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# The Maths Problem You Will Never Solve (Probably)! watch

1. (Original post by stephen_king-evans)
this is a simple way

Basically:

100a + 10b + 1c (HTU)

100+10+1=111

111/3 = 37 (so basically you can forget about place value)

if a+b+c is divisible by 3 then it will work

QED
But what about 4-digit numbers or 5-digit numbers or 7-digit numbers etc?
2. (Original post by DerBoy)
Sorry I stopped reading your proof halfway through, so I don't know if it is complete, but there is really easy way of doing it (not Unbounded's method).
What would that be? Other than 'look at 10^n mod 3', which makes it very obvious, and is just a short version of what I did [and I didn't say this, just because I shouldn't assume the OP is familiar with modular arithmetic], I can't think of anything particularly short.
For all:
3. (Original post by DerBoy)
But what about 4-digit numbers or 5-digit numbers or 7-digit numbers etc?
Hmm good point, that's that theory out the window
4. (Original post by Unbounded)
What would that be? Other than 'look at 10^n mod 3', which makes it very obvious, and is just a short version of what I did [and I didn't say this, just because I shouldn't assume the OP is familiar with modular arithmetic], I can't think of anything particularly short.

Want me to PM you?
5. Any number written in base 10 is congruent modulo 3 to the sum of its digits.
6. FFS why do people keep doing this?

What isn't a proof;

213 = 200 + 10 + 2. Now split everyhing up as follows; 200 = 66*3 + 2, 10 = 3*3 + 1 and 2 = 2.

Now everything is divisible by three in that sum if 2 + 1 + 2 is divisible by 3. It is.

Remarks; this only works for one number. Similary if we wrote abc for a three digit number and repeated we may add some "bbumf" but there are only finitely many numbers that can be expressed that way. So we might as well check them.

Why we need a proof? Well we want to know it works for all numbers. So we have to dispense with things like special cases. But we are still dealing with numbers and we can represent all numbers in the form a(n)10^n + a(n-1)10^(n-1) + ... + a(1)10 + a(0) with the a(i)'s between and including zero and less than 9 and n is just some number.
7. (Original post by stephen_king-evans)
this is a simple way

Basically:

100a + 10b + 1c (HTU)

100+10+1=111

111/3 = 37 (so basically you can forget about place value)

if a+b+c is divisible by 3 then it will work

QED
What? All you showed here is that when a, b and c all equal one the sum is divisible by 3. Also this is only for three digit numbers.
8. Ok so every number can be written as (if it is an integer)
X=a+10b+100c+....(10^n)*z
For all n, 10 congruent to 1 [3] so 10^n congruent to 1 [3] for all n
Therefore, X congruent to a+b+c+...+z
9. consider the 3 times table, 3,6,9,12,15,18,21,24,27,30,33,36 .

note the transition from 9-12, the sum of digits of 12 is 3, which is a multiple of 3. We do not need to consider the numbers in the sequence above BETWEEN 12 and 21 because it will just be the sum of digits of 12 in addition with a multiple of 3. EG 15 = sum of digits of 12 + 3, 18 = sum of digits of 12 + 6. The additional value will always be a multiple of three.

Now after consider the tens and units, we make a transition into a different tens digit. note that up to 30, the yens digits change when we add 9. eg 3,12,21,30. Observe that the units digits in these sequence decrease by 1 each time, 03,12,21,30. Also note that as we are adding 9, the tens digits increase by 1 each time. the sums of the digits of each number in that sequence is each mutually 3 because for ever time the units decreases by 1, the tens increases by 1 and we start with 3. meaning all digits make a sum of a multiple of 3.

I don't need to consider the numbers in between due to the paragraph preceeding the above. Also, i don't need to consider numbers above 30 because the two paragraphs preceeding this one will be a factor of every multiple of 3 after 30, for example 33 is just 30+3 and we've proved for these values.eg 96 is just 30 + 30 +30 +6 and we've proved for this values.

This was a very badly explained proof and i'm open to criticism, this is what makes us mathematicians better.
10. (Original post by DeanK22)
FFS why do people keep doing this?

What isn't a proof;

213 = 200 + 10 + 2. Now split everyhing up as follows; 200 = 66*3 + 2, 10 = 3*3 + 1 and 2 = 2.

Now everything is divisible by three in that sum if 2 + 1 + 2 is divisible by 3. It is.
No it's not. Anyway, the answer is obvious: they keep doing things like that because they're not university maths students. Educate them rather than going "FFS WTF RAAAGE". In fact, these people are doing something brilliant that most non-university maths students don't do: looking at a problem they can't solve, and trying things. You should be encouraging them, not berating them.

