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It is one digit which identifies one from two or more identical subsystems in the overall plant. The prefix number is used in cases when for instance in compass of one power unit appear two or more identical subsystems two boiler plants, two turbosets, etc. Adoption of a. Data character G as well as data character F0 can be omitted in KKS if the further part of designation will remain complete univocal. Process-Related Identification Process-related identification of systems and items of equipment according to their function in mechanical, civil, electrical and control and instrumentation engineering.

System code System code consists of three letters and two digits. No one of these symbols can be omitted. Coding letters and designations of the main groups F1 as given in the Function Key: A Grid and distribution systems B Power transmission and auxiliary power supply C Instrumentation and control systems D Instrumentation and control systems E Conventional fuel supply and residues disposal F Handling of nuclear equipment G Water supply and disposal H Conventional heat generation J Nuclear heat generation K Nuclear auxiliary systems L Steam, water, gas cycles M Main machine sets N Process energy supply for external users e.

Numbering starts anew when one of the preceding code elements changes. System numbering must consist of two digits. Redundant zeros must be written e. Generally valid rules on the use of FN numbering are not expedient.

It should be decided on a case- to-case basis which numbering convention is best suited to the extend and structure of the systems and plants to be identified. Equipment Unit Code Equipment Unit Code consists of two letters, three digits and sometimes one additional letter data character A3 which can be omitted. The Equipment Unit identified in breakdown level 2 belongs always to the System designated in breakdown level 1. The AN number of breakdown level 2 is used to number the items of equipment classified in A1 and A2.

The AN number must consist of three digits. Redundant zeros must be written. AN numbering may be consecutive or grouping. It is a common practice to apply s grouping: in this case the first digit points the kind of equipment unit.

The additional code can be omitted. Further applications are subject to agreement between the parties to the project. The additional code is not an alternative code for the components identified under breakdown level 3. Component Code A component designated in breakdown level 3 is a part of equipment unit identified in breakdown level 2.

The first digit is often used to point the kind sort of component designated in B1 B2. Point of Installation Identification Identification of points of installation of electrical and instrumentation equipment in installation units e.

This is subject to agreement between the parties to the project. Installation Unit Code Serial no. Details of application are subject to agreement between the parties to the project. Installation Space Code Serial no. The data characters in parentheses may be omitted if the code remains unique. Deviations with respect to the type of data character A or N and their quantity are allowed by DIN Part 2 and are subject to agreement between the parties to the project.

There are also cases of deviations consisting in using the dividing symbol twice, and AN, e. Examples Examples of the identification of installation spaces in panels To identify the installation space, draw on front of the panel coordinate system origin - so called grid.

Such established grid after addressing can state the coordinate system. AL According to cubicles, divisions, sections etc.

AK over a separate position code is to be established, e. X and Y zones are outside the grid. A breakdown level 2 is used to identify the installation space. Installation zone X Grid zone Installation zone Y. Location Identification Identification of locations in structures, on floors and in rooms and also of fire areas. Structure Code Serial no. Numbering begins anew for each structure. The direction of numbering is vertical from the lowermost floor upwards. Room Code The room code serves to identify physically separate and fictitious rooms in structures and in the outdoor area.

Physically separate rooms in structures are identified by numbering, grid squares in the outdoor area by coordinates. Fictitious rooms in structures may be identified by numbering or coordinate systems. Room Identification by Numbering Serial no.

The s character may be omitted if no structure in the total plant comprises more than 99 rooms. If this is not the case, all floor numbering codes of the total plant must be written uniformly with three data characters in AN. Room Identification by Coordinates Serial no. The grid sizes and the meaning and type of the data characters A or N are subject to agreement between the parties to the project.

Remark: The outdoor area is designated in the breakdown level 1 by symbols: UZT Notation of Codes The code may be written with or without spacing but in such a way that it cannot be misinterpreted.

