[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 281
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Mathemat писал(а) >>
Next:...........
I'm confused so far.
Consideration about the problem with 5^1000:
If you can prove that no powers of five can have two zeros in a row, then the answer is e.g. (5^1000)*11
MetaDriver писал(а) >>
If we can prove that no powers of five can have two zeros in a row, then the answer is for example (5^1000)*11
No, it won't work with 11. Some zeros will disappear, others will appear. But there is something to it.
Yes, at first the 5^1000 problem is confusing. But then you start to think. Try to consistently construct numbers divided by an increasing degree of five. I almost learned how to do it, just haven't proved it yet.
Well, I'm off to bed, Volodya. At the same time I'll think about the last problem.
Ладно, я ушел спать, Володя. Заодно о последней задачке подумаю.
Okay, good night. I'm gonna crash, too.
Ужыс. Я пока запутался.
Соображение нащёт задачи с 5^1000:
Если удастся доказать что ни в каких степенях пятёрки не может стоять два нуля подряд, тогда ответом будет например (5^1000)*11
the thing is, the entry 5^1000 has exactly those two zeros in a row - checked on a calculator, so it's a dead end:)
Oh, what a creepy calculator you have, alsu. Care to share?
Oh, yeah. If it counts the first 30 significant digits correctly, then yes, there are two zeros in a row.
Ой какой у тебя жуткий калькулятор, alsu. Не поделишься?
А, ну да. Если считать, что первые 30 значащих цифр он считает верно, то да, есть два нуля подряд.
Exactly. If you consider that.
Baba Yaga is against it! Once it starts rounding up, it accumulates such an error that you can only believe the first three or four digits on the left. :)
OK, let's construct a number since the methods of proof of pure existence don't work directly.
If we have a number consisting of one digit that is divisible by 5 (that's 5), then we can add a digit to its left side so that it becomes divisible by 5^2. This digit is either 2 or 7 (this is the base of induction).
Assertion of induction:
Suppose we already have a number of n digits that is divisible by 5^n. Then we can add a non-zero digit to its left side so that the resulting (n+1)-digit is divisible by 5^(n+1).
Proof:
The original number is A*5^n. After adding the digit b to the left we get the number
b*10^n + A*5^n = (2^n*b + A) * 5^n
Thus we have to find such a digit b that the bracket is divisible by 5. Then the induction statement will be proved.
We have to solve the comparison:
2^n*b = -A (mod 5)
Here b is the digits from 1 to 9 (zero is not allowed, it is forbidden), which span the complete system of deductions modulo 5. Since 2^n is not divisible by 5, the expression on the left also covers it. Hence, there will always be at least one digit b that is exactly equal to -A (mod 5).
That's it.
ОК, конструируем число, раз уж методы доказательства чистого существования напрямую не работают.
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Всё.
Sounds about right.
By the way, here is the solution to the problem of 5 numbers (and not just 5) given in the problem book: