[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 426
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I couldn't find the formula. At school we used to get out of situations like that - I don't remember how. But it was something very simple. Yeah, getting old...
Ah, well, here's the formula - in the Mile-Roo answers a^x±a^y=a^x-(1±a^(y/x)). Only it doesn't give us anything :(
X/60 is the length of wall Z.
And then you have to throw out the common walls somehow :)
By the way, Rambler and Yandex seem to have gone bust.
Indeed they have!
Anything will go bust from tasks like this!
X=2*Z*(A^2+A)
We have all amicably forgotten that A has to be natural. But this is the second one. The first derives from the original way of solving the quadratic equation: you have to find the complete square. But it seems that fifth graders don't know how to do this anyway, except for the smartest ones.
X/(2*Z) = A^2 + A = ( A + 1/2 )^2 - 1/4
Hence A is calculated.
P.S. And then we proceed from Richie's note: "all spent materials went to make a grid". This means that the parity is absolutely accurate, i.e. there is no surplus left. If so, what can be said about X/(2*Z)? I don't know yet, I think. Oh, right, it's natural too.
Yes. That's the thing, we have to come up with a solution for fifth grade. And they don't know quadratic equations either. the solution should probably be in the spirit of reasoning.
Or it's really some kind of Olympiad problem for the smart ones.
A solution for the fifth grade. Let's think about it.
What do we have? AA is the number of cells. Z is the length of the side of the square of one cell. X is the pagone metre of wire.
Reasoning.
To calculate the total amount of X you have to add the length of the horizontal bars to the length of the vertical bars. The first thing that catches your eye is the fact that there are 1 more horizontal rod than A. The same goes for the vertical rods. The total number of bars is (A+1)+(A+1). Now you need to find the length of one rod. It will be equal to A*Z. In total:
Х=((А+1)+(А+1))*(А*Z).
X=(2A+2)* (A*Z)
X=2A*AZ + 2*AZ
X=2Z*(A~2+A)
X/2Z=A~2+A
A~2 + A - X/2Z = 0
An equation of the second degree. It is not a problem for fifth grade. In Soviet times, the discriminant was taught in either 7th or 8th grade. It seems that we won't be able to find a solution for the fifth grade.
Let's try a different approach. How much rod will it take for 1 cell and how many cells in total?
Calculate the bottom row. The first grid will use 4Z of rod (perimeter of the grid). The second and all subsequent cells - 3Z bars (one side of the square is already built by the previous cell). Since we have A cells, the first row takes 4Z + (A-1)*3Z rods.
Consider the second row. The first cell will take 3Z of rod. The second and each subsequent one takes 2Z bars. So the second row takes 3Z+(A-1)*2Z
Likewise, each successive row will require a rod = 3Z+(A-1)*2Z. In total the total number of bars will be equal to:
X= [4Z + (A-1)*3Z]+[(4Z + (A-1)*3Z)*(A-1)] Let's try to simplify.
X= [4Z + 3AZ - 3Z] + [4Z + 3AZ - 3Z]*(A-1)
X= [4Z + 3AZ - 3Z] + [4AZ - 4Z + 3*(A~2)*Z - 3AZ - 3AZ + 3Z]
X= 4Z + 3AZ - 3Z + 4AZ - 4Z + 3*(A~2)*Z - 3AZ - 3AZ + 3Z
X=(4Z - 3Z - 4Z + 3Z) + (3AZ + 4AZ -3AZ - 3AZ) + 3*(A~2)*Z
X=AZ + 3Z*(A~2)
X=AZ + 3Z*A*A
X=AZ(1+3A)
X/Z= A(1+3A)
X/Z = A+3*A~2
Again we come to the quadratic equation 3A~2 + A - X/Z = 0
A friend once asked me to think about a problem about wise men. Here is the text of the problem.
"One wise man said to two other wise men A and B: 'I have conceived two
natural numbers. Each of them is greater than one, but the sum of them is less than
one hundred. To sage A I will now tell - in confidence from B - the product of these
and to wise man B I will tell, in confidence from A, the sum of the numbers. After that
he asked them to guess the numbers. A and B had
the following dialogue
A: "I cannot guess the numbers".
B: "I knew beforehand that you could not identify the numbers".
A: "Then I know the numbers".
B: "Then I know.
What kind of numbers did the wise man conjure up?"
I wonder if anyone has solved this problem and how? I solved it then.... :)
drknn, such long and complicated calculations - for fifth graders, even Olympians? I don't believe it :)
But ValS's problem is more interesting.
A friend once asked me to think about a problem about wise men. Here is the text of the problem.
What are the numbers the wise man riddled with?"
I wonder if anyone has solved this problem and how? I solved it then.... :)
The first thing that comes to mind is that the sage told both opponents the same number = 4. The product of 2 and 2 gives 4 and the sum is also 4. There is no rigid restriction in the condition that the numbers originally conceived were different. He could have intended X = two and Y = two.