[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 230
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А никто их и не обманывает. Здесь люди с мозгами, сами думать умеют.
Sorry, some people only use their backs.
Звиняй, некоторые только спинным пользуются.
I mean, you have to work hard and finally get a result.
and a result is a result
Логично мыслишь, но в рихметике подкачал. Там все проще получается.
С функцией я что-то не понял. y = 0? Но это частный случай нечетной функции, я уже о нем написал.
Exactly, 1980 is not the square of the whole.
3/1 + 5/2 +...87/43 + 44/44
86+1/1+1/2+...1/43 + 1
87+(1/1+1/2+...1/43)
How to calculate the sum of fractions still can't remember %(
With the function, it's just a joke, but you can rotate it to any angle at all.
Once again, check the rhythmics. The correct answer is 88 even. And prove the pattern, of course :)
Еще раз - проверь рихметику. Правильный ответ - 88 ровно.
That's it, I give up.
How do you calculate the nearest integers? If not by rounding, but by trimming the fractional part, then
from a^2 to (a+1)^2 we have 2a+1 numbers, i.e. for the natural number of squares 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15.... we get a natural series of "nearest whole" roots corresponding to it
1,1,1,2,2,2,2,2,3,3,3,3,3,3,3...
The nearest square to 1980 is 44^2 = 1936, so up to and including 1935 the square root is at most 43. And then another 44 times 44.
So I got this: 3/1 + 5/2 +...87/43 + 44/44 == 86+1/1+1/2 +...1/43 + 1
There's no way I can make 88.
And if you round it up, i.e. >1.5=2, you'll get a problem that can't be explained in normal language. Or certainly not in the language of an 8th grader.
Uh, no, that's not good in the Olympiad. For such a "solution" you would get 1, maximum 1.5 points out of five. That is, roughly speaking, somewhere somehow saw the pattern, but not so clear as to give an accurate, but unsubstantiated answer. If I had given an exact answer (88) without justification, I would have received at most 3. That's not bad.
Strictly between adjacent squares a^2 and(a+1)^2 there are exactly 2*a numbers (from a^2+1 to a^2+2*a). You get the pattern: somewhere in the middle, halfway to the next square, the integer part becomes greater than 0.5, and the nearest integer goes from a to a+1.
A direct check on small numbers confirms it and even allows to put forward hypotheses:
1. The nearest integer to sqrt(a^2+a) is a,
The nearest integer to sqrt(a^2+a+1) equals a+1.
We try to prove: sqrt(a^2+a) = sqrt((a^2+a+ 1/4) - 1/4 ) = sqrt((a+1/2)^2 - 1/4 ) < a+1/2, i.e. the nearest integer is a.
Further, sqrt(a^2+a+1) = sqrt((a^2+a+1/4) + 3/4 ) = sqrt((a+1/2)^2 + 3/4 ) > a+1/2, i.e. the nearest integer is a+1.
Great, now count how many integers for the root equal exactly to a. This is a numbers greater than a^2, the square of a itself and another a-1 numbers smaller than a^2(they remain from the previous square of a-1). The total is exactly 2*a numbers.
That is, the same fraction 1/a ideally occurs exactly 2*a times and gives a contribution to the sum equal to 2.
Now we look at 1980. The calculator says that its root is 44.497, i.e. it is probably the last number before increasing the nearest integer from 44 to 45. But in 1978 calculators were hardly given out at the Olympiads, you had to do everything by hand. In fact, 1980 = 44^2 + 44, i.e. the number 1980 exactly closes the group of 88 numbers with the nearest to the root equal to 44.
And then everything is clear.
Er, no, that's not how it works in the Olympiad. For such a "solution" you would get 1, maximum 1.5 points out of five. That is, roughly speaking, somewhere somehow saw a pattern, but not so clear as to give an accurate, but unsubstantiated answer. If I had given an exact answer (88) without justification, I would have received at most 3. Already not bad.
Strictly between adjacent squares of a^2 and(a+1)^2 are exactly 2*a numbers (from a^2+1 to a^2+2*a). You get the pattern: somewhere in the middle, halfway to the next square, the integer part becomes greater than 0.5 and goes from a to a+1.
A direct check on small numbers confirms this and even allows you to propose hypotheses:
1. The nearest integer to sqrt(a^2+a) is a,
The nearest integer to sqrt(a^2+a+1) equals a+1.
We try to prove: sqrt(a^2+a) = sqrt((a^2+a+ 1/4) - 1/4 ) = sqrt((a+1/2)^2 - 1/4 ) < a+1/2, i.e. the nearest integer is a.
Next, sqrt(a^2+a+1) = sqrt((a^2+a+1/4) + 3/4 ) = sqrt((a+1/2)^2 + 3/4 ) > a+1/2, i.e. the nearest integer is a+1.
Great, now count how many closest integers for the root equal exactly a. These are a numbers larger than a^2, the square of a itself and a-1 numbers smaller than a^2(they were left over from the previous square of a-1). The total is exactly 2*a numbers.
That is, the same fraction 1/a ideally occurs exactly 2*a times and gives a contribution to the sum equal to 2.
Now we look at 1980. The calculator says that its root is 44.497, i.e. it is probably the last number before increasing the nearest integer from 44 to 45. But in 1978 calculators were hardly given out at the Olympiads, you had to do everything by hand. Actually, 1980 = 44^2 + 44, i.e. the number 1980 exactly closes the group of 88 numbers, which is the closest to the root equal to 44.
The rest is clear.
I should have found a problem and published it before I regretted not doing it.
Actually, the tasks are serious. This one is one of the easiest for eighth graders. I don't post the really hard ones here.
Why don't you post something with your favorite Fibonacci numbers? I mean, they have a lot of unexpected properties. You guys, post it if you can find it. Even if you don't know the solution.
Just please don't say anything about trading, OK?
Ээ нет, так не пойдет на олимпиаде. За такое "решение" ты получил бы 1, максимум 1.5 балла из пяти. Т.е., грубо говоря, где-то как-то увидел закономерность, но не настолько четко, чтобы хотя бы выдать точный, но необоснованный ответ. Если бы дал точный ответ (88) без обоснования, получил бы от силы 3. Уже неплохо.
Строго между соседними квадратами a^2 и (a+1)^2 ровно 2*а чисел (от a^2+1 до a^2+2*а). Закономерность ты уловил: где-то в серединке на полпути к следующему квадрату целая часть становится больше 0.5, а ближайшее целое переходит от а к а+1.
How about posting something with your favourite Fibonacci numbers?
>> That's a great suggestion!