What do the numbers on transport tickets mean. lucky tickets

AT modern society to this day, people believe in omens, but there are few such superstitions that would come true almost always and for almost everyone. One of these is the sign of a lucky ticket. Buying a ticket to public transport, the person hopes to get exactly happy! What kind of ticket is this, and what to do with it?

How to determine the lucky ticket

A lucky ticket is one in which the sum of the first three numbers in the six-digit number is equal to the sum of the last three numbers. This method of determining the lucky ticket is the most common. It is also called Moscow.

There is a St. Petersburg way to determine the lucky ticket. According to old Leningrad signs and superstitions, that ticket will bring good luck, the sum of the even numbers of which will be equal to the sum of the odd numbers in its serial number.

Such lucky tickets come across very often. However, there are also special, rare tickets, which, according to signs, bring even greater luck. For example, a lucky ticket is a ticket whose six-digit number consists of same digits. Also, the ticket brings good luck, in which the first three digits and the last three are the same. For example: 321132.

In addition to lucky tickets, there are also counter tickets. They are those tickets, the sum of the first three numbers of which is one unit more or less than the sum of the last three numbers. If you come across such a ticket, then on this day you will meet a person whom you have not seen for a long time.

What to do with a lucky ticket

An old sign says that if you suddenly get a lucky ticket, you must immediately eat it, and then luck will come. However, as practice shows, apart from indigestion, this ritual does not carry anything. However, there is some wisdom in this ritual. Probably, the ticket was recommended to be eaten so that no one could see the location of the numbers on it. Most likely, the numbers on the lucky ticket or their sum help to win the lottery or solve some problem.

If you have found a lucky ticket, then save it. According to popular belief He will certainly bring good luck. But at the same time, do not forget to remember either its numbers or their sum - this can lead you to the right decision in a difficult situation. Be happy and don't forget to press the buttons and

30.07.2014 09:13

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How many ways are there to pay 50 cents? We think you can pay in 1 penny, 5 nickels, 10 dimes, 25 quarters, and 50 half dollars. György Pólya popularized this problem by demonstrating an instructive way to solve it using generating functions.

Let's write an infinite sum representing all possible ways of exchanging. The easiest place to start is when there are fewer varieties of coins, so let's start by saying we don't have any coins other than a penny. The sum of all ways to pay a certain amount of pennies (and only pennies) can be written as

since each payout option includes some nickels selected from the first multiplier and some pennies selected from P. (Note that N does not equal 1 + 1 + 5 + (1 + 5 ) 2 + (1 + 5 ) 3 + ..., since this amount includes many types of payments more than once. For example, the term (1 + 5 ) 2 = 1 1 + 1 5 + 5 1 + 5 5 treats 1 5 and 5 1 as if they were different, but we want to enumerate all coin sets once, regardless of their order. )

Similarly, if we also allow dimes, we get an infinite sum

Our task is to find how many terms in C cost exactly 50 cents.

The problem is solved with a simple trick. Let's replace 1 with z, 5 per z 5, 10 on z 10, 25 on z 25 and 50 on z fifty . Each term is then replaced by z n, where n the cost of the original term in pennies. For example, the term 50 10 5 5 1 will turn into z 50+10+5+5+1 = z 71 . Each of the four possible ways pay 13 cents, namely, 10 1 3 , 5 1 8 , 5 2 1 3 and 1 13 , comes down to z 13 ; therefore, the coefficient at z 13 after z-substitutions will be 4.

Let P n N n D n Q n and C n denote the number of ways to pay the amount in n cents, if you can use coins no older, respectively, 1, 5, 10, 25 and 50 cents. Our analysis has shown that these numbers are coefficients at z n in the corresponding power series

P =	1 + z + z 2 + z 3 + z 4 + ... ,
N =	(1 + z 5 + z 10 + z 15 + z 20 + ...)P,
D =	(1 + z 10 + z 20 + z 30 + z 40 + ...)N,
Q =	(1 + z 25 + z 50 + z 75 + z 100 + ...)D,
C =	(1 + z 50 + z 100 + z 150 + z 200 + ...)Q.

It's obvious that P n= 1 for all n≥0 . On a brief reflection it is easy to prove that N n = [n/5] + 1: to add up to n cents from pennies and nickels, we should take 0, or 1, or..., or [ n/5] nickels, after which there is only one way to choose the required number of pennies. So the values P n and N n easy to calculate, but D n , Q n and C n the matter is much more complicated.

One of the approaches to the study of these formulas is based on the observation that 1 + z m + z 2m+ ... there is just 1/(1 z m). Therefore, we can write

Now, equating the coefficients at z n in these equations, we obtain recurrent relations, from which the desired coefficients are easily calculated:

For example, the coefficient at z n in D= (1 z 25)Q equals Q n – Q n 25; so it should be Q n – Q n 25 = D n, as written above.

It would be possible to uncover these relations and express Q n, for example, in the form Q n = D n + D n 25+ D n 50+ D n 75 + ..., where the sum ends when the indices go negative. However, the original, non-iterative form is convenient in that each coefficient is calculated using just one addition, as in Pascal's triangle.

We use these relations to find C fifty . Firstly, C 50 = C 0 + Q 50 so we need to know Q fifty . Further, Q 50 = Q 25 + D 50 and Q 25 = Q 0 + D 25; so we are also interested D 50 and D 25 . These values D n in turn depend on D 40 , D 30 , D 20 , D 15 , D 10 and D 5 and from N 50 , N 45 , ..., N 5 . Thus, to determine all the necessary coefficients, it is enough to perform simple calculations:

n	0	5	10	15	20	25	30	35	40	45	50
P n	1	1	1	1	1	1	1	1	1	1	1
N n	1	2	3	4	5	6	7	8	9	10	11
D n	1	2	4	6	9	12	16		25		36
Qn	1					13					49
C n	1										50

The answer is at the bottom of the table. C 50: There are exactly 50 ways to tip 50 cents.

What can be said about the closed form for C n? Multiplying all the equations gives us a compact expression for the generating function

which is a rational function of z, whose denominator has a power of 91. Thus, we can factorize the denominator into 91 factors and express C n in a "closed form", consisting of 91 terms. But such a terrible expression does not climb into any gates. Is it possible to find something better in this particular case, rather than apply the general method?

And here is the first glimmer of hope: if in C(z) replace 1/(1 z) to (1 + z + z 2 + z 3 + z 4)/(1 z 5):

= (1 + z + z 2 + z 3 + z 4)Č (z 5), Č (z) =

then the degree of the denominator of the "compressed" function Č (z) is already only 19, so this function is much better than the original one. New expression for C(z) shows, in particular, that C 5n = C 5n+1 = C 5n+2 = C 5n+3 = C 5n+4 ; indeed, this ratio is easy to explain: there are exactly as many ways to tip 53 cents as there are to tip 50 cents, since the number of pennies modulo 5 is known in advance.

However, even for Č (z) there is no simple expression based on the roots of the denominator. Probably, simplest way coefficient calculations Č (z) is obtained by noting that each factor in the denominator is a divisor of 1 z ten . Therefore, we can write

Here, for the sake of completeness, is an expanded expression for A(z):

(1 + z + ... + z 9) 2 (1 + z 2 + ... + z 8)(1 + z 5) =
= 1 + 2z + 4z 2 + 6z 3 + 9z 4 + 13z 5 + 18z 6 + 24z 7 +
+ 31z 8 + 39z 9 + 45z 10 + 52z 11 +57z 12 + 63z 13 + 67z 14 + 69z 15 +
+ 69z 16 + 67z 17 + 63z 18 + 57z 19 + 52z 20 + 45z 21 + 39z 22 + 31z 23 +
+ 24z 24 + 18z 25 + 13z 26 + 9z 27 + 6z 28 + 4z 29 + 2z 30 + z 31 .

And finally, using the fact that

we obtain the following expression for the coefficients Č n at degrees z n in the expansion of the function Č (z), wherein n = 10q + r and 0≤ r<1 0:

Č 10q+r =	∑	A j	(	k + 4 k	)	=
	j, k 10k+j=n

= A r

(

q + 4 q

)

+ A r+10

(

q + 3 q

)

+ A r+20

(

q + 2 q

)

+ A r+30

(

q + 1 q

)

.

This actually contains 10 different cases, one for each value r; but it's still a good closed formula compared to alternatives involving powers of complex numbers.

Using this expression, we can find out, for example, the value C 50q = Č 10q. Here r=0 and we have

for an amount of 1 dollar, it turns out

(

6 4

)

+ 45

(

5 4

)

+ 52

(

4 4

)

= 292 ways;

and for a million dollars this number will be

(

2000004 4

)

+ 45

(

2000003 4

)

+ 52

(

2000002 4

)

+ 2

(

2000001 4

)

=

= 66666793333412666685000001.

In the section on the question How to determine a lucky ticket or not and do I need to have lucky tickets? given by the author marriage the best answer is more tickets - more luck

Answer from Alyonka[guru]
how can you eat a ticket that the conductor had previously touched with her dirty hands

Answer from Andrey Chervov[active]
If the sum of the first three digits is equal to the sum of the second three, then the ticket is lucky. You don't have to eat =(

Answer from Neurosis[guru]
No longer needed.

Answer from tractor building[newbie]
There are 7 numbers! Which ones to fold?
let's say there 5072937
so we add 5 0 and 7 we get 12 these are the first 3 numbers
the second 3 numbers that add up 2 9 and 3 get 14
what to do with seven?

Answer from Victoria fidotova[newbie]
I do not advise you to have a ticket because the conductor touched it, and his hands are not sterile! If the first three digits match the second, then this is a lucky ticket. To be happy, just carry it with you.

Answer from Irina Chikineva[newbie]
It is not enough just to find a ticket with the right combination of numbers in order to become happy. According to legend, such a coupon must be kept and carried everywhere with you - in a wallet or bag pocket. If you do not want to collect useless things, there is another option. The lucky ticket must be eaten. And then you will certainly be lucky))

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Most students are well aware of what a "lucky ticket" is. Yes, and schoolchildren often too. True, what exactly they are and what to do with them - here opinions most often differ.

Primarily, "student happy" a ticket is considered, the answers to which you know. Don’t even go to your grandmother here - you were lucky on the exam, you pulled out a lucky ticket and passed it the first time, even though out of a hundred questions, only these two managed to learn. Yes, he answered so briskly that the teacher, tired of "be-kane and me-kane", did not even listen to you to the end - he sent you with a five in the record book and with instructions to the remaining ones: "Here! Watch and learn how to pass the subject! Take an example from this good man!"
This is what I understand - "happy ticket"!

But there are tickets, they are also travel coupons, which are considered either happy or beautiful. The second is extremely rare. Most often they are called precisely "happy"! What kind of tickets are considered as such?
Firstly, and this is an extremely rare case, a ticket is considered lucky, the digits of which are the same or symmetrical.
For example: 555555 or 252252 . There is complete symmetry.
But sometimes the symmetry is incomplete or mirror. For example like this: 251251 - the numbers are arranged symmetrically here, but the numbers are not.
In any case, the above examples are really "happy" tickets. Are there many of them? Well, I think you can easily calculate that very, very little - a thousand per million, or every thousandth ticket. The probability of such a ticket falling into the hands of a passenger is extremely small. So far, I have only received two such tickets in my life, although I travel by public transport quite often,
Do you want happiness? Therefore, dodgy and quick-witted passengers in the boredom of the road immediately came up with other options for "happiness". For example, just the same numbers in the number, arranged in random order: 251521 , for example. There is no symmetry here, but all the numbers are present. Further more. A ticket was considered lucky, the sum of the triplets of digits of which is the same. For example, 474195:

