[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 173
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A fraction 10/97 is written on the blackboard. It is allowed to add the same number to the numerator and denominator, or multiply the numerator and denominator by the same number. Is it possible to obtain a fraction equal to a) 1/2; b) 1 as a result of several such operations?
And what to do here ? Just solve the equation (10+x)/(97+x)=1/2. Or am I missing something ?
Your equation cannot be solved in integers, Yuri.
I guess the whole trick is that you can only add and multiply - and not subtract and reduce. Reduction can only be done by these manipulations, determining whether the fractions are 1/2 or 1.
Pardon, it is solved. x=77. But the question about the unit puts me in a baffling oddity.
The next one, in which you first have to understand the condition:
There are only two letters A and B in the alphabet of the Mumbu-Yumbu tribe. Two different words denote the same concept if one of them can be derived from the other using the following operations:
Can a tribal savage count all the fingers on his hand? And the days of the week?
Да я тоже к этому выводу пришел, что невозможно. Но, кажется. alsu, думает иначе, или я ошибаюсь.
No, I just missed the post with the answer.
In my spare time I found a nice solution to the triangle construction problem by two sides and bisector. Deeply analytical:)
I'll draw it and lay it out.
It would be interesting to look at this 'formula'.
Why look at it? It says it 's complicated. Everyone knows that there are no accidents, the question is how to derive the complex formula. What's God got to do with it?
No, I just missed the post with the answer.
In my spare time I found a nice solution of the problem of constructing a triangle by two sides and a bisector. Deeply analytical:)
I thought for a long time. I thought there was a catch. But there is no trick.)
Did you mean deep geometric?
P.S. How many concepts do the Mumbu-Yumbu tribe have?
At least A, B, AB, BA, BAB. The concepts AA and BB do not exist, for they are empty (right?). So we have five words, i.e. we count the fingers of our hand.
Is it possible to make a 4-letter word out of ABA? We add the letters on the right. If it is ABAA, it equals AB, i.e. it is not new. If it is ABAB, it equals (ABA)B = BABB = BA, i.e. again not new.
Similarly with the addition of letters on the left and with BAB.
So there are only 5 concepts in their language and they cannot count the days of the week.
Твое уравнение в целых не решается, Юрий.
Наверно, весь прикол в том, что можно только прибавлять и умножать - и не вычитать и сокращать. Сокращение можно сделать только этих манипуляций, определяя, равны ли дроби 1/2 или 1.
Пардон, решается. х=77. А вот вопрос о единичке ставит меня в тупик своей странностью.
Следующая, в которой надо вначале понять условие:
В алфавите племени Мумбу-Юмбу есть лишь две буквы А и Б. Два разных слова обозначают одно и то же понятие, если одно из них может быть получено из другого с помощью следующих операций:
Может ли дикарь племени сосчитать все пальцы на своей руке? А дни недели?
Fingers can. days of the week no.
There are only 5 words: A, B, AB, BA, BAB.
All others are archived in these 5.
Next. A walk is a walk (the last few problems are from the theme of invariants).
There are 16 glasses on the table. Fifteen of them are standing correctly and one is upside down. It is allowed to turn any four glasses upside down simultaneously. Is it possible, repeating this operation, to place all the glasses correctly?
Может ли дикарь племени сосчитать все пальцы на своей руке? А дни недели?
No. There can only be four different words in their dictionary: A, B, AB, BA. All others are abbreviated and reduced to one of the four specified.
--
// While I was writing, you were already ahead of me. And he answered correctly, because I also yawned that the transformation ABA => BAB is one-sided.
MetaDriver, еще слово БАБ есть.
Already on it. ;)
Mathemat писал(а) >>
There are 16 glasses on the table. Of these, 15 are correctly positioned and one is upside down. It is allowed to turn any four glasses upside down at the same time. Is it possible, by repeating this operation, to place all the glasses correctly?
Ricci certainly could.