# The Naked Scientists Forum

### Author Topic: Fun maths problems  (Read 9659 times)

#### alancalverd

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##### Re: Fun maths problems
« Reply #25 on: 15/11/2014 00:11:44 »
If you haven't already seen it, may I recommend The Imitation Game? I've just got back from the cinema, deeply impressed by a story well told (even though it's been told many times before).

#### CliffordK

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##### Re: Fun maths problems
« Reply #26 on: 15/11/2014 09:17:48 »
The answer to the ten-digit problem is:

6210001000

True enough, but is there a logical or algebraic approach to the solution?
You can arrive at the solution through logic.

Consider that you have 10 digits.  9 of them zeros.

9,000,000,000   However, in this case you must have one - 9.
9,000,000,001   However, here you only have 8 zeros, and one 8.
8,000,000,010   However, you also have one 1 (which would leave you with 7 zeros).
7,100,000,100   However, in this case, you have two - ones.
7,200,000,100   However, you now have one two, and thus 6 remaining zeros.

Now, the question would be whether this answer is unique.
I believe that it is.  Now, proving that it is unique is more difficult, but one seems to get nowhere by adding in a three.

Algebra is difficult because it doesn't deal with whole numbers as smoothly.

#### alancalverd

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##### Re: Fun maths problems
« Reply #27 on: 15/11/2014 10:42:22 »
That is the trial and error procedure I was doing in my head when I should have been concentrating on driving, though it would have taken me longer as I was working upwards from 1000000000. Pretty much the crunching approach of Colossus!

My interest is in whether a "direct solution" exists, and then if it is (a) unique and (b) generalisable. Any number theorists out there?

#### CliffordK

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##### Re: Fun maths problems
« Reply #28 on: 15/11/2014 11:30:27 »
Certainly one of the formulas you could apply is that all the digits must add up to 10.
Given that X0 is the first digit, X1 is the second digit, ... X9 is the last (10th) digit, then one gets the two equations (I think).

X0 + X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 = 10.

0*X0 + 1*X1 + 2*X2 + 3*X3 + 4*X4 + 5*X5 + 6*X6 + 7*X7 + 8*X8 + 9*X9 = 10

Given that all of your integers are positive, then you'll be limited very quickly.

Looking at the two formulas with the answer (6,210,001,000), one gets:
6+2+1+(0+0+0)+1+(0+0+0) = 10
0+2+2+(0+0+0)+6+(0+0+0) = 10

It then reduces to 6 + 2 + 2
Introducing a 3 or a 4 would throw it out of kilter.  I'm having troubles thinking of a better way to put it.

Hmmm....  So, if one was doing Octal:
01234567
42101000

And Hex
0123456789ABCDEF
C210000000000C000

Hmmm, Base 7
0123456
3211000

Now Base 6
012345
Hmmm, I think one gets stuck with this one.

#### jeffreyH

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##### Re: Fun maths problems
« Reply #29 on: 15/11/2014 23:52:43 »
For 0 + 1*X1 + 2*X2 + 3*X3 + 4*X4 + 5*X5 + 6*X6 + 7*X7 + 8*X8 + 9*X9 = 10 certain positions are limited to the value of 1. That is 9, 8, 7, and 6 can only appear once. For 5 we still could't have 2 because we would need a 1 in the twos position etc. Summing to 10 cannot take this into account.

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##### Re: Fun maths problems
« Reply #29 on: 15/11/2014 23:52:43 »