(Original post by DeanK22)
Bumph? At least make your spelling look plausible...
11. (Original post by tanakataku7)
consider the 3 times table, 3,6,9,12,15,18,21,24,27,30,33,36 .

note the transition from 9-12, the sum of digits of 12 is 3, which is a multiple of 3. We do not need to consider the numbers in the sequence above BETWEEN 12 and 21 because it will just be the sum of digits of 12 in addition with a multiple of 3. EG 15 = sum of digits of 12 + 3, 18 = sum of digits of 12 + 6. The additional value will always be a multiple of three.

Now after 18, we make a transition into a different tens digit. note that up to 30, the yens digits change when we add 9. eg 3,12,21,30. Observe that the units digits in these sequence decrease by 1 each time, 03,12,21,30. Also note that as we are adding 9, the tens digits increase by 1 each time. the sums of the digits of each number in that sequence is each mutually 3 because for ever time the units decreases by 1, the tens increases by 1 and we start with 3. meaning all digits make a sum of a multiple of 3.

I don't need to consider the numbers in between due to the paragraph preceeding the above. Also, i don't need to consider numbers above 30 because the two paragraphs preceeding this one will be a factor of every multiple of 3 after 30, for example 33 is just 30+3 and we've proved for these values.eg 96 is just 30 + 30 +30 +6 and we've proved for this values.

This was a very badly explained proof and i'm open to criticism, this is what makes us mathematicians better.
Good thoughts. Few criticisms, then. What about 99 + 3? 999 + 3? Can you find some general way of explaining what happens when we have to take up a new 'column'? Also, as far as I can see, you've attempted to explain here that every multiple of 3 has digits that add up to a multiple of 3, but not conversely; can you prove that everything that has digits that add up to a multiple of 3 is a multiple of 3? (Equivalently, can you prove that if something isn't a multiple of 3 then its digits don't add up to a multiple of 3?)
12. (Original post by Unbounded)
Oh dear.
Spoiler:
Show
Well, if you really want a proof...

Note that for all natural numbers n: which is divisible by 3.

Write a number N in its digits:

So let us consider N minus the sum of its digits:

, which is divisible by 3. Hence N leaves the same remainder on division by 3 as the sum of its digits (as their difference is a multiple of 3), and so N is divisible by 3 if and only if the sum of its digits is divisible by 3.
noob.
Nice to meet you Unbounded.
13. OP, you've angered the mathmos.
14. (Original post by generalebriety)
Good thoughts. Few criticisms, then. What about 99 + 3? 999 + 3? Can you find some general way of explaining what happens when we have to take up a new 'column'? Also, as far as I can see, you've attempted to explain here that every multiple of 3 has digits that add up to a multiple of 3, but not conversely; can you prove that everything that has digits that add up to a multiple of 3 is a multiple of 3? (Equivalently, can you prove that if something isn't a multiple of 3 then its digits don't add up to a multiple of 3?)

Erm sorry i was a bit vague but the solution to the '9999' is consideredthe concept if we have a number with all digits 9, eg 9999. then it will make a transition into a number with 1 as the preceeding digit followed by 2 as the last digit, this will always results into sum of 3. I see what you mean about the converse. I enjoy talking about this to you but eastenders is on so maybe in 30 minutes., thnx!!!
15. The concept of proof is one of the most difficult things for young people to begin to understand. A very large majority of the population of this country would agree that if you show something is true for five cases, then it must always be true. The idea that this might not be true for the sixth (or any subsequent case) would simply not occur to them.

Similarly, most people would be satisfied that drawing a triangle, measuring the angles and adding them together constitutes a proof that the angles in a triangle sum to 180 degrees.
16. The way i'd do it (without using modulus stuff) is as follows:

(using 4 digit numbers as my example)
any 4 digit number abcd can be written as 1000a + 100b + 10c + d (for larger numbers add more powers of 10.)

seperate these to a + b + c + d + 999a + 99b + 9c, then factorise
a+b+c+d + 3(333a + 33b + 3c)

3(333a + 33b + 3c) is a multiple of 3 and hence divisible by three. For the whole thing to be divisible by 3, a + b + c + d must also be divisible by 3. Exactly the same thing works with pulling out 9, only a + b + c + d must be divisible by 9.

You can change this slightly to check for divisibility by 11 -

1000a + 100b + 10c + d = -a + b - c + d + 1001a + 99b + 11c = -a + b -c + d + 11(91a + 9b +c)

Hence if -a +b -c +d is divisible by 11, then so is abcd. (the pattern continues so for a 7 digit it would be a - b + c - d + e - f + g)
17. (Original post by JAKstriked)
noob.
Nice to meet you Unbounded.
Nice to meet you too
18. I think the answer is 4.

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EDIT: Who is/What is PMT? Thanks anyway!
19. (Original post by stephen_king-evans)
Hmm good point, that's that theory out the window
20. (Original post by milliondollarcorpse)
It's even in the wrong forum...

Biggest fail thread of the day
ahahah totally agree BIGGEST FAIL THREAD OF THE DAY

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