n is a number and n=(p1^c) * (p2^d).Here p1 and p2 are prime. Let a=p1^c and b=p2^d.
gcd(i,n)= 
gcd(i,a) * 
gcd(i,b)
How to prove this? Any explanation? Thanks in advance.
→ Pay attention
Before contest
CodeTON Round 9 (Div. 1 + Div. 2, Rated, Prizes!)
30:50:32
Register now »
CodeTON Round 9 (Div. 1 + Div. 2, Rated, Prizes!)
30:50:32
Register now »
*has extra registration
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3823 |
| 3 | Benq | 3738 |
| 4 | Radewoosh | 3633 |
| 5 | jqdai0815 | 3620 |
| 6 | orzdevinwang | 3529 |
| 7 | ecnerwala | 3446 |
| 8 | Um_nik | 3396 |
| 9 | ksun48 | 3390 |
| 10 | gamegame | 3386 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2024 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Nov/22/2024 10:44:29 (l3).
Desktop version, switch to mobile version.
Supported by
Each divisor of n looks like p1i × p2j. The sum on the right is
(φ(p1a) + p1 + p12 + p13 + ... + p1a) × (φ(p2b) + p2 + p22 + p23 + ... + p2b)
Using CRT you can make a bijection between terms on the right and and i on the left. For example, there are φ(n) values for which gcd(i, n) = 1, and there is φ(p1a)φ(p2b) = φ(n) on the right. Also, you can take φ(p1a) values which are coprime to p and then you can take only one of p2, p22, ... to make i for which gcd(i, n) = p2, or p22, etc.
Please explain briefy. I am too weak in math.