The Maths Problem You Will Never Solve (Probably)!

213 = 200 + 10 + 2.

If i've proved that EVERY number which is a multiple of 3 has sum of digits which is a multiple of 3. Then that will mean every number which doesn't have a factor of 3 doesn't have sum of digits with a factor of 3.

Not quite, actually. For example: every number which is a multiple of 6 has digit sum a multiple of 3, but this doesn't prove that every number which is not a multiple of 6 has digit sum which is not a multiple of 3.

I was expecting the Riemann Hypothesis to come up in this topic.

I am throughly disappointed.

That's not what i said, I said every number which isn't a multiple of 3 will not have sum digits with a multiple of 3.

I know, but you didn't prove it! You proved that every number which is a multiple of 3 will have digit sum a multiple of 3, you didn't prove the converse. It's not hard to prove (in fact, you might even think it's obvious from here - it kind of is!), but strictly speaking, you haven't proved it.

I know, but you didn't prove it! You proved that every number which is a multiple of 3 will have digit sum a multiple of 3, you didn't prove the converse. It's not hard to prove (in fact, you might even think it's obvious from here - it kind of is!), but strictly speaking, you haven't proved it.

Ok, i don't understand what i've done wrong, but i'm guessing you've attained a calibre of an experienced and capable mathematician. So i dismiss my proof is invalid. I don't understand why, but thnx......... can you explain plz? i'm dumb........

...

Great.. and for once i wanted to plant in my mathematica seed.....

I get why it works, but I don't get how you would prove it with proper mathematical formula.

Essentially each digit is a measure of how far it is above a multiple of 3.

E.G. 10 is one unit away from a multiple of three (9).
20 is two units away from a multiple of three (18).
30 is three units away from a multiple of three (27).
40 is four units away from a multiple of three (36).
50 is five units away from a multiple of three (45).
60 is six units away from a multiple of three (54).

etc...

This works as you go up by tens too

100 is one unit away from a multiple of three (99).
200 is two units away from a multiple of three (198).
300 is three units away from a multiple of three (297).
etc..

1000 is one unit away from a multiple of three (999).
2000 is two units away from a multiple of three (1998).
3000 is three units away from a multiple of three (2997).

So, essentially. Each digital can be taken as a measure of how far the whole number is away from a multiple of three.

E.g. take 4285

(4 units away from a multiple of three) + (2 units away from a multiple of three) + (8 units away from a multiple of three) + (5 units away from a multiple of three)

=19

Therefore we can deduce that 4285 is 19 digits away from a multiple of three (4266, of course) therefore, as this 19 unit difference is not divisible by three, 4285 itself cannot be a multiple of three

Another example: 4575

(4 units away from a multiple of three) + (5 units away from a multiple of three) + (7 units away from a multiple of three) + (5 units away from a multiple of three)

= 21

This means 4575 is 21 units away from a multiple of three (4554) and as the difference itself is also dividible by three we know 4575 must be a multiple of three. (Adding two multiples of any given number will always give you another multiple).

I don't know how to write a mathematical proof, but that's why.

Prelims in a couple of days. It means I can pretend that I'm doing something math-y and preparational

No, your proof was actually very good (and led to some interesting ideas, in particular where you stated that adding 9 was the same as adding 1 in the tens column and subtracting 1 in the units column - have you ever heard of "modular arithmetic"?) - probably not entirely watertight, but not far off.

My complaint is more general. Take this as an example. It is quite easy to prove the following statement:

1. "If a number $a^b - 1$ is prime (with a and b integers greater than 1), then b is prime and a = 2."

What this doesn't prove is the converse statement (a kind of opposite), which can be expressed in two equivalent ways:

2a. "If $a^b - 1$ is not prime, then it is not true that b is prime and a = 2."
2b. "If b is prime and a = 2, then $a^b - 1$ is prime."

In fact, the converse isn't true! Indeed, if we set a = 2 and b = 11 (which is prime), then 2^11 - 1 = 2047 isn't prime (it's 23 * 89). So a statement and its converse are not always necessarily related (though of course sometimes they are). Now, how does this relate to your problem? You proved the statement:

3. "If a number x is divisible by 3, then the digit sum of x is divisible by 3."

but you did not prove:

4a. "If x is not divisible by 3, then the digit sum of x is not divisible by 3."
4b. "If the digit sum of x is divisible by 3, then x is divisible by 3."

In this case, the statement and its converse are both true (but certainly not in my example when I swapped "3" for "6[noparse]")[/noparse], but as you can hopefully gather from my primes example, they both need separate proof. (Incidentally, primes of the form a^b - 1, or equivalently 2^(prime) - 1, are called Mersenne primes.)

Ah, thank you, i now understand. I must prove the converse for the conjecture to be proved.

erm, i'm not yet familiar with odular arithmetic as i'm a year 10 gcse student

but i think i get the jist of what you've said. I must say, you're a very concise and coherent mathematician, thanks mate.

Cracking explanation Grommit!

The GCSE/A-level syllabus won't teach you it, I'm afraid. It's a very interesting (and very accessible) part of number theory - you might want to look into it at some point if you're interested. Here's a couple of links, the first rather simplistic and the second rather advanced, that eventually lead up to solving this problem using modular ("clock[noparse]")[/noparse] arithmetic: [1], [2] - and of course, although it might take you an hour or two to get your head round this solution, you should see it from the point of view of someone who already knows a bit about modular arithmetic. Once you're armed with modular arithmetic, the problem becomes really easy - in fact, I solved it myself in half a line earlier in this thread.

Sorry, but modular arithmetic is not a clock. The clock analogy is at best weak. Also, lol modular is not a hard word, so why even mention clock. Sorry, but lol I read through a book called music of the primes and yeah had to read clock arithmetic and quantum drum a thousand times.

Because it is sometimes called clock arithmetic, and gives the person I was responding to something else to google that might potentially yield more accessible results. I don't care what book you had to read or how much the term annoys you, it is a term that is used. I didn't say it was one I particularly liked.

Lol, I have never read a real maths book that used clock arithmetic. But, lol hmm modular arithmetic is not that hard to understand. I don't know but in a way talking about clocks is like sort of talking down to someone like they are a idiot(it would be like talking to someone about fractions and then talking about cutting cakes). Its if you can understand division then you can understand modular arithmetic, as its just the remainder after divison(I hope to god that yr 10s can understand division).

P.S. Also, a clock is misleading, as I don't see what division has to do with clock. The only clock thing about it is that if you have say x mod m, then you can add and subtract m and they will still be equivalent mod m. But, lol mod is not a clock