Cho số tự nhiên n. Hãy tính tổng các ước số nguyên tố của n!
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
uses crt;
var i,n,t,j,kt:integer;
begin
clrscr;
readln(n);
t:=0;
for i:=2 to n do
if n mod i=0 then
begin
kt:=0;
for j:=2 to trunc(sqrt(i)) do
if i mod j=0 then kt:=1;
if kt=0 then t:=t+i;
end;
write(t);
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
uses crt;
var i,j,t,kt:integer;
begin
clrscr;
t:=0;
for i:=1 to 100 do
if i>1 then
begin
kt:=0;
for j:=2 to i-1 do
if i mod j=0 then kt:=1;
if kt=0 then t:=t+i;
end;
writeln(t);
readln;
end.
Câu 2:
*Viết chương trình:
uses crt;
var n,p,t:integer;
begin
clrscr;
write('Nhap n='); readln(n);
t:=0;
for p:=2 to n-1 do
if n mod p=0 then t:=t+p;
writeln('Tong cac uoc that su cua ',n,' la: ',t);
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
program bai_1;
uses crt;
var i,n,j,d,dem:word;
begin
clrscr;
repeat
write('nhap n:');readln(n);
if (n<=0)or(n>=10000)then writeln('so ban nhap khong hop le, ban hay nhap lai:');
until (n>0)and(n<10000);
writeln('cac uoc so la so tu nhien cua ',n,' la:');
for i:=1 to n do
if n mod i=0 then write(i,' ');
writeln;
dem:=0;
for i:=2 to n do
begin
d:=0;
for j:=2 to i div 2 do
if i mod j=0 then inc(d);
if (d=0)and(n mod i=0)then inc(dem);
end;
if dem>0 then writeln('cac uoc so la so nguyen to cua ',n,' la:');
begin
d:=0;
for j:=2 to i div 2 do
if i mod j=0 then inc(d);
if (d=0)and(n mod i=0)then write(i,' ');
end;
if dem=0 then write(0);
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
uses crt;
var i,n,m,k,d:integer;
{---------------chuong-trinh-con-tim-ucln--------------------}
function ucln(x,y:integer):integer;
var t:integer;
begin
while y<>0 do
begin
t:=x mod y;
x:=y;
y:=t;
end;
ucln:=x;
end;
{------------chuong-trinh-con-kiem-tra-so-nguyen-to-------------------}
function nt(b:longint):boolean;
var j:longint;
begin
nt:=true;
if (b=2) or (b=3) then exit;
nt:=false;
if (b=1) or (b mod 2=0) or (b mod 3=0) then exit;
j:=5;
while j<=trunc(sqrt(b)) do
begin
if (b mod j=0) or (b mod (j+2)=0) then exit;
j:=j+6;
end;
nt:=true;
end;
{---------------chuong-trinh-chinh---------------------}
begin
clrscr;
write('Nhap N: '); readln(N);
write('Nhap M: '); readln(M);
d:=0;
k:=ucln(N,M);
for i:=1 to k do
if nt(i) then d:=d+1;
if d>0 then writeln('2 so nay tuong duong voi nhau')
else writeln('2 so nay khong tuong duong voi nhau');
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
#include <bits/stdc++.h>
using namespace std;
long long n,i,t,j;
bool kt;
int main()
{
cin>>n;
for (i=2; i<=n; i++)
{
kt=true;
for (j=2; j*j<=i; j++)
if (i%j==0) kt=false;
if (kt==true) cout<<i<<" ";
}
cout<<endl;
t=0;
for (i=1; i<=n; i++)
if (n%i==0) t+=i;
cout<<t;
return 0;
}
![](https://rs.olm.vn/images/avt/0.png?1311)
n=p.q
mà p;q là các số nguyên tố
=>Ư(n)={1;p;q;n)
=>n=1+p+q=pq
=>1+q=pq-p
=>1+q=p(q-1)
=>2+q-1=p(q-1)
=>p(q-1)-(q-1)=2
=>(p-1)(q-1)=2
2=1.2
=>p-1=1 hoặc p-1=2
p-1=1=>p=2;q-1=2=>p=3
=>n=3+2+1=6
p-1=2=>p=3;q-1=1=>q=2
=>n=3+2+1=6
Vậy n=6
uses crt;
var n,i,s,kt,j,t:longint;
begin
clrscr;
write('Nhap n='); readln(n);
s:=1;
for i:=1 to n do
s:=s*i;
t:=0;
for i:=2 to s do
if s mod i=0 then
begin
kt:=0;
for j:=2 to i-1 do
if i mod j=0 then kt:=1;
if kt=0 then t:=t+i;
end;
writeln('Tong cac uoc nguyen to cua n,'! la: ',t);
readln;
end.
ta dinh nghia nhu sau
n!=n*(n-1)*(n-2)*...*1;
vs 4!=1*2*3*4;
vi the ban chi can tinh tong cac so nguyen to tu 1 den n thoi
ko can tinh n! lam j cho mat tg ,do phuc tap la O(n) nhe