[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 287
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А покажи, Володь, как ты доказал, что разность не могёт быть равной, скажем, 14.
every 3 of its terms is divisible by 3
each of its 5 terms is divisible by 5
each of its 9 members is divisible by 9
each of its 11 members is divisible by 11
each of its 13 terms is divisible by 13
and only 2 and 7 and 14 (and possiblylarger numbers) do not divide any or all at once. All at once cannot be divided if at least one of them is prime.
// This is not exactly a proof, but how to prove it is hopefully clear.
Let's think further.
Okay:
We cross out multiples of 2. That leaves us with numbers like 2k+1.
Now cross out multiples of 3 from the rest. These can only be numbers of the form 2(3t) + 3 = 6t + 3. This leaves us with 6t+1 and 6t+5.
Then we cross out multiples of 5 from the remaining ones. Therefore we remove only 2*3*5*t + 5, 25. That leaves 30t + 1, 7, 11, 13, 17, 19, 23, 29. Note that the remainders all do not divide by any prime up to and including 5.
Same for 7: the remainder is 210t + 1, 11, 13, 17, 19, 23, etc. (then all the lesser 210s and not multiples of 2, 3, 5, or 7; there may be compounds there - say, 121).
And so on up to and including simple 13.
This leaves only numbers 2*3*5*7*11*13*t + some remainders, not divisible by any prime up to 13.
And then I'm stumped. I've made a mess of things.
The picture is called "Oral Counting"
State Tretyakov Gallery. Art of the 12th - early 20th century.
The picture shows a 19th century village school during a lesson on oral arithmetic. The teacher is a real person, Sergei Alexandrovich Rachinsky. He was a professor at Moscow University, a botanist and mathematician. On the wave of nationalism in 1872, Rachinsky returned to his native village of Tatevo, where he created a school with a dormitory for peasant children, developed a unique method of teaching oral counting. Bogdanov-Belsky, himself a former pupil of Rachinsky, dedicated his work to an episode of school life with the creative atmosphere that prevailed in the classroom.
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Hardly anyone learned by heart the squares of two-digit numbers then
( 14*(14+1)(14+2) - 9*(9+1)(9+2) ) / (6*365) = (14*15*16 - 9*10*11)/ (6*365)
No, I can't do it verbally.
Just add up the squares of the sums, memorise 5*10^2, then 21+44+69+96 - realistically for a schoolboy with an impaired memory, pizot to 230 that 730, the result is a favourite score...?
it's easier to add than multiply
вустно раскладываем квадраты суммов, запоминаем 5*10^2, далее 21+44+69+96 - реально для школьника с непропитой памятью, пицот да 230 того 730, в результате получаем любимую оценку...?
складывать вроде проще чем помножать
All this on the condition (I wrote at the end) that two-digit squares were learnt by heart at the time, and if they weren't
All this on the condition (I wrote at the end) that two-digit squares were learned by heart at the time, and if not
so there are two-digit squares only 1010*10 + (10*10 + 2*10*1 + 1*1) + (10*10 + 2*10*2 + 2*2) +... there is only simple multiplication of 1-digit
To my surprise, it turned out that I remember the first four squares, the only thing left to do is to calculate and remember the fifth. Now if you add up the first three and the second two separately the answer to this problem and a twist in it becomes clear.
I think, by the way, that in those days the average schoolboy was working with his head much more than he is now.
I remember when I was in 8th grade I used to crack brackets like this on the fly, now it takes time =)