I got something interesting on this Problem.Generating X values it shows ϕ(N) for N.But i can't evaluate why this picture follows phi function.Can anyone explain me theory behind it? How co-primes are related with skipping X on this problem?
I got something interesting on this Problem.Generating X values it shows ϕ(N) for N.But i can't evaluate why this picture follows phi function.Can anyone explain me theory behind it? How co-primes are related with skipping X on this problem?
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | Radewoosh | 3415 |
| 8 | Um_nik | 3376 |
| 9 | maroonrk | 3361 |
| 10 | XVIII | 3345 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 164 |
| 2 | adamant | 150 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 144 |
| 5 | errorgorn | 141 |
| 6 | cry | 139 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 9 | TheScrasse | 134 |
| Название |
|---|



You number the mirrors with the point of the lazer as 0, next mirror as 1 and so on. Now if you fire the lazer at mirror numbered x it passes through the mirrors x, 2x mod(n), .., nx mod(n) which are all multiples of x taken modulo n. Now let g = gcd(x, n) if g > 1 the light comes back to the source mirror after hitting n / g mirrors (as x * (n / g) mod(n) = n * (x / g) mod(n) = 0). So they are not correct choice and the avaliable mirrors left are those for which gcd(x, n) = 1. Suppose we fired at one of those mirrors and the light comes back to source mirror at kth hit, so we have k * x mod(n) = 0 as (x, n) = 1 inverse of x modulo n exist, multiplying by its inverse we get k = 0 mod(n) which means k = n.
So all those mirrors for which we have gcd(x, n) = 1 are the right choices.
"light comes back to the source mirror after hitting n / g mirrors" ....thanks a lot.