General Chat Thread, Solved: My Son Went To A Maths Master-Class Today... in General; While he was there he was set this problem:
Write down six numbers [from 1 to 9]
Starting from the ...

Solved: My Son Went To A Maths Master-Class Today...

While he was there he was set this problem:

Write down six numbers [from 1 to 9]

Starting from the left add consecutive numbers together in pairs. If the result is bigger than nine record the right-most digit.

Continue to do this until there are no more numbers to add together.

Record the last number when the addition is complete.

A worked example:

Starting numbers:

1, 2, 3, 4, 5, 6

1st stage:

3, 5, 7, 9, 1

2nd stage:

8, 2, 6, 0

3rd stage:

0, 8, 4

4th stage:

8, 4

5th stage:

2

Another worked example [for clarity]

Starting numbers:

4, 7, 9, 8, 2, 1

1st stage:

1, 6, 7, 0, 3

2nd stage:

7, 3, 7, 3

3rd stage:

0, 0, 0,

4th stage:

0, 0

5th stage:

0

The problem: Be able to predict the answer [at the 5th stage] without having to go through the stages in between.

The tutor called this 'Modular Maths' and he said that when you see the solution it is very simple. I can't see the pattern at the moment ['A' level maths is a long time ago for me. However my son is 11!]

Does anyone know how to explain this before it drives me crazy?

Thanks.

Last edited by DaveP; 8th November 2011 at 10:40 PM.
Reason: Correct thread content.

AFAICS This doesn't match with your description of the problem as there would be no "lone" number if pairing 1+2, 2+3, 3+4 etc

You only end of with "lone" numbers if the idea is to add 1+2, 3+4, 5+6 (e.g 3,7,1) then next round you end up with 3+7,1 (e.g 0,1)

Unless I've missed something?
Si

No I think you are correct. I was explained to me in somewhat of a rush. Your interpretation is correct. Having said that it does not shed any light anew for me.

Assuming you can ignore the modulus thingy at all stages and just do a final modulus then 5th stage equals

a + 5b + 10c + 10d + 5e + f

therefore you can ignore 10c and 10d as they will always produce 0.

so answer is a + 5b + 5e + f

i.e example 1

1 + 10 + 25 + 6 = 42 equiv to 2

example 2

4 +35 + 10 + 1 = 50 equiv to 0

QED (or maybe just lucky and it won't work for all cases )

Si

Looks good from here. However I am still curious to know if there is a more formal proof of this. There seems to be a proof for other 'trivial' problems like this. I can't believe that doesn't already have someone's name attached to it as a theorem/proof/conundrum/...

I don't get it either - I was bored silly in maths A Level and that was some years ago

What is it about this that you don't get?

The question or the answer?

I ask because I thought I had understood what was required initially but, as pointed out by @SimpleSi my original explanation of the problem was not correct.

Well I can reliably get the right answer, but I've utterly failed to get a working equation out of it, and I did GCSE Maths and Statistics, A level maths, physics and Chemistry, along with both Chemistry and Maths courses at degree level.

@SimpleSi: has pretty much got it.
If you run with Simple's explanation, the middle two numbers have no effect on the final outcome, so literally ignore them.

The next pairing of numbers (the 2nd and 5th numbers in the chain) provide a modifier according to the difference. If the difference between these two numbers is odd, then you will have a modifier of 5, but will have a modifier of 0 if the difference is even.

The final pair of numbers (1st and 6th) is a straight forward addition.

The simplest explanation I can give:

1st number + 6th Number + 5x(2nd + 5th) = XX, and take the right hand number as your answer.

What I couldn't get was an equation that could remove the left hand number reliably.