The allowable notations are fixed. Their use for any intended purpose is subject to agreement between the parties to the project. Spaced notation promotes the comprehensibility and mnemonic quality of codes. General Notes I - 5. Assignment of System Codes to Mechanical Equipment and Measuring Circuits Equipment units are identified according to the system within the interfaces of which they are located.

Supply systems which serve only one equipment unit are identified on an equal footing as subsystems of the system in which they are installed. Where an equipment unit serves as an interface between two or more systems, its system code is assigned on the originator principle, i. Measuring circuits are identified according to the system within the interfaces of which the sensor is installed.

Special Rules for Mechanical Engineering I - 5. Valves Valves are identified on the equipment unit level by the letters AA regardless of their type, design or actuator. Safety equipment, consisting of safety valves and their associated piping, are assigned to the system to which they are connected.

Drains and vents are identified as far as the final isolation element — including double valves — or as far as the open outlet as part of the drained or vented system originator principle. This applies analogously to supply systems. System isolation valves for measuring circuits are identified on the level of the equipment unit code by the classifying letters AA. Subsequent isolation and pressure equalization valves in the measuring circuit are components of the measuring circuit and therefore classified as KA.

Mechanical Supports Supports may be identified using the system code or the structure code. On breakdown level 2 supports should be identified by BQ. Structure-related identification is expedient where different systems are held by shared supports. System-related identification is expedient where the supported items are positively assignable to one system. Mechanical Service Systems Mechanical service systems which serve more than one main group identified on breakdown level 1 are identified as auxiliary or ancillary systems in separate main groups F1 by means of the following coding letters: G Water supply and disposal K Reactor auxiliary systems Q Auxiliary systems S Ancillary systems Example: User system Supply system Code Designation Code Designation L..

Steam, water, gas cycles Q.. Auxiliary systems M.. Main machine sets common to L. Where several mechanical systems identified differently in data characters F2 and F3 are connected to one mechanical service system, the following code letters are used for the service systems on breakdown level 1 in the data characters concerned as follows: V Lubricant supply system W Sealing fluid supply system X Fluid supply system for control and protection equipment.

Feed water system Sealing fluid supply system for LB. Steam system LW. If several process-related systems are situated in one structure we designate this structure according to the main, most important system.

Where individual structures for which separate structure codes exist are combined to form one structure, that structure is given the code of the most important individual structure. Special structures, ducting structures are designated in main group F1 with coding letter U and in subgroup F3 with following coding letters: X special structures, Y bridge structures, Z ducting structures.

Shared equipment Combined electrical and control and instrumentation equipment Combined electrical and control and instrumentation equipment, with or without a power unit e. If necessary, further grouping can be performed by using FN. Transducer Racks, Supports, Frames This auxiliary equipment is identified on breakdown level 1 using the code for the equipment location structure code , with the exception of auxiliaries for main machine sets and heavy machinery system code.

On breakdown level 2, transducer racks, supports and frames are identified by means of GZ. Every point of this spiral must be a different color from all points of the circle C1. Hence every circle center O with radius smaller than C1 must include a point of different color to those on C1.

Suppose there are n colors. Then by taking successively smaller circles C2, C3, So the result is true. Keeping A fixed and summing over X gives n - 3 cases. The ends of a diagonal AX are connected by r sides and n-r sides. The idea of the upper limit is that its length is less than the sum of the shorter number of sides. Evaluating it is slightly awkward. We consider n odd and n even separately. A2 n and k are relatively prime, so 0, k, 2k, So k, 2k, So in any case r and s have the same color.

By giving i values from 1 to n-2 this establishes that all the numbers have the same color. A3 If i is a power of 2, then all coefficients of Qi are even except the first and last. We use induction on in. So suppose it is true for smaller values of in. So we have established the result for in. B1 Suppose we have a set of at least 3. So each number can be written as p1r We classify each ri as even or odd.

That gives 2n possibilities. Remove those two and look at the remaining numbers. We must find two pairs with the same classification. The product of these four numbers is now a fourth power.

