Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 17
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Yeah, I see. I hadn't thought in that direction, although it really is a more universal method. Using only problem conditions ("different shoulders"), that's how I solved it.
2 MD: I don't want to waste my brain on problems with difficulty less than 3 :) It seems that a proof is not required here. But if you want, you may think about uniqueness.
Here's another one (4 points). This one is serious:
Find all natural numbers which, when multiplied by 4, turn into their mirror image. (A mirror image is when the numbers in it go in reverse order.)
I've found a lot of these, but I don't know if they're all there yet. These are numbers of the form: 21(9)78. Where the digit in brackets repeats any number of times. Starting with zero.
Yeah, I checked up to 11 nines in Excel, it doesn't have enough digit capacity beyond that. But I don't see any obstacles, the sequence is obviously infinite.
.
A little bit more than all of them. A computational search shows others. For example, 21782178 and 217802178.
I am not squeamish about it - it allows me to see and formulate sensible schizoteas.
A little bit more than all of them. A computational search shows others. For example, 21782178 and 217802178.
I'm not squeamish about it - it allows you to see and formulate sensible schizoteas.
Well, then the others are already obvious:
217821782178217821782178[ 2178]
2178(0)2178(0)2178(0)2178(0)[ 2178(0)] 2178 // as long as the zeros are the same everywhere
21(9)7821(9)7821(9)7821(9)7821(9)7821(9)78[ 21(9)78] // as long as there are the same number of nines everywhere
21(9)78(0)21(9)78(0)21(9)78(0)21(9)78(0)21(9)78(0)[ 21(9)78(0)] 21(9)78// similarly for zeros and nines
I have the same number. Couldn't find the second one, although the singularity is not obvious yet. Any thoughts on the proof?
Let's designate this number by QWERTYUIOP :)
According to the conditions, the equation must be fulfilled:
Q+W+E+R+T+Y+U+I+O+P=10 (1)
Then we look at different variants ( 1) like Q+1, Q+2, Q+1+1
But if there are two ones among the summands, then there must be a two (which will denote this). If three ones, then a three.(2)
If there is one 2, then there must also be a 1, i.e. the number of repetitions of each digit (3)
If there is only one unit among the summands, then it has to be a 2 (except Q=9, W=1, but that does not fit) (4)
I.e. from (2) (3) (4) it follows that variations are possible:
Q+2+1 (does not fit, because only at Q=7, W=2,E=1, (1) is satisfied, and W=2 and there must be one more digit besides E)
Q+2+1+1
Q+3+2+1+1 (cancel it, because for 3 there is no realization - only one Q is free)
Q+3+2+1+1+1 (cancel because for 2 no realization - only one Q is available).
Only Q+2+1+1 =10
--------------------------------------------
P.s. in general, truncated overkill and probably could be simpler
Starts with 21, then any number of 9 (including 0) and ends with 78.
2199999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999978
Any number of sequences 2178.
217821782178217821782178217821782178217821782178217821782178217821782178217821782178217821782178217821782178
Well then the others are already obvious:
217821782178217821782178[ 2178]
2178(0)2178(0)2178(0)2178(0)[ 2178(0)] 2178 // as long as the zeros are the same everywhere
21(9)7821(9)7821(9)7821(9)7821(9)7821(9)78[ 21(9)78] // on condition that nines are the same everywhere
21(9)78(0)21(9)78(0)21(9)78(0)21(9)78(0)21(9)78(0)[ 21(9)78(0)] 21(9)78// similarly for zeros and nines
I went through 13 characters by hand. In addition to those listed, a new one was found:
It turns out that it is necessary to present a generator of such numbers. As the number of digits increases, new combinations will pop up. Although it is not so new 2178 21(9)78 2178
So far this works out for me:
If the numbers a and b have this property, then the numbers have:
1) a(0)a
2) a(0)b(0)a - here we have the same number of zeros
So far we have found one elementary number 21(9)78. The rest are obtained according to the suggested rules. They are all such numbers.
The proof is a pain in the ass. Prove one by one the following statements: where x is a sequence of digits, possibly empty.
1. All numbers have the form 21x78
2. After digits 21, there are digits 7 or 9
3. The digits 78 are preceded by the digits 1 or 9
4. If 219x78 is such a number, then 21x78 is such a number
5. If 21x978 is such a number, then 21x78 is such a number
Get rid of the nines.
6. If the first three digits of a number are 217, then the fourth digit is 8.
Then we remove the level according to rules 1) or 2), until we get the elementary combination 21(9)78 or an empty set, getting rid of zeros, of course.
Whoever is interested can do this
Yes, we need some general approach, from which any possible combination is naturally obtained.
Another number problem (weight 5):
There are 32 natural numbers (not necessarily distinct) written in a string. Prove that between them one can place brackets, signs of addition and multiplication so that the value of obtained expression is divisible by 11000.
Note from me: 11000 = 11 * 2^3 * 5^3.
32 = 11 + 2 + 2 + 2 + 5 + 5 + 5.
It remains to prove the auxiliary statement: between any n numbers it is possible to place brackets and signs (*, +) so that the expression is divisible by n.
You can't concatenate numbers (you can't get 79 from 7 and 9).
Yes, we need some general approach, from which any possible combination is naturally obtained.
Another number problem (weight 5):
There are 32 natural numbers (not necessarily distinct) written in a string. Prove that between them one can place brackets, signs of addition and multiplication so that the value of obtained expression is divisible by 11000.
Note from me: 11000 = 11 * 2^3 * 5^3.
32 = 11 + 2 + 2 + 2 + 5 + 5 + 5.
It remains to prove the auxiliary statement: between any n numbers it is possible to place brackets and signs (*, +) so that the expression is divisible by n.
You can't concatenate numbers (you can't get 79 from 7 and 9).