# What is half of frac 99 69

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There's news from our pirates Black pearlafter they had disappeared from the scene for a while: They were submerged! Captain Sperling came up with the idea that, thanks to the curvature of the earth, you have to cover less distance in a ship if you dive a little in between, and that's how it came about Black pearl converted into a submarine that can go up to 500 m deep. (Side effect: It is easier to escape the pursuers!) For comparison: The earth's radius is about 6366 km.

On the demonstration voyage, Captain Sperling would like to take the really shortest route through the English Channel between the ports of Dover and Calais, which is usually 41 km of `` earth-curved '' route. How deep will he go?

Finally it's off to a long journey: The team takes the (on measured by the water) 6431 km long stretch through the Atlantic from Porto (in Portugal) to - who would have thought! - the island of Tortuga in the Caribbean. How many kilometers can Sperling save by being able to dive?

Solution:

For the first trip we first consider an average of the globe along the route Calais-

Be there C. Calais and D. Dover. The shortest route possible for the submarine is the straight route CD, provided their deepest point is not deeper than 500m below the
Sea level is. This
Point is the center P. the way CD. The center perpendicular to CD runs
through the center of the earth
M.. The diving depth in the point P. then results as the difference between the earth's radius r = 6636km and | MP|.

Be the angle DMC. Then the trigonometric relation cos results = .

On the other hand, that is true to 360 behaves like the arc of a circle to the circumference of the earth 2r. So it applies = .

Substituting into the above equation gives

The maximum diving depth is then the difference to the earth's radius, i.e. about 33m.

Comment: The distance saved compared to driving on the water surface is only 7cm.

We are now trying the same approach for the Porto-Tortuga route. Inserting into the above formula with a distance of 6431km results in the maximum diving depth

thus a far greater value than possible for the submarine.

To find the shortest possible route, we consider - similar to above - an average of the earth along the route from Porto (P.) to Tortuga (T). We also draw an inner circle around the center of the earth M. with a radius of 6365.5km, which indicates the maximum diving depth.

We are now looking for the shortest path from P. to T inside the outer annulus. Vividly speaking, this is the path that a rubber rope describes when you pull it off P. to T tense and when the sea floor is exactly 500 meters below the surface of the water. In mathematical terms, this means that you go through a tangent at the beginning and at the end of the path P. or. T at the inner circle (with tangent points A. or. B.) moves. The optimal path between the tangential points is the arc of a circle .

There in A. and B. a right angle each PAM or. MBT is present, results from the Pythagorean theorem

The two triangles PAM and TBM are therefore congruent and we denote the angle AMP or. TMB as .

Be = BMA. Then applies = .

It follows with sin = = , so = arcsinthat

= - 2 arcsin
is.

On the other hand is = = .

Insertion of results

The total path then has the length | PA| + + | BT| = 6270, 9319km + 2 . km = 6430, 5033km. The distance saved is thus approx. 497m.

Comment: The calculations in this solution come close to the calculation accuracy of a pocket calculator: In the first part of the exercise, for example, a similarly large number (| MP|) deducted. In this way, many `` valid '' (i.e. not rounded) digits are deleted at the beginning, and as a result only a few digits are really precisely calculated. The calculations in the second part of the exercise are also partly sensitive to rounding for a similar reason. In such cases, it is best to use a "complete formula" to avoid rounding in between. If Dover and Calais were only a few feet apart, that is not enough (on normal pocket calculators) either; then you have to look for other tricks. (For the initiated: With the help of addition theorems, the formula can be converted into a more calculable one.)

exercise 2

Which five-digit numbers, in which no digit is zero, have the property that their last digit divides the number formed from the last two digits, that the number formed from the last two digits divides the number formed from the last three digits, that the number formed from the last three digits divides the number formed from the last four digits and finally that the number formed from the last four digits divides the whole number itself?

Solution:

There are exactly three solutions: 53125, 91125 and 95625.

Proof: be the five-digit number abcde a solution to the task. Then share in particular bcde the number abcde and with it the number abcde - bcde = a0000 = a . 24 . 54. There a should be a digit (not equal to zero), this means that bcde a product of a factor less than or equal to 9 (a factor of a), a power of two 2i With i 4 and a power of five 5j With j 4 must be. Also, according to requirement e 0, therefore bcde not divisible by 10, or in other words: does not contain prime factors 2 at the same time and 5.