The key is to find the 4th power in two stages, by first finding lots of squares. If we try to go directly to a 4th power, this type of argument does not work we certainly need more than 5 numbers to be sure of finding four which sum to 0 mod 4, and 59 is far too big. Let X be the point of intersection. The argument depends slightly on how the points are arranged. Now XM. Also Sn x has non-negative coefficients, so it is strictly increasing in the range [0,1].

Thus an is an increasing sequence and bn is a decreasing sequence with all an less than all bn. So we can certainly find at least one point x1 which is greater than all the an and less than all the bn. So the resulting series xn satisfies all the required conditions. It remains to consider uniqueness.

Suppose that there is an x1 satisfying the conditions given. We show uniqueness by showing that bn - an tends to zero as n tends to infinity.

IMO A1 Consider residues mod A perfect square must be 0, 1, 4 or 9 mod However, if d is 1 or 13, then 13d - 1 is not one of these values. If d is 5 or 9, then 5d - 1 is not one of these values. So we cannot have all three of 2d - 1, 5d - 1, 13d - 1 perfect squares. Alternative solution from Marco Dalai Suppose 2d-1, 5d-1, 13d-1 are all squares. Squares mod 4 must be 0 or 1, considering 2d-1, so d must be odd.

So b must be even, so k must be even. A2 The product of three successive rotations about the three vertices of a triangle must be a translation see below. It is now easy to show that it can only be the identity if the triangle is equilateral. So A1A2A2 is equilateral. Note in passing that it is not sufficient for the triangle to be equilateral. We also have to take the rotations in the right order. If we move around the vertices the opposite way, then we get a net translation.

It remains to show that the three rotations give a translation. Define rectangular coordinates x, y by taking A1 to be the origin and A2 to be a, b.

Let A3 be c, d. Thus after the three rotations we will end up with a linear combination of x's, y's, a's, b's, c's and d's for each coordinate. But the linear combination of x's and y's must be just x for the x-coordinate and y for the y-coordinate, since three successive degree rotations about the same point is the identity. It makes things slightly easier to take the triangle to be 0, 0 , 1, 0 , a, b. So the third vertex must make the triangle equilateral and it must be on the correct side of the line joining the other two.

This approach avoids the need for the argument in the first paragraph above, but is rather harder work. Then the operation reduces S, but S is a greater than zero, so the process must terminate in a finite number of steps. So see that S is reduced, we can simply write out all the terms. S is not the only expression we can use. Take coordinates with origin at A, x-axis as AB and y-axis directed inside the n-gon. Let YZA be t. Let the coordinates of X be x, y. Thus the locus of X is a star formed of n lines segments emanating from O.

We have thus established that if a function f meets the conditions then it must be defined as above. It remains to prove that with this definition f does meet the conditions. B3 Answer: yes. We prove the result by induction on the number n of points. Suppose it is true for all numbers less than n.

Pick an arbitrary point P and color it red. Now take a point in the same row and color it white. Take a point in the same column as the new point and color it red.

Continue until either you run out of eligible points or you pick a point in the same column as P. The process must terminate because there are only finitely many points. Suppose the last point picked is Q. Let S be the set of points picked. If Q is in the same column as P, then it is colored white because the "same row" points are all white, and the "same column" points are all red. Now every row and column contains an equal number of red points of S and of white points of S.

By induction we can color the points excluding those in S, then the difference between the numbers of red and white points in each row and column will be unaffected by adding the points in S and so we will have a coloring for the whole set. This completes the induction for the case where Q is in the same column as P. In other words, pick a point in the same column as P and color it white. Then pick a point in the same row as the new point and color it red and so on. Continue until either you run out of eligible points or you pick a point to pair with Q.

If Q was picked as being in the same row as its predecessor, this means a point in the same column as Q; if Q was picked as being in the same column as its predecessor, this means a point in the same row as Q.

Again the process must terminate. Suppose the last point picked is R. Let S be the set of all points picked. If R pairs with Q, then we can complete the coloring by induction as before.

