Question

Реши задачу и запиши ответ В обсуждении докладов принимали участие все философы. Некоторые выступали по два раза. Какое наименьшее количество выступлений могло бы быть, если это число при делении на 2, 5, 6, 7 в остатке дает 1? Ответ ​

Answer 1

Ответ:

31 доклад.

Пошаговое объяснение:

Если из этого числа вычесть 1, то получится число, которое нацело делится на 2,3 и 5.

Наименьшее такое число 2*3*5 = 30.

Значит, число докладов равно 31.

И неважно, что некоторые выступали 1 раз, а некоторые по 2 раза.

Answer 2

Problem Analysis

We are given that all philosophers participated in a discussion and some of them spoke twice. We need to find the minimum number of speeches that could have taken place if the number, when divided by 2, 5, 6, and 7, leaves a remainder of 1.

Solution

To find the minimum number of speeches, we need to find the smallest positive integer that satisfies the given conditions. We can use the Chinese Remainder Theorem to solve this problem.

The Chinese Remainder Theorem states that if we have a system of congruences of the form: ``` x ≡ a1 (mod n1) x ≡ a2 (mod n2) ... x ≡ ak (mod nk) ``` where `n1, n2, ..., nk` are pairwise coprime (i.e., they have no common factors), then there exists a unique solution for `x` modulo `n1 * n2 * ... * nk`.

In our case, we have the following congruences: ``` x ≡ 1 (mod 2) x ≡ 1 (mod 5) x ≡ 1 (mod 6) x ≡ 1 (mod 7) ```

To find the solution, we can calculate the product of the moduli: `N = 2 * 5 * 6 * 7 = 420`. Then, for each modulus `ni`, we can calculate `Ni = N / ni`. Finally, we can calculate the solution `x` using the formula: ``` x = a1 * N1 * inv(N1, n1) + a2 * N2 * inv(N2, n2) + ... + ak * Nk * inv(Nk, nk) (mod N) ``` where `inv(Ni, ni)` is the modular multiplicative inverse of `Ni` modulo `ni`.

Let's calculate the solution step by step:

1. Calculate `N = 2 * 5 * 6 * 7 = 420`. 2. Calculate `N1 = N / 2 = 210`, `N2 = N / 5 = 84`, `N3 = N / 6 = 70`, `N4 = N / 7 = 60`. 3. Calculate the modular multiplicative inverses: `inv(N1, 2) = 1`, `inv(N2, 5) = 4`, `inv(N3, 6) = 5`, `inv(N4, 7) = 1`. 4. Calculate the solution `x`: ``` x = 1 * 210 * 1 + 1 * 84 * 4 + 1 * 70 * 5 + 1 * 60 * 1 (mod 420) = 210 + 336 + 350 + 60 (mod 420) = 956 (mod 420) = 116 ```

Therefore, the minimum number of speeches that could have taken place is 116.

Answer

The minimum number of speeches that could have taken place, given the conditions, is 116.