https://www.youtube.com/channel/UCrvbojLGIWZot6_JUWWoMuw
đắng kí giúp mình

#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;

}

a) Phân tích thành tích các thừa số nguyên tố: \(3072=2^{10}.3\)

Tổng các ước của \(3072\)là: 

\(\sigma\left(3072\right)=1+3+\left(2^1+2^2+...+2^{10}\right)+3\left(2^1+2^2+...+2^{10}\right)\)

\(=4\left(1+2+2^2+...+2^{10}\right)\)

Ta có: \(A=1+2+...+2^{10}\)

\(2A=2+2^2+...+2^{11}\)

\(2A-A=\left(2+2^2+...+2^{11}\right)-\left(1+2+...+2^{10}\right)\)

\(A=2^{11}-1\)

Suy ra \(\sigma\left(3072\right)=4\left(2^{11}-1\right)=2^{13}-4\)

b) Tương tự. 

\(4608=2^9.3^2\)

\(\sigma\left(4608\right)=1+3+3^2+\left(2^1+2^2+...+2^9\right)+3\left(2^1+2^2+...+2^9\right)+3^2\left(2^1+2^2+...+2^9\right)\)

\(=\left(1+3+3^2\right)\left(1+2^1+2^2+...+2^9\right)\)

\(=13299\)

b, \(3+12+48+...+3072+12288\)

\(=3\left(4+4^2+...+4^6\right)\)

\(=3.\frac{\left(4+4^2+...+4^7\right)-\left(1+4+...+4^6\right)}{3}=3\times\frac{4^7-1}{3}=4^7-1\)

c, 2 + 5 + 7 + 12 + 19 + 31 + 50 + 81 + 131 

= [ (2+5) + 7 ] + [ ( 12+19) + 31] + [ ( 50+81)+131]

= 2 ( 7+31+131)

= 338

a, 2 + 5 + 11 + 20 + 32 + 47 + 65 

= 2 + ( 5 + 65 ) + ( 11 + 32 + 47 ) + 20 

= 2 + 70 + 90 + 20 = 182

ko bạn ơi đừng thêm mũ vào

program tim_uoc;

uses crt;

var i,n,tong:integer;

begin

clrscr;

write('nhap so n:');readln(n);

i:=1;tong:=0;

writeln('cac uoc cua ',n,' la:');

while i<=n do

if n mod i=0 then

begin

write(i:3);

inc(i);

end;

writeln;

i:=1;writeln('cac uoc chan:');

while i<=n do

begin

if n mod i=0 then 

begin

if i mod 2=0 then write(i:3);

tong:=tong+i;

end;

end;

writeln;

write('tong cac uoc chan:',tong);

readln;

end.

uses crt;

var n,i,t:integer;

begin

clrscr;

write('Nhap n='); readln(n);

i:=1;

writeln('Cac uoc cua ',n,' la: ');

while i<=n do

  begin

if n mod i=0 then write(i:4):

i:=i+1;

end;

writeln;

writeln('Cac uoc chan cua ',n,' la: ');

t:=0;

i:=1;

while i<=n do 

  begin

if (n mod i=0) then

begin

t:=t+i;

write(i:4);

end;

inc(i);

end;

writeln('Tong cac uoc chan cua ',n,' la: ',t);

readln;

end.

u30={1;2;3;5;6;10;15;30}

Ư(30)=(1;2;3;5;6;10;15;30)

b) 120 = 2 3  . 3 . 5 có ( 3 + 1 ) . ( 1 + 1 ) . ( 1 + 1 ) = 16 (ước)

Ư(120) = {1; 2; 3; 4; 5; 6; 8; 10; 12; 15; 20; 24; 30; 40; 60; 120}