Suppose S does not pair with Q. Then there is a line meaning a row or column containing Q and no uncolored points. There is also a line containing R and no uncolored points.

These two lines have an excess of one red or one white. All other lines contain equal number of red and white points of S. Now color the points outside S by induction. This gives a coloring for the whole set, because no line with a color excess in S has any points outside S.

So we have completed the induction. This follows immediately from the law of inclusion and exclusion: let Ni be the number which fix i, Nij the number which fix i and j, and so on.

Then N0, the number with no fixed points, is n! We use induction on n. Suppose it is true for n. Comment This is a plodding solution. If you happen to know the result for no fixed points which many people do , then it is essentially a routine induction. Clearly, if we fix x, then there are n-1! So the total count is n!. But if we count the number of permutations s with exactly k fixed points, then we get the sum in the question. A3 This is an application of the pigeon-hole principle. Assume first that all xi are non-negative.

Divide this into kn-1 equal subintervals. Two values must lie in the same subinterval. Take their difference. Its coefficients are the required ai. Finally, if any xi are negative, solve for the absolute values and then flip signs in the ai.

Comment This solution is due to Gerhard Woeginger, email 24 Aug We show this leads to a contradiction. Suppose they have r1 and r2 respectively. Hence if m has a different residue r mod k, then f m cannot have residue r1 or r2. Thus the residues form pairs, so that if a number is congruent to a particular residue, then f of the number is congruent to the pair of the residue. But this is impossible for k odd. A better solution by Sawa Pavlov is as follows Let N be the set of non-negative integers.

Obviously B is a subset of f N and if k belongs to B, then it does not belong to f f N since f is injective. Similarly, a member of f f N cannot belong to B. Clearly A and B are disjoint. We show that the distance between any two points is irrational and that the triangle determined by any 3 points has non-zero rational area.

Hence its square root is irrational. So we have a prime p such that p divides N, but p2 does not divide N. Now p must divide r, hence p2 divides r2 and so p divides s2. Hence p divides s. Hence non-squares have irrational square roots. Let B be the point b, a2 , C the point c, a2 , and D the point c, b2.

Hence c is also. Let BC meet the small circle again at Q. Let O be the center of the circles. If we shrink the small circle by a factor 2 with P as center, then Q moves to M, and hence the locus of M is the circle diameter OP. A2 Answer: n even. Each of the 2n elements of Ai belongs to at least one other Aj because of iii. But given another Aj it cannot contain more than one element of Ai because of ii. There are just 2n other Aj available, so each must contain exactly one element of Ai.

Hence we can strengthen iii to every element of B belongs to exactly two of the As. This shows that the arrangement is essentially unique. We may call the element of B which belongs to Ai and Aj i,j. Then Ai contains the 2n elements i, j with j not i. Hence n must be even. This clearly has the required property. My original solution was a pedestrian induction: We show by induction that a labeling is always possible for n even. For example, we may assign 0 to 1,2 , 1,3 , 2,4 , 3,5 , 4,5.

Now suppose we have a labeling for n. A3 Answer: So we also have all 32 numbers in the range to except for and , giving another 30, or 92 in total. It remains to prove the assertions above. Again we use induction.

So suppose it is true for 1, 2, We use induction on r. So assume it is true for smaller values. Hence f x tends to plus infinity as x tends to n from above. Similarly, f x tends to minus infinity as x tends to n from below. Hence f x decreases monotonically from plus infinity to zero over the range [70, infinity]. The sum of the roots is minus the coefficient of x69 divided by the coefficient of x We do this indirectly.

Hence it is also the incenter of AK'X. Hence we can identify K and K', and L and L'. Let M be the midpoint of BC. Finally, we may notice that this can be used to go down as well as up. So a2 must divide k. Note that jumping straight to the minimal without the infinite descent avoids some of the verification needed in the infinite descent.

