(n-2) + ... + 10^(n-1)*x1.
For example you could write 35,662 as: 2 + 10*6 + 100*6 + 1000*5 + 10000*3.
Now rewrite this as x
👎 + 9x(n-1) + x(n-1) + 99x(n-2) + x(n-2) + 999x(n-3) + x(n-3) + ...
You get x
👎 + x(n-1) + ... + x2 + x1 + 9x(n-1) + 99x(n-2) + 999x(n-3) + ...
Obviously all the terms with the 9s in them are divisible by 3. And we know that x
👎 + x(n-1) + ... + x2 + x1 is divisible by 3 because that's just the sum of the digits (remember that the condition here is that the sum of the digits is divisible by 3). So when the sum of the digits is divisible by 3, the number itself is divisible by 3.
That applies to this problem because you need a quick way to figure out what's divisible by 3 and what isn't.
Since we're working with 3 digit numbers, if you have the same digit 3 times it will certainly be divisible by 3. Also since you have 3 consecutive numbers, using each once will give you a number divisible by 3. No other combination of the digits will work. It HAS to be those 2 cases.
Since you can only use 222, 333, and 444 as those 3 numbers, they only count as 3 possibilities. But 234 can be arranged in 6 different ways: you can pick any of the 3 for the 1st digit, any of the remaining 2 for the second digit, and you're stuck with the 3rd for the 3rd digit. That's 3*2*1 = 6.
