[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 219

New comment
 
Mathemat >>:
Наверно, хитрющая Кристина специально запутала Ганса, чтобы легче было брать его тепленьким. Или Ганс сам на это повелся.

No, no! You have to look for a Third Party here. That green, soft, defiantly erotic grass is to blame for the lie ;)

 

This all reminds me of three muziks crawling towards each other on a stretchy rubber band with three ends...

// Ehm... That's how we live.

 

Next (8th):

Given several distinct naturals, enclosed between the squares of two consecutive naturals. Prove that all their pairwise products are also different.

 

TheXpert писал(а) >>

.....................

.............360д

Doubtful. It's unlikely the goose will survive Christmas...

:)

 
Mathemat >>:

Следующая (8-й):

Даны несколько различных натуральных, заключенных между квадратами двух последовательных натуральных. Доказать, что все их попарные произведения также различны.

Well, it's quite trivial:

From the contrary. Suppose there are two pairs of numbers from the specified range that have matching products.

Then they can be represented as (k*a1)*b1=a2*(k*b2), where k is the same natural quotient and the numbers in brackets are also natural.

The minimum number k can be early 2.

But this is impossible, because no two consecutive squares of natural numbers are more than twice as different from each other.

// The exceptions are 0 and 1. But there is no hole between them for inserting anything else natural. ;)

Proved.

 
MetaDriver >>:

Но это невозможно, так как никакие два подряд идущих квадрата натуральных чисел не отличаются друг от друга более чем вдвое.

// Исключение 0 и 1. Но меж ними нет места для вставки ещё чего-либо натурального.

Here's a counterexample: 1^2 = 1, а 2^2 = 4.

Or 2^2 = 4 and 3^2 = 9. Demonstrate your reasoning on pairs (4,9) and (5,7). Where did you get this k, which should be natural?

 

Actually, Richie, there's not enough information to say anything. I've long resisted installing Eight, and I don't work with IE at all.

In fact, you might find something here.

 

I understand your idea in principle, MetaDriver. It just needs a little more care. Numbers from different pairs don't have to be multiples, because a given product can be split into 2 multipliers in different ways.

 
Mathemat >>:

Вот контпример: 1^2 = 1, а 2^2 = 4.

Или 2^2 = 4, а 3^2 = 9. Продемонстрируй свои рассуждения на парах (4,9) и (5,7). Откуда у тебя взялся этот k, который должен быть натуральным?

Ugh, shit! Yawned again, relaxed at the end. Indeed there are exceptions at the beginning of the series. Namely :

0, 1, 4, 9. That's it, the rule works.

Then we check by examining the beginning of the row directly.

0-1 -- there are no elements between.

1-4 -- spacing 2 and 3. The only variant of the pairwise product, no variants.

4-9 -- interval 5,6,7,8. The only pair of mutually incomplete numbers is 6 and 8. There is no third even number, so there is no rebuttal.

I think that's it now.

 
Mathemat >>:

В принципе твою идею я понял, MetaDriver. Просто ее надо поаккуратней оформить. Числа из разных пар не обязаны быть кратными, т.к. заданное произведение можно раскидать на 2 множителя разными способами.

You can spread it out, but if you spread it, it's all the way to the end.

Let's see. If we decompose a product into multipliers, then no multiplier can occur more than 2 times in the expansion.

Otherwise, we would have to admit that this set occurs at least twice in one of the numbers. But then ....

Further on by yourself?

1...212213214215216217218219220221222223224225226...628
New comment