[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 315

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The last digit of a number in binary is not equal to the last digit in decimal. This is where the problem lies.
 
Mathemat >>:
Последняя цифра числа в двоичной не равна последней в десятичной. Тут вся и проблема.

If the sequence of low bits of a number is non-periodic, then the sequence itself is non-periodic.

If D1,D2, ...,Dn is a periodic sequence

then the sequence D1 mod 2, ... Dn mod 2 is periodic.


 
Yes, but that does not mean that the sequence of the least significant digits in a decimal notation is non-periodic either.
ihor, do you have a formula for calculating the last digit of a number in decimal by its representation in binary?
Your answer is correct (and so I suspected), but the proof is a bit thinner:

It is not clear why gamma_2n+1 = 1.
 
Mathemat >>:
Да, но это не означает, что последовательность младших разрядов в десятичной записи - тоже непериодическая.
ihor, у Вас есть формула, позволяющая вычислить последний разряд числа в десятичной по его представлению в двоичной?

(N mod 10) mod 2 = N mod 2 ;
(the least significant bit of the last decimal digit = the least significant bit of the number)

 
Convinced, ihor.
Next:
 
in my opinion, it's quite simple. All the healthy ones will visit their sick friends on the first day. If no one was immune, on the second day they will all get sick and their previously sick friends, who have already recovered and, moreover, are immune, will come to visit them. That is, after such a visit no one will fall ill and on the third day, when all the sick people have recovered, the epidemic will cease.
If someone initially had immunity, not all healthy runts will catch the disease on the first day, but only those who haven't been vaccinated. As a result, on the second day, those who were sick on the first day will recover and be immune, those who were not immune will get sick and those who were immune will stay healthy. As a result, we have the same picture as on the first day: all three groups of short-stemmers are present, and if this continues, they will all simply cycle into each other on a daily basis. Consequently, the epidemic will never end.
 
Here's the solution:


Next. Problem for grade 8 - so they are unlikely to know the formulas for solving recurrence equations:
 
The first sequence is the Fibonacci numbers 1,2,3,5,8,13,21 etc. The second is the same sequence, but as the first two are rearranged, starting with b4,b5,... will be missing until a4,a5,... first 1, then another 1, then the sum of those 1 (=2), then the sum of 1 and 2, and so on, i.e. all members of bn are decreased consecutively by 1,1,2,3,5,8 etc.: 4=5-1,7=8-1,11=13-2,18=21-3, 29=34-5,47=55-8, i.e. the same Fibonacci sequence, but shifted to the right by 3 positions. Since the i-3rd term of the Fibonacci sequence is always strictly smaller than the difference of the i-th and i-1st of its terms, it turns out that bn sequence starting from the 4th number cannot contain Fibonacci numbers. Consequently, the answer is that there are only 3 such numbers: 1, 2 and 3.
 
Yes, the answer is the same, three numbers. Solution: "By induction it is proved that a(n-1) < b(n) < a(n) when n>=4".
That's what induction is in the 8th grade!
Next (8th):
 
Take any point with number C and L lines pass through it

1 : C+ci+...=0
.............
L : C+cj+..=0
added, we obtain L*C+the sum of all numbers (S) except C =0
L*C+S-C=0
S=C(1-L)

S=C1(1-L1)
S=C2(1-L2)

1-L is always < 0
It turns out S has the opposite sign to each number.
Since C1+C2+=0 => S=0;

0=Ci*(not 0) => Ci=0 (all numbers are 0)
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