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

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. 

\(4608=2^9.3^2\)

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

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

\(=13299\)

#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;

}

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

Ư(30)=(1;2;3;5;6;10;15;30)

a) 72 = 2 3 . 3 2  có (3 + 1) . (2 + 1) = 12 (ước)

Ư(72) = {1; 2; 3; 4; 6; 8; 9; 12; 18; 24; 36; 72}

a) 72 =  2 3 .  3 2 có (3 + 1) . (2 + 1) = 12 (ước)

Ư(72) = {1; 2; 3; 4; 6; 8; 9; 12; 18; 24; 36; 72}

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}