[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 277
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The last five are abvgd.
First abv, then bvgd.
Now multiply: abv*bvg = ag.
And then, the third one is bwd.
Well, yes, the idea is essentially the same.
It will be harder to prove that it's not less than 12.
P.S. The trick is that we can clear 64 for 21 + 1 = 22 questions, but we can't clear 32 for 11 yet...
P.S. Прикол в том, что 64 мы можем прояснить за 21 + 1 = 22 вопроса, а вот 32 за 11 пока не можем...
Come on. Let's try the next one.
// I don't know how to prove it, although the impossibility of reduction is obvious.
// all residues modulo 3 = 1 need 1 move increment (1), all %3 = 2 need two moves, and zero is zero anyway.
//(1) exception at the beginning of a natural number - = 4, we need four moves.
Have you seen the 5^1000 divisibility problem? Thoughts e or lazy so far?
No, wait. There is one more point in the card problem.
Now the cards are on a circle, 50 pieces. The cards have the same numbers, +1 and -1. For one question, you can find out the product of three in a row. How many questions do you need?
Теперь карточки - по окружности, 50 штук. На карточках - те же числа, +1 и -1. За один вопрос можно узнать произведение трех идущих подряд. Сколько нужно вопросов?
Well, that's rubbish. It takes exactly fifty questions.
And there are three ways of doing it. In one round, in two and in three.
Wow, that's three. Come on, if it's not too much trouble. Although... suit yourself. The simplest solution is to multiply consecutive triples, shifting by one in position. The most interesting part is the proof, of course.
Ты видел задачку с делимостью на 5^1000? Мысли е или лень пока?
The thought is there. But it's complicated. You'll probably have to look into Pascal's signs, or something like that.
And look for some regularity there, like "no number divided by (...above five*5) contains zeros in its record.
Then conclude that (5^1000)*...above zero is the required one.
Ого, целых три. Ну давай, если не в лом. Хотя... как хочешь. Самое простое решение - перемножить последовательные тройки, сдвигаясь на единицу в позиции. Самое интересное - доказательство, разумеется.
Well the options are simple - with a shift of 1, 2, or 3 positions per move. You could do more, but...
It doesn't matter though, if the sequence of questions doesn't matter - there's only one solution.
The proof here is simple, I think.
All the numbers will have to participate in the dance at least 3 times. (But no more than that.)
You can't "archive" the solution, because the condition of not dividing the cards into non-adjacent groups interferes.
MetaDriver писал(а) >>
The solution cannot be "sarchived", because the condition of non-separation of cards into non-adjacent groups prevents it.
Of course, if you drop it, you can solve it in 18.
OK, the answer is correct, 50. Let's not prove it. Although the proof is curious.
With a five in a thousand, I offered a variant by induction, but you apparently didn't see it. I'll try to finish it off. Next:
A game of naval combat - on a 7 by 7 field. How many minimum shots are enough to hit a four-decker (a) linear, (b) of adjacent squares with sides adjacent to each other for sure? 444
Fuck her, incomprehensible condition. Another one:
Find a set of five different naturals such that any two are mutually prime, but any few numbers give a composite number in sum. 449
Mathemat писал(а) >>
A game of naval combat - on a 7 by 7 field. How many minimum shots are enough to hit a quadruped (a) linear, (b) from adjacent squares with sides adjoining each other for sure ? 444
Could you be more specific? like draw something or describe it unambiguously. So that you don't get a glitch in the conditions. (:I'm good at it:)