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If we have four two-digit prime numbers $p_1$, $p_2$, $p_3$ and $p_ 4$ such that they all end with a different number. For example 11, 13, 17 and 19. What can be the last three digits of $p_1p_2p_3p_4$? It is pretty obvious that the last digit has to be $9$ since $1\cdot3\cdot7\cdot9 \equiv 9$ (mod $10$), but I have a tough time figuring out that this number can be mod $1000$.

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  • $\begingroup$ There are only a few two-digit prime numbers. $\endgroup$ – Dietrich Burde Apr 14 at 18:51
  • $\begingroup$ By my count there are $750$ possible products. Can you enumerate them? $\endgroup$ – Oscar Lanzi Apr 14 at 22:47
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Here is a program in Ocaml that checks it I think:

let whatweneed=[[11;31;41;61;71];[13;23;43;53;73;83];[17;37;47;67;97];[19;29;59;79; 89]]

let rec makelists ll =
  match ll with
  |q::w -> [q]::makelists w
  |[] -> []

let rec appendele l1 l2=
  match l1 with
  |q::w -> (List.map (fun x -> q::x) l2) @ (appendele w l2) 
  |[]->[]


let rec oneineach ss=
  match ss with
  |q::w ->appendele q (oneineach w) 
  |_->[[]]

let combinations = oneineach whatweneed

let multall= List.map (fun y -> (List.fold_left (fun acc x->acc*x) 1 y)) combinations

let multallmodthousand = List.map (fun x -> x mod 1000) multall

let rec removeDuplicates lst =
  match lst with
  | [] -> []
  | x :: xs -> x :: (List.filter (fun u -> not (u = x)) (removeDuplicates xs));;

let finalresult = removeDuplicates multallmodthousand

Here are the results:

$ [189; 499; 429; 49; 359; 529; 439; 169; 989; 899; 699; 909; 539; 959; 39; 849; 279; 709; 549; 259; 389; 809; 519; 719; 729; 759; 779; 789; 859; 469; 299; 129; 929; 839; 569; 69; 579; 109; 639; 689; 919; 739; 149; 419; 239; 649; 559; 589; 629; 249; 769; 979; 599; 409; 309; 749; 969; 79; 819; 619; 659; 679; 159; 509; 369; 879; 939; 949; 489; 219; 229; 289; 89; 269; 99; 319; 209; 339; 19; 29; 59; 139; 479; 669; 329; 999; 829; 349; 399; 199; 459; 889; 799; 9; 379; 449; 869; 119; 609; 179]$

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