Would bcde not divisible by 5, so it would be bcdea . 24 9 . 16 = 144 <1000 contradicting the requirement that too b 0 should apply. So is bcde = . 5j With j 4 and a divider of a, and because 9 . 52 = 225 <1000 is j 3. Because of 8 . 53 = 1000 comes in case j = 3 only the choice = 9 and thus also a = 9 possible. It's 9 . 125 = 1125 a number without zeros, so 91125 is a possible solution. Because of 1125 . 81 = 91125, 125 . 9 = 1125, 25 . 5 = 125 and 5 . 5 = 25, 91125 is actually a solution.

There may be other solutions just in case j = Give 4, so it is then bcde = . 625, where again a divisor of a and must be odd. 1 . 625 = 0625 returns a zero; 3rd . 625 = 1875 leads just like 7 . 625 = 4375 leads to no solution, because obviously 875 is not a divisor of 1875 (or 1000) (not even 75 is a divisor of 875) and a little less clearly 375 is not a divisor of 4375 (because 3 divides 375, but not 4000). One chooses = 5, so is bcde = 55, so must because a0000 = 16 . 54 . a is a also be divisible by 5. There a The only option is to have a digit not equal to zero a = 5. Because of 53125 = (16 + 1) . 3125 (as constructed) and 3125 = 25 . 125, 53125 is another solution.

One finally chooses = 9, so it is bcde = 54 . 9, and so must as well a = 9 can be selected. It's 9 . 625 = 5625, and because of 95625 = (16 + 1) . 5625 and known 625 = 25 . 25, 95625 is a third and final solution to the problem.

A couple of further comments (without evidence): As you may have already seen from the evidence, the requirement that none of the digits should be a zero is an essential requirement. If you allow zeros, you get a number of other solutions.

If you already allow zeros, you can also investigate the problem for any number of digits. You then automatically get an infinite number of solutions, because you can generate more from one solution by inserting any number of zeros between the first and second digits. You can also add any number of zeros at the end.

According to the above, only solutions are interesting in which the first digit is not a zero (this is reasonable in any case), the last digit is also not a zero and in which the second digit is only a zero if there is no solution results if you omit the zero. We want such a solution as essential solution describe. After all, one can show that there are exactly 148 essential solutions. These include 9 single-digit, 41 double-digit, 30 three, 25 four, 17 five, 20 six, 5 seven-digit solutions and the longest one is an eight-digit solution. It is 90703125 and shows that the somewhat cumbersome formulation above for the second digit is necessary for an essential solution, because 703125 is 9 . 57so that the next digit not equal to zero, which has to be nine because no power of ten is divisible by three, only occurs on the 10th7-er-place can stand.

After all, there are 90 essential solutions without zeros, and the largest of them are exactly the three five-digit ones. All 14 four-digit solutions are - this can be shown exactly as in the solution above - with one exception (9225), divisible by 125. After the digits 7 or 9, among the essential solutions without zero, only the trivial solutions 7, 9, 77 and 99 end.

Most of the zeros in a substantial solution have the solutions a000064 with a = 1, 3, 5, 7 or 9.

It would certainly be interesting to examine the problem in other place value systems - but that would speak the scope here, so we leave it to the willing reader to do this himself.

When filing the following system of equations, it was unfortunately punched in the wrong places ...

 x + 4y = 2x =

The missing numbers are still found: , , , but you no longer know which number was in which gap. But Sjaan remembers that there was a clear solution and that x was positive. Can the system of equations be reconstructed from this?

Solution:

Instead of the holes we use variables. The system of equations

 x +  *-4y = a1 (1) 2x + a2y = a3 (2)

should be clearly solvable according to the prerequisite. You would a2 = 8, the left side of the second equation would be twice the left side of the first equation. This means that the system of equations is either not solvable at all (if a3 2a1 is) or that the two equations are equivalent. Then, in truth, you have only changed one equation, namely: x = a1 - 4y, and you can see right away that they are available for any choice of y is solvable and also different xValues. That would be a contradiction to the requirement. Therefore must a2 8 apply.

We now calculate the solutions of the system of equations in general; the transformations are in any case consequences of (if not equivalences to) the previous equations.

First we form 2 . () - ():

 (8 - a2)y = 2a1 - a3 y = .

Dividing by 8 - a2 is generally possible because we are a2 8 had determined.

This in () is used together with the prerequisite that when the task is solved x is positive

 0 < x = a1 - 4y = a1 - .

For a2 = - 4 and also for a2 = - 8 applies 8 - a2 > 0. This further follows from the inequality
 8a1 - a1a2 - 8a1 + 4a3 > 0 - a1a2 + 4a3 > 0 a1a2 - 4a3 < 0.

Accepted, a2 = - 4 leads to a solution. Then
 -4a1 - 4a3 < 0 a1 + a3 > 0.

