## A Conversation for Diophantine Equations

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Tyrian Started conversation Mar 11, 2003

Since you obviously seem to know a bit about maths I was wondering if you could explain the question I asked in http://www.bbc.co.uk/dna/h2g2/alabaster/A987771

Thanks

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Gnomon - time to move on Posted Mar 11, 2003

It's a good question. I know that it is because 9 is one less than the base in which you write the numbers. The same thing applies in other bases:

In Base 8, adding the digits of a multiple of 7 will eventually lead to 7.

But the original property goes further than you have stated: the single digit number you get by adding all the digits repeatedly is called the 'digital root' of the number. It is the remainder after dividing the number by 9.

I'll have to do some mathematics to work out why. I'll get back to you.

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Recumbentman Posted Mar 11, 2003

Congratulations Gnomon! Great stuff altogether!

I also attempted an answer to Tyrian's question, and funnily enough like you I used an example from base 8. What does that say?

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NAITA (Join ViTAL - A1014625) Posted Mar 11, 2003

What you have is the property where in a number with the digits xy you eventually get 9 if you add the digits together. This is because:

If y is zero, adding nine to the number adds nine to the sum.

If y is more than zero, x will increase by one, while y will decrease by one.

If the number has three digits the first digit is only affected when the number is on the form x9y. x increases by one, y decreases by one, and the 9 goes to zero. The same is true for other multiple digit numbers.

Not a real mathematical proof, but I think you can make one if you work on it.

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Gnomon - time to move on Posted Mar 11, 2003

If you have a number N with digits a b c d e so that it looks like this: abcde, then what that actually means is:

N = a*10000 + b*1000 + c*100 + d*10 + e

But you can rewrite this as:

N = (a*9999 + a) + (b*999 + b) + (c*99 + c) + (d*9 + d) + e

= a*9999 + b*999 + c*99 + d*9 + a+b+c+d+e

= 9*(something) + (a + b + c + d + e)

So N will have the same remainder after division by 9 as a+b+c+d+e.

Repeating the process will continue to give the same remainder after division by 9. Eventually you will get a one digit result. It will be the remainder after division by 9 of the original number.

So if the original number is divisible by 9, the final result of the process must be 9.

Does that make any sense to you?

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Tyrian Posted Mar 11, 2003

Thanks for your attempt at an explanation Gnomon. It very nearly made sense when it suddenly made a kind of 'whoosh' noise and flew by me......

The reason some of you used octal as another example is because you are probably computer people - but you should have been brave and pointed out that the same rule applies to the number E in hex rather than 7 is octal

I always thought it was something to do with subtracting 1 from 10 to get the number 9 (in base 10, of course). Note the lack of any other digits apart from 1 and 0 to obtain this number. Maybe thats why I never worked it out.

I had a similar question about multiples of 11 with 4 digit numbers where digits 1&2 are the same and 3&4 are the same - or 1&4 are the same and 2&3 are the same but I doubt I'd understand the answer. I'm sure stuff like this was easier when I was younger!

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Recumbentman Posted Mar 11, 2003

In the early days of pocket calculators I discovered the pleasure of dividing things by seven. The bit after the decimal point always falls into the pattern 142857 repeating, or 428571 or 285714 and so on, which is easy to remember by 14, 28, and (just-over-)56, all multiples of 7, as indeed are 42 and (just-over-)84. I never got round to figuring out a story to account for why it was just over 56 and 84, but a nice bit of symmetry all the same. Try it!

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