A number theory problem by Rocco Dalto

I used Free Pascal to write the code below. $$

program for_brillant2;

type arraytype = array[1 .. 300] of longint;

var myfile:text; {Not necessary. Just adding a file to list all possible sums.}

$$ function lintpower(base,exponent:longint):longint;

var n,product:longint;

begin

 product:= 1;

for n:= 1 to exponent do

begin

product:= base * product;

end;

    lintpower:= product;

end; $$

function minnum(a:arraytype; last:longint):longint;

var j,n:longint;

small:longint;

  begin

        small:= a[1];

for j:= 2 to last do

begin

  if a[j] < small then

        small:= a[j];

end;

minnum:= small;

    end;

$$

function maxnum(a:arraytype; last:longint):longint;

var j,n:longint;

large:longint;

  begin

        large:= a[1];

for j:= 2 to last do

begin

if a[j] > large then

        large:= a[j];

end;

maxnum:= large;

    end;

$$

procedure getlist2;

var o,n,e,m,t,r,i,answer,a1,a2,a,b,total,j:longint;

      sum:arraytype;

      bool:boolean;

begin

assign(myfile,'brillant2.txt');

rewrite(myfile);

bool:= false;

j:= 1;

for o:= 1 to 9 do

begin

for n:= 0 to 9 do

begin

for e:= 0 to 9 do

begin

for m:= 1 to 9 do

begin

for t:= 1 to 9 do

begin

for r:= 0 to 9 do

begin

for i:= 0 to 9 do

begin

if (o <> n) and (o <> e) and (o <> m) and (o <> t) and (o <> r) and (o <> i)

and (n <> e) and (n <> m) and (n <> t) and (n <> r)

and (n <> i) and (e <> m) and (e <> t) and (e <> r)

and (e <> i) and (m <> t) and (m <> r) and (m <> i)

and (t <> r) and (t <> i) and (r <> i)

then

begin

a1:= o * lintpower(10,2) + n * 10 + e;

a2:= m * lintpower(10,3) + o * lintpower(10,2) + r * 10 + e;

sum[j]:= t * lintpower(10,3) + i * lintpower(10,2) + m * 10  + e;

 if a1 + a2 = sum[j] then

begin

bool:= true;

writeln(myfile, a1,' + ',a2,' = ',sum[j]); {not needed- just listing sums}

 j:= j + 1;

end;

end;

end;

end;

end;

end;

end;

end;

end;

if bool then

begin

total:= j - 1;

a:= maxnum(sum,total);

b:= minnum(sum,total);

answer:= a - b + total;

 writeln('max = ', a);

 writeln('min = ', b);

 writeln('n = ', total);

 writeln('answer = ', answer);

  end;

close(myfile);

end;

$$

begin

getlist2;

readln;

end.

$$

Running the program above we obtain: $$

$a = 9480, \: b = 2310, \: n = 42 \implies \boxed{a - b + n = 7212}.$

$$

Although it's not necessary to this problem, the list of all possible sums are:

$$

540 + 2580 = 3120

560 + 3570 = 4130

570 + 3560 = 4130

580 + 2540 = 3120

610 + 4630 = 5240

610 + 8670 = 9280

630 + 4610 = 5240

630 + 7640 = 8270

630 + 8650 = 9280

640 + 1670 = 2310

640 + 7630 = 8270

650 + 8630 = 9280

670 + 1640 = 2310

670 + 8610 = 9280

720 + 5730 = 6450

720 + 8760 = 9480

730 + 1780 = 2510

730 + 5720 = 6450

730 + 8750 = 9480

740 + 2780 = 3520

750 + 8730 = 9480

760 + 8720 = 9480

780 + 1730 = 2510

780 + 2740 = 3520

810 + 3820 = 4630

810 + 4830 = 5640

820 + 3810 = 4630

830 + 4810 = 5640

850 + 1860 = 2710

860 + 1850 = 2710

910 + 3920 = 4830

910 + 4930 = 5840

910 + 5940 = 6850

910 + 6950 = 7860

920 + 3910 = 4830

920 + 5930 = 6850

920 + 6940 = 7860

930 + 4910 = 5840

930 + 5920 = 6850

940 + 5910 = 6850

940 + 6920 = 7860

950 + 6910 = 7860