program abcdef;

uses Crt;

var

k, lowerLimit, upperLimit, i, j, reversed, temp, remainder: integer;

isPrime, isPalindrome: boolean; 

begin

clrscr;

write('Nhap so chu so k (1<=k<=9): ');

readln(k);

 lowerLimit := 1;

for i := 1 to k - 1 do

lowerLimit := lowerLimit * 10; 

upperLimit := lowerLimit * 10 - 1; 

writeln('Cac so nguyen to doi xung co ', k, ' chu so la:');

for i := lowerLimit to upperLimit do

begin

_{// Kiểm tra số nguyên tố}

isPrime := True;

if i < 2 then

isPrime := False

else

for j := 2 to trunc(sqrt(i)) do

if i mod j = 0 then

beginisPrime := False;break;end;

_{// Kiểm tra số đối xứng}

if isPrime then

begin

reversed := 0;

temp := i;

while temp <> 0 dobeginremainder := temp mod 10;

reversed := reversed * 10 + remainder;

temp := temp div 10;

end;

isPalindrome := (i = reversed);

 if isPalindrome then

writeln(i);

end;

end; 

readln;

end.

#include <bits/stdc++.h>
using namespace std;
long long x;
//chuongtrinhcon
bool ktnt(long long n)
{
    for (int i=2; i*i<=n; i++)
      if (n%i==0) return(false);
    return(true);
}
//chuongtrinhcon
bool ktso(long long n)
{
    int a=n%10;
    int b=n/10; b=b%10;
    int c=n/100; c=c%10;
    int d=n/1000; d=d%10;
    if (a+b==c+d) return(true);
    else return(false);
}
//chuongtrinhchinh
int main()
{
    for (x=1000; x<=9999; x++)
      if ((ktnt(x)==true) and (ktso(x)==true)) cout<<x<<" ";
    return 0;
}

Program HOC24;

var i: integer;

function nt(x: integer): boolean;

var j: integer;

begin

nt:=true;

if (x=2) or (x=3) then exit;

nt:=false;

if (x=1) or (x mod 2=0) or (x mod 3=0) then exit;

j:=5;

while j<=trunc(sqrt(x)) do

begin

if (x mod j=0) or (x mod (j+2)=0) then exit;

j:=j+6;

end;

nt:=true;

end;

Begin

for i:= 10 to 99 do if nt(i) and i mod 2=0 then write(i,' ');

readln

end.

uses crt;

var n,i,dem,j,t:integer;

kt:boolean;

begin

clrscr;

readln(n);

t:=0;

for i:=2 to n do 

begin

kt:=true;

for j:=2 to i-1 do 

  if i mod j=0 then kt:=false;

if kt=true then 

begin

write(i:4);

t:=t+i;

end;

end;

writeln;

writeln(t);

readln;

end.

program bai_toan;

var

      N, i, sum: integer;

begin

      write('Nhap so N: ');

      readln(N);

      write('Cac uoc cua ', N, ' khong ke ', N, ' la: ');

      for i := 1 to N - 1 do

            if N mod i = 0 then

                  write(i, ' ');

      writeln;

      sum := 0;

      for i := 1 to N - 1 do

      begin

            if N mod i = 0 then sum := sum + i;

      end;

      if sum = N then writeln(N, ' la so hoan hao')

      else writeln(N, ' khong phai la so hoan hao');

      writeln;

      writeln('Tat ca so hoan hao trong pham vi 1 -> ', N, ' la:');

      for i := 1 to N do

      begin

            sum := 0;

            for j := 1 to i - 1 do

            begin

                  if i mod j = 0 then sum := sum + j;

            end;

            if sum = i then writeln(i);

      end;

      readln;

end.

số cần tìm là 1979

#include <bits/stdc++.h>
using namespace std;
int diaphuong(int n) {
if (n<2) {
    return 0;
}
int squareRoot = (int) sqrt(n);
int i;
for(int i=2; i<=squareRoot; i++) {
    if(n%i==0) {
        return 0;
    }
}
    return 1;
}

int main() {
int i, N;
cin >> N;
for(int i=0; i<=N; i++) {
    if(diaphuong(i)) {
    cout << i << endl;
}
}
}

Chúc bn học tốt!

Đáp án cần chọn là: A

#include <bits/stdc++.h>
using namespace std;
const long long maxint=1000000;
bool a[maxint];
long long n,i,j;
int main()
{
    cin>>n;
    for (i=2; i<=n;i++)
        a[i]=true;
    for (i=2;i<=n; i++)
    if (a[i]==true)
    {
    for (j=i*i;j<=n;j+=i)
        a[j]=false;
    }
    for (i=2; i<=n; i++)
        if (a[i]==true) cout<<i<<" ";
    return 0;
}