Очень плохо что не видно какие темы будут рассказаны лекторами на других неделях обучения(возможно я не нашел как это посмотреть). Хотелось бы видеть все, чему научат на этом курсе.
Вопрос по видеопрактике, неделя 1, задача 2, в модуле задача 3. Если к трехзначному числу приписать такое же число, по полученное число будет делиться на 7, 11, 13. Мне не понятен переход от записи abcabc к abc* 1001
EugeneM, попробую ответить. Возьмём конкретный пример, так, мне кажется, наглядней. Допустим есть два числа: 715 и 715715 . Как из первого получить второе?
В практической части было разобрано много любопытных задач. Были такие, которые отличались сложностью постановки вопроса (имею ввиду, что увидев дробь или выражение, приведенное в примере, я бы точно не задалась вопросом доказать, что оно обладает свойством делимости на 24 или 7, 13 итд). Каким образом происходит постановка данных вопросов: метафизики древности тратили жизни на перебор вариантов и поиск эффектных задач? Или есть какие-то пока еще невидимые мне закономерности, на основе которых ученый понимает, какой вопрос необходимо поставить задаче? Может быть, идут наоборот: взяв за основу алгоритм доказательства, раскручивают его в обратном направлении, приходя к условию задачи?
Спасибо за интересный вопрос. Некоторые задачи придумывают в результате поиска алгоритма для практических целей, некоторые составляют для демонстрации признака или алгоритма.
Опять поймал себя на мысли, что для меня ход мыслей лектора. В практике "Второй способ нахождения линейного представления НОД" я не понял, как получить xr = xa - qxb.
Сначала находим НОД(a, b), используя алгоритм Евклида. Расписываем a и b через их представления вида
a = a * x_a + b * y_a
b = a * x_b + b * y_b
Находим x_a, y_a, x_b, y_b, их значения будут лежать в интервале {-1, 1}. Далее смотрим на остатки x_r, полученные на этапе нахождения НОД(a, b). Например, при a > b:
a = b * q + x_r1
b = x_r1 * q + x_r2
.....
В отличие от первого способа, мы не выражаем x_r1 = a - q * b на последнем шаге, а идем по остаткам, начиная с x_r1 (т.е. сверху вниз). Причем
x_r1 = x_a - x_b
x_r2 = x_b - q * x_r1
......
Для "игреков" аналогично. Надеюсь, что доступно изложил. Пересмотрите еще раз практику по второму способу.
В дополнительном материале второй недели, а именно - в примере про фокусника, есть такая строка. Почему такой переход равносилен и каким образом мы получили выражение справа?
В примере про пифагоровы тройки в процессе замены мы получили некоторое значение для u. Мы домножили его на -1, т.к. m > n?
Тогда к правой или к левой части можем добавить число, кратное 12, от этого сравнение окажется верным. К левой части добавим 36y, затем к правой части добавим 12.
В формуле для пифагоровых троек домножение на -1 было не нужно, но при подстановке в формулу с суммой квадратов и получением итоговой формулы мы действительно используем, что m > n, так как числа x, y, z натуральные.
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Очень плохо что не видно какие темы будут рассказаны лекторами на других неделях обучения(возможно я не нашел как это посмотреть). Хотелось бы видеть все, чему научат на этом курсе.
Павел, здравствуйте! На странице курса есть полная программа. :)
Вопрос по видеопрактике, неделя 1, задача 2, в модуле задача 3. Если к трехзначному числу приписать такое же число, по полученное число будет делиться на 7, 11, 13. Мне не понятен переход от записи abcabc к abc* 1001
EugeneM, попробую ответить. Возьмём конкретный пример, так, мне кажется, наглядней. Допустим есть два числа: 715 и 715715 . Как из первого получить второе?
715715 = 715000 + 715 = 715*1000 + 715 = 715*(1000 + 1) = 715*1001.
покажу в общем виде: abc*1001=abc*1000+1)=abc*1000+abc=abcabc
Добрый день!
В практической части было разобрано много любопытных задач. Были такие, которые отличались сложностью постановки вопроса (имею ввиду, что увидев дробь или выражение, приведенное в примере, я бы точно не задалась вопросом доказать, что оно обладает свойством делимости на 24 или 7, 13 итд). Каким образом происходит постановка данных вопросов: метафизики древности тратили жизни на перебор вариантов и поиск эффектных задач? Или есть какие-то пока еще невидимые мне закономерности, на основе которых ученый понимает, какой вопрос необходимо поставить задаче? Может быть, идут наоборот: взяв за основу алгоритм доказательства, раскручивают его в обратном направлении, приходя к условию задачи?
Добрый вечер!
Спасибо за интересный вопрос. Некоторые задачи придумывают в результате поиска алгоритма для практических целей, некоторые составляют для демонстрации признака или алгоритма.
С уважением, Сергей Иванов, составитель курса.
Опять поймал себя на мысли, что для меня ход мыслей лектора. В практике "Второй способ нахождения линейного представления НОД" я не понял, как получить xr = xa - qxb.
Сначала находим НОД(a, b), используя алгоритм Евклида. Расписываем a и b через их представления вида
a = a * x_a + b * y_a
b = a * x_b + b * y_b
Находим x_a, y_a, x_b, y_b, их значения будут лежать в интервале {-1, 1}. Далее смотрим на остатки x_r, полученные на этапе нахождения НОД(a, b). Например, при a > b:
a = b * q + x_r1
b = x_r1 * q + x_r2
.....
В отличие от первого способа, мы не выражаем x_r1 = a - q * b на последнем шаге, а идем по остаткам, начиная с x_r1 (т.е. сверху вниз). Причем
x_r1 = x_a - x_b
x_r2 = x_b - q * x_r1
......
Для "игреков" аналогично. Надеюсь, что доступно изложил. Пересмотрите еще раз практику по второму способу.
В дополнительном материале второй недели, а именно - в примере про фокусника, есть такая строка. Почему такой переход равносилен и каким образом мы получили выражение справа?
В примере про пифагоровы тройки в процессе замены мы получили некоторое значение для u. Мы домножили его на -1, т.к. m > n?
Добрый вечер!
В примере про фокусника имелось в виду следующее.
-35y сравнимо с -10 по модулю 12.
Тогда к правой или к левой части можем добавить число, кратное 12, от этого сравнение окажется верным. К левой части добавим 36y, затем к правой части добавим 12.
В формуле для пифагоровых троек домножение на -1 было не нужно, но при подстановке в формулу с суммой квадратов и получением итоговой формулы мы действительно используем, что m > n, так как числа x, y, z натуральные.
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