4+7+4=15= 1+9+5

1. Examples of tickets, "happy in sum":

Again, everyone knows that such tickets are found, although not every day, but still quite often. Approximately every 18th ticket is "happy in terms of amount." And if you travel constantly, then they meet at least once a week. Somehow I conducted a small experiment: I didn’t throw it away, but put these tickets in the pocket of my bag to count them at the end of the month. It was a long time ago, I don’t remember exactly how many, but in a month I had at least ten of them. Considering that I travel by municipal transport on average two or three times a day (the rest of the time is minibuses, and for some reason it is not customary to issue tickets there), it turns out that every 6-9 trip is "rewarded" with such simple happiness . Well, or one ticket in three days. But this, you see, I just got a good month, because every 18th ticket should come across as if less often.
Indeed, there are times when not a single one is caught in a month. So what to do? And the need for inventions is cunning. For example, there are tickets "happy in Moscow"(they are - "in Leningrad") is when not triples of numbers are counted, but their pairs. For example, the sum of every even number with odd ones: 6 3 49 86 . Here:

3+9+6= 18= 6+4+8

What do you think, is it possible, in addition to addition, to apply the operation subtraction? Of course you can! The main thing is to decide for yourself how to subtract - in order or from larger to smaller: 720821 . Here:

7-2-0=5= 8-2-1

But ... it is not customary for us to "subtract happiness" somehow. It is better when it is added or even multiplied!

So I came up with another kind of lucky tickets for myself: "lucky by multiplication"!
It is enough to multiply the numbers in triplets to get yourself an additional "multiplier" cheerfulness. For example: 338924. Here:

3*3*8=72= 9*2*4

Use on health! But why do you sum everything up and sum it up ... You can also multiply!

Upd: Moreover, you can not just multiply! Here in the comments docbrowns I noticed that you can also raise to a power! For example 261812 :

(2^6)^1 = 64 = (8^1)^2

And this still many times increases both the chances of "finding happiness" and the entertainment of the trip.

2. Ticket example, "lucky by multiplication" a la:

If you use public transport, take a closer look at the passengers. Very, very often you can see how, when they receive a ticket, they begin to study its numbers. Everyone is looking for happiness ... And then what to do with it? Once I heard a conversation between two girls who were going to the test: "Wow! I have a lucky ticket!" one exclaimed. "Eat it! Then you'll pass the test!!!" - immediately responded the second. Right, I laughed. Better they hoped for that happy "student" the ticket I mentioned at the beginning. And even better - so that all fifty course tickets are happy for them. But... they prefer to eat trolleybus than to teach lectures.
Guys! No need to eat coupons! It's not even useful at all. And it won't bring you happiness. Treat lucky tickets easier - time he fell for you, then happiness will not come, no - you already happy or, more simply, lucky human! That's all. This is just an excuse to slightly improve your mood. Do not believe in signs - they are far from always based on facts, and often they can also bring harm, especially if you start eating four-leaf flowers from the ground or recycled paper coupons on the bus! As in that joke: ate a lucky ticket, and then happiness came over - the controller came in!

Treat "lucky tickets" as a way to pass the time of the trip a little with arithmetic exercises, and as an additional reason to rejoice in it.

By the way, note to dads and moms: it’s very useful to tell children about such exercises. They don’t really like mental counting at school, so let them at least have fun in trolleybuses, summing up or multiplying numbers. And it won’t hurt adults either: both in a row and through one, assimilating the concepts of parity, symmetry, multiplicity ... And you can also not forget about subtraction with division. In any case, for the development of the child, such fun puzzles will not hurt.

And if you are unlucky with a ticket - do not be discouraged! There are so many cars with "lucky numbers" driving down the street!

Good luck to you, and happiness!