[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 211
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I'm stuck on TheXpert's problem (page 207 of the thread). I feel it's not hard to set a limit on the number of digits of the largest number (unlikely to be much more than 10).
In the mean time, here's the proof:
Prove that if n is odd, then 46^n + 296*13^n is divisible by 1947.
P.S. 1947 = 3*649.
Что-то застрял я на задаче TheXpert'a (стр. 207 ветки). Чувствую, тут несложно установить предельное количество цифр самого большого числа (вряд ли намного больше 10).
Probably just the opposite :) -- I have that suspicion. I haven't looked at the answer yet -- I'm guessing that the maximal number is 1 less than some prime number.
Math. induction rules :) .
Alexei, did you know that you can do complex calculations in your head without computers?
It turns out there are different kinds of multiplication:
. (point) - surface multiplication.
x (cross) - spatial multiplication
* (star) - spatio-temporal.
Video lessons on arithmetic
Наверное как раз наоборот :) -- есть у меня такое подозрение. Ответ я пока не смотрел -- есть предположение что макс. кол-во на 1 меньше какого-то простого числа.
The further on, the fewer options are found for numbers that satisfy the conditions. After ten, assuming only zero, the real hitches begin.
Math. induction rules :) .
Too simple again, damn!
Video lessons in arithmetic
We'll see, thanks, Ilya.
Чем дальше, тем меньше находится вариантов для цифр, удовлетворяющих условиям. После десятки, предполагающей только нуль, начинаются реальные затыки.
Thanks, Andrew, but I hope I can somehow avoid this mess :)
OK, this one can be solved without induction:
Prove that you can always choose several (at least one) positive integers from n such that their sum is divisible by n.
P.S. Pardon, the problem is trivial.
P.P.S. No, it's non-trivial.
Спасибо, Андрей, но все же надеюсь, что можно будет как-то обойтись без этой каши :)
It's from RSDN, and it's highly appreciated -- which means it can't be solved easily -- I spent most of my time on RSDN in the branch where such problems are asked :)
Prove that you can always choose several (at least one) positive integers from n such that their sum is divisible by n.
Yeah, it's more interesting :)
Задачка с RSDN
in which case are you sure the problem can be solved analytically?
It is probably still analytically proving the existence of a maximal number. But how it is constructed is a dark matter. I don't want to get into all these mazes of divisibility... Besides it would be necessary to count the number of such numbers.
Вероятно, все же аналитически доказывается существование максимального числа. А вот как оно конструируется - темный лес. Как-то не хочется лезть во все эти дебри признаков делимости... К тому же еще нужно будет и считать количество таких чисел.
Digging in slowly, too. Picked at twelve and shut up. For 11 digits max number = 98765456405. Divide by 12 with the next addition does not work.
In this connection I doubt, that process will shut up necessarily before prime number.
// I thought of making a program that would try to find all solutions, and the maximal one at that.
// But then I realized that the simple number will not work - the long will not hold more than fifteen decimal places.
// But assembling numbers from pieces is too boring... :))