Task: find all prime numbers not exceeding the given one. Below is the code. Could you please check if there is an error. I’m not entirely sure if it works correctly. I would be grateful.

Program Lab_4;
var
 i, n, range: integer;
 Flag: boolean;
Begin
 i: = 0;
 n: = 0;
 write ('Enter range:');
 read (range);
 while i & lt; range do
 Begin
  Flag: = True;
  n: = 0;
  i: = i + 1;
  while n & lt; i - 1 do
  Begin
   n: = n + 1;
   if ((i mod n = 0) and (n & gt; 2)) or ((i mod 2 = 0) and (i & lt; & gt; 2)) then
    Flag: = False;
  end;
  if Flag then
   writeln (i, '- Prime number!');
 end;
end.

Answer 1, authority 100%

Here, take a look. Something like this should be.

program Lab_4;
var
 i, n, range: integer;
 Flag: boolean;
begin
 i: = 1;
 n: = 0;
 write ('Enter range:');
 read (range);
 while i & lt; range do
 begin
  Flag: = True;
  n: = 1;
  i: = i + 1;
  if i & lt; & gt; 2 then
   while n & lt; = sqrt (i) do
   begin
    n: = n + 1;
    if i mod n = 0 then begin
      Flag: = False;
      break;
    end;
   end;
  if Flag then
   writeln (i, '- Prime number!');
 end;
end.

Answer 2, authority 97%

Sometimes it is much easier to rewrite the code. Then its readability will increase.

Why do you use the while loop when for is needed here, I did not understand. Well, as you rightly noted, you need to check divisibility not up to the number itself, but up to its square root.

Total:

program Lab_4;
var
 i, n, range: integer;
 Flag: boolean;
begin
 write ('Enter range:');
 read (range);
 for i: = 2 to range do begin
  Flag: = True;
  for n: = 2 to Trunc (Sqrt (i)) do begin
   Flag: = i mod n & lt; & gt; 0;
   if not Flag then
    break;
  end;
  if Flag then
   writeln (i, '- Prime number!');
 end;
end.

Answer 3, authority 100%

You can check your program for errors in an automated system. For example, here .
True, the I / O of the program will need to be adapted for the testing system, but it’s simple.

Answer 4, authority 100%

You can use the ideas Sundarama , the essence of which is simple and logical.

1. Sift only odd numbers. Those. the sieve element a [m] must contain the number m and match the number 2m + 1 .
2. Exclude from the array all numbers of the form i + j + 2ij , since 2 (i + j + 2ij) +1 = (2i + 1) ( 2j + 1) .
   This is sufficient since all composite odd numbers are the product of odd numbers.
3. Boundaries for the lower index i are determined by the condition (2i + 1) ² ∈ [3, 2M + 1] , where M – dimension of array a .
   Those. i ∈ [1, sqrt (2M + 1) / 2 -1]
4. Boundaries for the larger index j are determined by the condition i + j + 2ij ∈ [1, M] .
   Those. j ∈ [1, (M-i) / (2i + 1)] .
5. Enumeration by i and j is performed with step 1.
6. The m values ​​obtained from the selection should be converted to prime numbers using the formula p = 2m + 1 .