Options

# blank

Registered User regular
edited April 2016
blank

Napalm Donkey on

## Posts

• Options
Simply Barbaric Registered User regular
edited September 2009
What is this for, a combinatorics class? How much help are you allowed to get?

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.

Savant on
• Options
Registered User regular
edited September 2009
Yes, what he means is that if you add the digits up and the sum is divisible by 3, then the original number is divisible by 3.

That should make this pretty easy.

musanman on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Simply Barbaric Registered User regular
edited September 2009
My pre-calculus teacher gave it as extra credit assignment so don't tell me the answer, but I was hoping that somebody could give me some hints on solving it.

I looked up combinatorics and it seemed really complicated is there any basic concepts that would help me solve the problem?

Oh, 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.

Savant on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Registered User regular
edited September 2009
I guess no one has taught you the Secret to Math Homework Success yet:
write a mathematica script to do it for you 8-)

scrivenerjones on
• Options
Registered User regular
edited September 2009
musanman wrote: »
Yes, what he means is that if you add the digits up and the sum is divisible by 3, then the original number is divisible by 3.

That should make this pretty easy.

No shit? That really works?

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

Zombiemambo on
• Options
Registered User regular
edited September 2009
I'll lay a framework for you, this should be enough to get you started:
A 21
B 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?
The solution is an exercise left for the reader.
Because I'm a jerk.

Zoelef on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Simply Barbaric Registered User regular
edited September 2009
musanman wrote: »
Yes, what he means is that if you add the digits up and the sum is divisible by 3, then the original number is divisible by 3.

That should make this pretty easy.

No shit? That really works?

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

You 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.

Savant on
• Options
Arrrrrf. the ivory tower's basementRegistered User regular
edited September 2009
I guess no one has taught you the Secret to Math Homework Success yet:
write a mathematica script to do it for you 8-)

I 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.

ronya on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Registered User regular
edited September 2009
I got 144.

The 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 Duckling on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on
• Options
Registered User regular
edited September 2009
I got 144.

The 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.

Wont there be more because each number is different; like 210 could be made up of each different number even if there both congruent modulo 3.

That'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 Duckling on
• Options
Registered User regular
edited April 2016
blank

Napalm Donkey on