[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 314

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Yeah, that's a mistake.
In the last two equations, there should be a minus in the right-hand sides. But this does not change the essence of the solution, just that the red line will be under the abscissa axis, not over it.
 
Yes, thank you, alsu. But where do the twos under the sines come from? This, however, doesn't affect the essence of the solution either.
Any thoughts about (n+1) weights with total weight of 2n?
 
Mathemat >>:
Да, спасибо, alsu. Тока вот откуда двойки под синусами? Это, правда, на суть решения тоже не влияет.
Каике-нибудь мысли по поводу (n+1) гирек с общим весом 2n появились?

The number of kettlebells with weight 1 must not be less than the weight of the maximum kettlebell (maximum difference between the bowls).

 
First of all, maybe we should try to prove that the weights of the weights can be divided into two groups with equal total sums. Maybe from there an understanding of what to do next will come?
 

I'll try to describe it in more detail.


M - weight of the maximum weight (<=n)

2n-M - weight of n remaining weights.

Since the weight of a weight is a natural number, then

at least M of them should have weight 1.

When we decompose all weights > 1 we get weights A and B and A -B <=M

and M weights of 1 will remain.

Since the total weight is divisible by 2, adding M weights of 1

balance the weights.

 
Mathemat >>:
Да, спасибо, alsu. Тока вот откуда двойки под синусами? Это, правда, на суть решения тоже не влияет.
I'm not thinking straight today. The twos can all be removed.

Any thoughts on (n+1) weights with a total weight of 2n?

The infinite descent method is on the tip of my tongue, but I can't figure out how to turn it around...

 
Here's ihor's summary here, it even looks like proof. Right!


Yeah, we've got another one in the stash, with a quadruple number generator, 409. Here it is: https://forum.mql4.com/ru/29339/page309
P.S. Pardon me, I solved it on page 311 :)
Next one:
 

Sorry, busy today too.
-
Here's the program:
Dim M As Long
Dim N As Long
Private Sub Command1_Click()
For M = -100 To 100
For N = -100 To 100
If (5 + 3 * (2 ^ 0.5)) ^ M = (3 + 5 * (2 ^ 0.5)) ^ N Then Print "M=", M, "N=", N
Next N
Next M
End Sub
-
The answer is succinct, although I guessed it without the program, it must be a 4th grade problem :)))

 
No, the 10th, oddly enough. The most important thing here is not just to find one obvious solution, but to justify why it is the only one.
Follow-up (9th):

For the root of 10 it's kind of obvious, since with an even degree the last digit is always 0 (except for degree 0), and with an odd one (say, 7th)
[10^3 * 3.162277...] = [3162.27...] = 3162,
i.e., it turns out two is the 3rd decimal digit in the decimal expansion of the root of 10. Correspondingly, for powers of 2n+1, it is the nth decimal digit of the expansion of the root of 10. The sequence is non-periodic.
For the root of 2 it is more complicated.
 
Mathemat >>:
Вдогонку (9-й):

Для корня из 10 вроде как все очевидно, т.к. при четной степени последняя цифра всегда 0 (кроме степени 0), а при нечетной (скажем, 7-й)
[10^3 * 3.162277...] = [3162.27...] = 3162,
т.е. получается двойка - 3-я цифра после запятой в десятичном разложении корня из 10. Соответственно для степени 2n+1 это n-я цифра разложения корня из 10. Последовательность получается непериодической.
Для корня из 2 все сложнее.

For the root of 2 your proof is also valid, but only in binary. The answer is no.

But the author of the problem must have meant a different proof.


1...307308309310311312313314315316317318319320321...628
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