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Napalm Donkey
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My first thought would be to consider all of the numbers modulo 3. It doesn't find you the answer but may help clarify your thinking depending upon how much experience you have with this stuff.

SavantonThat should make this pretty easy.

musanmanonNapalm DonkeyonOh, pre-calc? Yeah, this is pretty tricky for that level, you probably have no idea what modular arithmetic is. I guess the best I could recommend without giving it all away is to try to think about the remainders of the numbers when you divide them by 3, and what happens to the remainders of sums of those numbers. That's sort of the basics of the modular arithmetic stuff.

You'll still figure out how to count the different viable combinations, however.

SavantonNapalm DonkeyonscrivenerjonesonNo shit? That really works?

I don't like math, but there are some neat tricks.

ZombiemamboonB 31

C 41

D 51

E 61

F 71

G 81

Now, taking the above information, the sum of the digits are:

18 - ABCD

21 -

24 -

27 -

30 - DEFG

ABCDEFG doesn't work, but ABCDGEF does. Why is this true?

ZoelefonNapalm DonkeyonNapalm DonkeyonYou can see why that works due to modular arithmetic. Modular arithmetic partitions numbers into equivalence classes based on their remainder modulo a particular natural number.

For example, 0 (mod 3) = 0, 3, 6, 9, ... (mod 3), because the remainder of the division of all the multiples of three by three is 0. Also, 1 (mod 3) = 1, 4, 7, 10, ... (mod 3), and 2 (mod 3) = 2, 5, 8, 11, ... (mod 3).

10 (mod 3) = 1 (mod 3), and 10 * 10 (mod 3) = 1 (mod 3), and 10^n (mod 3) = 1 (mod 3) for all positive integers n.

When you consider the decimal digits of numbers, it's equivalent to breaking it down to a sum of the multiples of powers of ten. So 945705 = 9 * 10^5 + 4 * 10^4 + 5 * 10^3 + 7 * 10^2 + 0 * 10^1 + 5 * 10^0.

If you consider that modulo 3, those powers of 10 are congruent to 1 mod 3, so

945705 (mod 3) = 9*1 + 4*1 + 5*1 + 7*1 + 0*1 + 5*1 (mod 3) = 0 + 1 + 2 + 1 + 0 + 2 (mod 3) = 0 (mod 3)

So, since the digits of 945705 sum to a multiple of 3 (which they do, to 30), it is congruent to 0 modulo 3, which means it is divisible by 3.

SavantonI feel obliged to point out that there are only roughly five thousand ways to arrange seven numbers, and as such a computer script will run through every possible combination in a few seconds.

At which point you write "which is obviously true by inspection" and then observe your teacher's reaction carefully.

ronyaonNapalm DonkeyonNapalm DonkeyonThe first 3 numbers and the last 3 numbers (in mod 3 form) have to be the same (e.g. 210, or 012), and the middle has to be a 0 (I'll let you reason this out, assuming I'm right), making things like 2100210 or 2010201.

So there are 3 * 2 * 1 different ways to choose the pattern that repeats.

Now there are 3 * 2 * 1 different ways to position the 0 numbers, and 2 * 1 ways of positioning each of the 2 numbers and the 1 numbers.

So the total number of ways of arranging the numbers is 6 * 6 * 2 * 2 = 144.

Smug DucklingonNapalm DonkeyonThat's the "Now there are 3 * 2 * 1 different ways to position the 0 numbers, and 2 * 1 ways of positioning each of the 2 numbers and the 1 numbers." part. There are 3 different numbers congruent to 0 mod 3, so those can be arranged 6 ways into the 0 spots, etc.

How did you get your number? It's quite possible that I'm missing something.

Smug DucklingonNapalm Donkeyon