C3:

uses crt;
var n:integer;
begin
clrscr;
write('nhap so n:');readln(n);
while n>86 do n:=n mod 100;
if (n=68) or (n=86) then write('n la so dep');
readln;
end.

C4:

uses crt;
var n,i,s,c,max,min,j,tam:integer;
a:array[1..100]of integer;
begin
clrscr;
write('nhap so ptu:');readln(n);
for i:=1 to n do
begin
 write('A[',i,']=');
 readln(a[i]);
 if (a[i] mod 3=0) and (a[i] mod 5=0) then c:=c+1;
 s:=s+a[i];
end;
writeln('trung binh cong:',s/n:2:4);
max:=a[1];min:=a[1];
for i:=1 to n do
begin
 if max<a[i] then max:=a[i];
 if min>a[i] then min:=a[i];
end;
writelN('so lon nhat la:',max);
writeln('so be nhat la:',min);
for i:=1 to n do
for j:=i+1 to n do
if a[i] >a[j] then
begin
 tam:=a[i];
 a[i]:=a[j];
 a[i]:=tam;
end;
write('sap xep theo chieu tang dan:');
for i:=1 to n do write(a[i]:3);
writeln;
write('sap xep mang theo chieu dan dan:');
for i:=n downto 1 do write(a[i]:3);
readln;
end.

(mong dc 5*hay nhất)

câu 3:

uses crt;
var n:integer;
begin
clrscr;
write('nhap so n:');readln(n);
if(n mod 100=68) or( n mod 100=86) then write('YES') else write('NO');
readln;
end.

câu 4:

uses crt;
var i,j,n,s,s2,max,min,tam:integer;
a:array[1..100] of integer;
begin
clrscr;
write('nhap so ptu:');readln(n);
for i:=1 to n do
begin
 write('A[',i,']=');
 readln(a[i]);
 s:=s+a[i];
 if(a[i] mod 3=0) and(a[i] mod 5=0) then s2:=s2+a[i];
end;
writeln('trung binh cong cua mang la:',s/n:4:2);
max:=a[1];
min:=a[1];
for i:=2 to n do
begin
 if max<a[i] then max:=a[i];
 if min>a[i] then min:=a[i];
end;
writeln('so lon nhat la:',max);
writeln('so nho nhat la:',min);
writeln('tong cac so chia het cho 3 va 5 la:',s2);
for i:=1 to n-1 do
for j:=i+1 to n do
 if a[i]>a[j] then
 begin
  tam:=a[i];
  a[i]:=a[j];
  a[j]:=tam;
 end;
write('mang sau khi sap xep tang dan:');
for i:=1 to n do write(a[i],' ');
writeln;
for i:=1 to n-1 do
for j:=i+1 to n do
 if a[i]<a[j] then
 begin
  tam:=a[i]; 
  a[i]:=a[j];
  a[j]:=tam;
 end;
write('mang sau khi sap xep giam dan la:');
for i:=1 to n do write(a[i],' ');
readln;
end.