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, 3, 7, 9 В остатке даёт 1? Ответ:

Answer 1

Искомое число должно быть четным, делиться на 9 и на 7.
Наименьшее число, делящееся и на 9 и на 7 равно 7•9 = 6:.
Но 63 нечетное.
Следующее число, делящееся и на 7 и на 9 и на 2 равно:
2•7•9 = 126.

Это число также делится на 3, поскольку по признаку делимости на 3 делятся числа, если сумма чисел, выраженных цифрами в его составе, делится на 3:
1 + 2 + 6 = 9 - делится на 3.

Ответ: 126 выступлений.

Answer 2

Problem Analysis

To find the minimum number of presentations, we need to find the smallest number that, when divided by 2, 3, 7, and 9, leaves a remainder of 1.

Solution

To solve this problem, we can use the Chinese Remainder Theorem (CRT). The CRT 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 3)
x ≡ 1 (mod 7)
x ≡ 1 (mod 9)

Since 2, 3, 7, and 9 are pairwise coprime, we can apply the CRT to find the solution.

Applying the Chinese Remainder Theorem

To find the solution using the CRT, we can follow these steps:

1. Calculate `N = n1 * n2 * ... * nk`, where `n1, n2, ..., nk` are the moduli (2, 3, 7, and 9 in our case). 2. For each `ni`, calculate `Ni = N / ni`. 3. For each `ni`, calculate the modular inverse `Mi` of `Ni` modulo `ni`. This can be done using the Extended Euclidean Algorithm. 4. Calculate `x` as the sum of `ai * Ni * Mi` for each `i`, where `ai` is the remainder in each congruence.

Let's calculate the solution step by step:

1. `N = 2 * 3 * 7 * 9 = 378`. 2. `N2 = 378 / 2 = 189`, `N3 = 378 / 3 = 126`, `N7 = 378 / 7 = 54`, `N9 = 378 / 9 = 42`. 3. `M2 = 189^-1 (mod 2) = 1`, `M3 = 126^-1 (mod 3) = 0`, `M7 = 54^-1 (mod 7) = 4`, `M9 = 42^-1 (mod 9) = 6`. 4. `x = 1 * 189 * 1 + 1 * 126 * 0 + 1 * 54 * 4 + 1 * 42 * 6 = 378`.

Therefore, the minimum number of presentations that satisfies the given conditions is 378.

Answer

The minimum number of presentations that could have taken place, given that the number leaves a remainder of 1 when divided by 2, 3, 7, and 9, is 378.