[Musicforawhile (OP)]

The answer to the ten-digit problem is:

6210001000

[Musicforawhile (OP)]

Is it a number puzzle or letter puzzle?

[alancalverd]

The answer to the ten-digit problem is:

6210001000

True enough, but is there a logical or algebraic approach to the solution?

[David Cooper]

Is it a number puzzle or letter puzzle?

FMAMJJA should not be approached as a mathematical problem.

[chiralSPO]

Is it easier if posed as JFMAMJJA... ?

[alancalverd]

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]

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.
6,210,001,000   Which is your answer.

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]

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]

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 (10^th) 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]

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.