- #1

Freyja

- 62

- 39

## Homework Statement

Find a number of 5 different digits that equals the sum of all numbers of 3 digits that can be made with said 5 digits.

## Homework Equations

V[itex]^{m}_{n}[/itex] = n∙(n-1)∙(n-2)∙(n-3)∙…∙(n-m+1) = [itex]\frac{n!}{(n-m)!}[/itex]

I'll clarify that the above expression means "

*Variations without repetition*of

*m*elements taken from a set of

*n*elements". I know in English you guys don't use the term "variation" as we do here in Spain for this particular context; said variations are the "ordered" combinations (so to speak) of

*m*elements from a set of

*n*elements, and obviously result of the product of "disordered" combinations for the permutations of

*m*. I hope that makes it clear

## The Attempt at a Solution

So the number we're looking for will have the form

**abcde**(there's no repetition of digits, since the problem states that they're all different) and it'll equal the sum of all the possible variations of 3 elements that can be done with the digits {a,b,c,d,e}

Such variations are 60 (according to the formula, V[itex]^{3}_{5}[/itex] = 5∙4∙3 = 60), as follows:

abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde,

acb, adb, aeb, adc, aec, aed, bdc, bec, bed, ced,

bac, bad, bae, cad, cae, dae, cbd, cbe, dbe, dce,

cab, dab, eab, dac, eac, ead, dbc, ebc, ebd, ecd,

bca, bda, bea, cda, cea, dea, cdb, ceb, deb, dec,

cba, dba, eba, dca, eca, eda, dcb, ecb, edb, edc.

Now let's calculate the sum of all the above numbers, which we know will be a 5-digit number that I'll call

**XMhtu**(

*u*be the column of the units,

*t*that of the tens,

*h*the hundreds,

*M*the thousands and

*X*the ten-thousands).

If we look at the series of

*variations*(3-digit numbers) above, we can see that every last number of them (the numbers in the units column) is repeated as many times as variations of 2 digits can be formed with the other four digits of the number we're trying to find (V[itex]^{2}_{4}[/itex] = 4∙3 = 12), so every number is repeated 12 times. Therefore, the sum of all the numbers in the column of the units (

**) will be:**

*u***Ʃu = 12a + 12b + 12c + 12d + 12e = 12(a+b+c+d+e)**

and the same reasoning applies to the other columns; of course, the numerical value of the column of the tens will be the same as that of the units multiplied by 10 and the numerical value of the column of the hundreds will be that of the units multiplied by 100:

**Ʃt = 10 ∙ Ʃu = 120(a+b+c+d+e)**

**Ʃh = 100 ∙ Ʃu = 1200(a+b+c+d+e)**

And that's as far as I got. Now, how do I relate the number XMhtu to the number abcde that's been asked?