But this leads to a contradiction because a1 + a3 the sum of -8 and 8, which is zero. So it must apply: a2 = - 8. This results in
 -8a1 - 4a3 < 0 2a1 + a3 > 0 2a1 > - a3.

This inequality is only true for a1 = 8 and a3 = - 4 (and not vice versa).

A sample confirms the solution.

The correct system of equations looks like this:

 x + 4y = + 8 (3) 2x - 8y = - 4 (4)

and has the solutions x = 3, y = .

The seven-person correspondence circle team sits in a circle and passes the time with the following game:

At the beginning everyone has a random number of apples in front of them, but not more than 20. In the middle of the round there is also a very large apple supply. A turn now consists of all players with an uneven number of apples taking an apple from the supply. Then all players give half of their apples to the person sitting next to them on the right at the same time.

The game is over as soon as the supply mountain is used up.

Show that the game never ends if you choose the supply big enough (but finally).

Depending on the starting situation, find the smallest possible supply that is sufficient for the game not to end.

Solution:

Part one: the game never ends.

At the beginning of the game, according to the task, no player has more than 20 apples in front of them. But this also applies after the first half of the move (,,add an apple if your number is odd``) still, because 20 is an even number.

Be now a the number of apples a player has X and be b the number of apples of his left neighbor (after the first half of the turn). Based on the last consideration, the following applies a, b 20. In the second half of the turn the player deals X keeps half of his apples, in particular keeps the (other) half of his own apples, and at the same time receives half of the apples of his left neighbor, so that after the turn he gets exactly + = Owns apples. With the above estimate for a, b follows for this number = 20.

As a result, each player has a maximum of 20 apples in front of him even after the entire turn. This also applies after every further move.

Together, the seven players need a maximum of 7 . 20 = 140 apples; and at the latest when everyone has 20 apples in front of them, no more apples from the supply are needed and the game can go on indefinitely.

Second part: how big does the stock have to be?

This number is very dependent on the given initial situation and is not yet fully known to us either.

We have already seen a first limitation of the number in the first part: In total, never more than 140 apples are needed, namely in the number of apples in front of the players at the beginning and in the supply combined.

We can improve this estimate by choosing the bounds from the first part more precisely: Be x1, x2,..., x7 the number of apples in front of each player at the beginning. Be M. the maximum of these numbers. If M. is straight, so we set m : = M., and otherwise - so if M. is odd - let's bet m : = M. + 1. In both cases, then m an even number and neither player starts with more than m Apples.

There m is again even, the result is, as above, that after the first half of the move no player is more than m Has apples. And even after the second half of the turn, the following applies to the number of apples of each player (with the corresponding designations as above): = m.

So there will be a maximum of 7 . m Apples needed. And there already x1 + x2 +...+ x7 Apples are in front of the players from the start, the supply needs a maximum of 7 . m - (x1 + x2 +...+ x7) To contain apples.

There are many nontrivial examples in which this estimate is sharp: For example, the starting situation (5, 19, 19, 13, 0, 0, 0) after 22 moves results in all players having 20 apples.

However, there are also examples in which the second estimate is not the best either:

If there are only two players who start with 4 or 8 apples, they do not need any additional supply, because: Both numbers are even, and after the first move they have = 6 or = 6 apples, so that after each of the following moves each of the two players has exactly 6 apples and no additional apple is required.

Addition: At the "end" everyone has the same number of apples.

So far we have only shown that only a finite supply is ever necessary. After the task has been set, the game is never over if the supply is large enough. But, as one sender correctly wrote: "The players will stop at some point anyway when they have to go back to work or don't feel like it anymore."

The question arises as to when the game gets boring. Certainly when everyone has the same even number of apples, because then nothing can change. Indeed, this will always happen.We only want to suggest the proof for this: One can show that the following applies: Let the situation be given that not all have the same number of apples. We specifically consider the smallest number of apples that occur in a pile. Then after the next step, either this minimum number has increased or the number of players who have a pile with the minimum number has decreased. This means that the game situation changes significantly with each move, until the situation is reached where all players have the same even number of apples in front of them.

Further remarks.

The game with a maximum of 20 apples ends after 22 rounds at the latest. And the example above with the starting situation (5, 19, 19, 13, 0, 0, 0) gives the maximum required supply: 84 apples.

If you allow more apples at the beginning, you can significantly increase the maximum supply required: The starting situation (0, 97, 99, 93, 99, 99, 60), for example, after 30 rounds leads to the final result that all players have 100 apples, So you need a supply of 153 apples. We don't know whether this is the maximum, our computer is still calculating ...

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