[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 288

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Of course, more. There were no calculators back then. Nowadays an ordinary schoolboy, unencumbered by knowledge of richmetic but armed with a calculator, cannot prove that "two plus two multiplied by two" can be 6, not just 8. But that's a joke. It's a matter of interpretation.
 
omgwtflol >>:


так там из квадратов двухзначных только 10

10*10 + (10*10 + 2*10*1 + 1*1) + (10*10 + 2*10*2 + 2*2) +... тут только простое умножение 1значных


for sure
But it needs more RAM)
 
Well, it doesn't take much RAM to memorise three or four numbers.
I was messing something up (they didn't give the sum of consecutive squares back then, and they don't give it now either). And I calculated it wrong :)
 
Candid >>:
Э-эх, зарекался же заглядывать в эту тему :)
Hard :)
Now if you add up separately the first three and the second two, both the answer to this problem and the twist in it become clear.

10^2 = 2*2*(11 + 12) + 2*2^2. So you really only have to count the first three squares. It's beautiful.

_______________________

By the way, about cigarettes:

 
TheXpert >>:

Кстати, про сигареты:

Wah beautiful picture...

 
Mathemat >>:
Что-то мне подсказывает, что решето Эратосфена может спасти отцов русской демократии...
Итак:
Вычеркиваем кратные 2. Остались числа вида 2k+1.
Теперь вычеркиваем кратные 3 из оставшихся. Это могут быть только числа вида 2(3t) + 3 = 6t + 3. Останутся 6t+1, 6t+5.
Дальше вычеркиваем кратные 5 из оставшихся. Вычеркнем, следовательно, только 2*3*5*t + 5, 25. Останутся 30t + 1, 7, 11, 13, 17, 19, 23, 29. Обращаем внимание на то, что остатки все не делятся ни на одно простое до 5 включительно.
То же для 7: остались 210t + 1, 11, 13, 17, 19, 23 и т.п. (дальше все меньшие 210 и некратные ни 2, ни 3, ни 5, ни 7; составные там могут быть - скажем, 121).
И т.п. до простого 13 включительно.
В результате останутся только числа 2*3*5*7*11*13*t + некие остатки, не кратные ни одному простому до 13.
А дальше я в ступоре. Что-то намудрил я.

And rightly so in a stupor. It's a dead-end road. I figured out the sieve yesterday. It won't help us, even though it's antique imported.

You have to think for yourself. By the way, your answer was correct yesterday (2*3*5*7*11). Just need to write out the proof. Now, maybe later.

Here is a puzzle, also simple numbers:

We have an algorithm and a box of dice. Algorithm:

Roll one die. If you get a prime number (1, 2, 3, 5), roll another one.

If the first one adds up to a prime number, roll another one.

If the sum of the first two results in a prime number, throw another one.

Continue in the same way, i.e. as long as the sum of all previous ones results in prime numbers - throw another one.

Finish adding the dice when the sum is a composite number (end of round). Write it down and start rolling again.

// If you run out of dice, go to the shop and buy another box. If we run out of money... mmm... don't know yet, I'll ask questions, then I'll deal with the money. :)

1) How much (maximum) money would it take to buy all the cubes if they cost exactly a ruble apiece ?

2) How many points on average would be on all dice of a completed round if there were an infinite number of rounds ?

--

// I suspect that this is not a very easy task. Although at the Olympiad the youngsters would still quack... ;)

 
The difference between 89 and 97 is 8, i.e. 89 is the maximum possible prime number that can be thrown this way. The rest of the calculation is generally not difficult.
Although (2) will be a bit tricky.
 
I don't get it, MetaDriver, which numbers you have are prime. I don't think 1 is prime (or composite). It's special.
So, 1 is included in the set, specified at the first cast (1? 2, 3, 5)?
P.S. Now I understand.
 
I thought "Why not write a script to solve this problem...".
Started to think about the algorithm, surprised - it turns out the archiver by Huffman..... well almost... :) :)
 
It looks like you need an exhaustive rounds tree. It's a bit cumbersome. Or is there a neat solution?
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