Find this 10-digit number

20 lines of brute force code solves it in under a second

First, let’s calculate digit j.

We know that the sum of all digits from 0 to 9 = 45,

We have a+b+c+d+e+f+g+h+i+j = 45

Calculating Modulo 9 for both sides:

Mod (a+b+c+d+e+f+g+h+i+j,9) = Mod(45,9) = 0
(Eq. 1)

We also have Mod(( abcd + efg + hi; 9 = Mod(2019,9)

That means Mod (a+b+c+d+e+f+g+h+i,9) = Mod(2019,9) = 3 (Eq. 2)

Subtracting (Eq. 2) From (Eq. 1) we get:

Mod(j,9) = Mod (-3,9) = 6

j = 6 (Eq. 3)

Now we have {a,b,c,d,e,f,g,h,i} = {0,1,2,3,4,5,7,8,9} in some order.

To solve this problem we have the following equations from the addition:

d + g + i = 10c1 + 9 (Eq. 4)

c1 + c + f + h = 10c2 + 1 (Eq. 5)

c2 + b + e = 10c3 (Eq. 6)

c3 + a = 2 (Eq. 7)

Where c1, c2 and c3 are the carry from col.1, col.2 and col.3 from the right.

We have the following restrictions on the carry from each column.

Restriction on c1:

From equation (4), the minimum value for the sum of 3 different digits is 0+1+2 = 3 and the maximum value for the sum of 3 different digits is 9+8+7 = 24. That means the right side can only be 9 or 19 that means

0<=c1<=1 (Eq. 8)

Restriction on c2:

From equation (5), the minimum value for the sum of 3 different digits is 0+1+2 = 3 and the maximum value for the sum of 3 different digits is 9+8+7 = 24. But using the minimum and maximum values of c1 as well we’ll have: The minimum value of the left side = 3 and the maximum value is 25. That means the right side can only be 11 or 21 that means

1<=c2<=2 (Eq. 9)

Restriction on c3:

From equation (6), the minimum value of the right side is 0+0+1 = 1 and the maximum value is 2+9+8 = 19. This will only leave us with a single value to the right side which is 10. That means c3=1 (Eq. 10)

Calculating a:

From equation 7 we get: 1 + a = 2. That means

a = 1 (Eq. 11)

This will leave us with the values for the remaining variables as:

{b,c,d,e,f,g,h,i} = {0,2,3,4,5,7,8,9} in some order. (Eq. 12)

Let us rewrite the equations (4), (5) and (6) with the results we found so far and after transferring all carries to the right side:

d + g + i = 10c1 + 9 (Eq. 13)

c + f + h = 10c2 + 1 - c1 (Eq. 14)

b + e = 10 - c2(Eq. 15)

Adding equations (13), (14) and (15) we get:

d + g + i + c + f + h + b + e = 10c1 + 9 + 10c2 + 1 - c1 + 10 - c2

Simplifying and reordering we get: b + c + d + e + f + g + h + i = 9c1 + 9c2 + 20 We can calculate the left side from equation (12) we’ll get: 9c1 + 9c2 + 20 = 38 9c1 + 9c2 = 18 That means c1 + c2 = 2.

We have 2 options: Option 1: c1 = 0 and c2 = 2. Option 2: c1 = 1 and c2 = 1.

Let’s see option 1:

d + g + i = 9 (Eq. 16)

c + f + h = 21 (Eq. 17)

b + e = 8 (Eq. 18)

Equation 16 gives us the following possibilities: {d,g,i} = {0,2,7} or {0,4,5} or {2,3,4} in some order

Equation 17 gives us the following possibilities: {c,f,h} = {4,8,9} or {5,7,9} in some order

Equation 18 gives us the following possibilities: {b,e} = {0,8} or {3,5}

We have now to choose the acceptable solutions i.e. the sets which make all digit appears once to be an accepted. In this case we’ll have 2 ways: {d,g,i} = {0 ,2,7} in some order and {c,f,h} = {4,8,9} in some order and {b,e} = {3,5} in some order

OR

{d,g,i} = {2,3,4} in some order and {c,f,h} = {5,7,9} in some order and {b,e} = {0,8} .

Each way of this will give us many solutions but since we are interested in the lowest number we must assign the digits to the letters in ascending order as much as possible. This will give us the following values:

b = 0, e = 8, c = 5, f = 7, h = 9, d = 2, g = 3, i = 4. We also found that a = 1 and j = 6, this gives us the number: 1052873946.

Let’s see option 2: d + g + i = 19 (Eq. 19) c + f + h = 10 (Eq. 20) b + e = 9 (Eq. 21)

Equation 19 gives us the following possibilities: {d,g,i} = {2,8,9} or {3,7,9} or {4,7,8} in some order

Equation 20 gives us the following possibilities: {c,f,h} = {0,2,8} or {0,3,7} or {2,3,5} in some order

Equation 21 gives us the following possibilities: {b,e} = {0,9} or {2,7} or {4,5} in some order

We have now to choose the acceptable solutions i.e. the sets which make all digit appears once to be an accepted.

In this case we’ll have 3 ways: {d,g,i} = {2,8,9} in some order and {c,f,h} = {0,3,7} in some order and {b,e} = {4,5} in some order OR {d,g,i} = {3,7,9} in some order and {c,f,h} = {0,2,8} in some order and {b,e} = {4,5} in some order OR {d,g,i} = {4,7,8} in some order and {c,f,h} = {2,3,5} in some order and {b,e} = {0,9} in some order.

Each way of this will give us many solutions but since we are interested in the lowest number we must assign the digits to the letters in ascending order as much as possible. This will give us the following values:

b = 0, e = 9, c = 2, f = 3, h = 5, d = 4, g = 7, i = 8. We also found that a = 1 and j = 6, this gives us the number: 1024937586. The lowest number is the one of option 2.

Answer is: 1024937586