Then abc - 1 is even, so all a, b, c are odd. So there are no solutions. Take any x. Leave the diagonals of each square uncolored, color the remaining edges of R red and the remaining edges of B blue.

Color blue all the edges from the ninth point X to the red square, and red all the edges from X to the blue square. Color RiBj red if i and j have the same parity and blue otherwise. Clearly X is not the vertex of a monochrome square, because if XY and XZ are the same color then, YZ is either uncolored or the opposite color.

There are three uncolored edges. Take a point on each of the uncolored edges. The edges between the remaining 6 points must all be colored. Take one of these, X. Similarly, for BC and CA.

Let L' be the line YZ. The locus is the open ray from Z along L' on the opposite side to Y. Let C' touch QR at Y'. We focus on the QORO'.

Hence P lies on the open ray as claimed. Gerhard only found this after Theo Koupelis, University of Wisconsin, Marathon had already supplied the following analytic solution. So we have two simultaneous equations for A and B. So we have shown that the locus is a subset of the line YZ. In the general case, take a horizontal constant z plane dividing S into two non-empty parts T and U. Suppose we could express N as a sum of N - 13 squares. Let the number of 4s be a, the number of 9s be b and so on.

Hence c, d, But neither 13 nor 8 is a multiple of 3, so there are no solutions. A little experimentation shows that the problem is getting started. Most squares cannot be expressed as a sum of two squares. Then grouping four 4s into a 16 gives us 38, 35, Grouping four 16s into a 64 gives us 8 and 5. We now show how to use the expressions for to derive further N.

For any N, the grouping technique gives us the high k. Continuing in this way gives N3 as a sum of 1 thru k1k2 squares. So when combined with the top down grouping we get a complete set of expressions for N3. We show that all the a's are divisible by 3 and use that to establish a contradiction.

First, r and s must be greater than 1. We can now proceed by induction. Assume a1, So considering the coefficients of x, x2, Hence the factorization is not possible. In other words the tangents to the circumcircles at D are perpendicular.

Hence, by symmetry reflecting in the line of centers the tangents at C are perpendicular. Theo Koupelis, University of Wisconsin, Marathon provided a similar solution about 10 minutes later!

Marcin Mazur, University of Illinois at Urbana-Champaign provided the first solution I received about 10 minutes earlier!

A3 We show first that the game can end with only one piece if n is not a multiple of 3. X The key technique is the following three moves which can be used to wipe out three adjacent pieces on the border provided there are pieces behind them: X X X X X. We now show how to wipe out the rectangle.

First, we change the 2 x 2 rectangle at one end into a single piece alongside the now 2 x 3n rectangle: X X. Finally, we use a parity argument to show that if n is a multiple of 3, then the square side n cannot be reduced to a single piece.

Let suppose that the single piece is on a red square. Let A be the number of moves onto a red square, B the number of moves onto a white square and C the number of moves onto a blue square. A move onto a red square increases the number of pieces on red squares by 1, reduces the number of pieces on white squares by 1, and reduces the number of pieces on blue squares by 1.

Then there are initially m pieces on red squares, m on white and m on blue. But the number of moves of each type must be integral, so it is not possible to reduce the number of pieces to one if n is a multiple of 3. B1 The length of an altitude is twice the area divided by the length of the corresponding side. The claim follows from the following lemma. Proof: let H be the foot of the perpendicular from F to DE.

One of D and E must lie on the opposite side of Y to H. Suppose it is D. Consider the three sides of ABX. It remains to consider X outside ABC.

If the former, then it is longer than the perpendicular from B to k, which equals the altitude from A to BO. If the latter, then it is longer than the altitude from B to AO. We show that it is between 0 and 1. This satisfies the required conditions. We have to show that every n has a unique expression of this type. We show first by induction that it has at least one expression of this type. Then by induction we have an expression for n - ur. So adding ur to the expression for n - ur gives us an expression of the required type for n, which completes the induction.

Hence it is true for all r. Now we can prove that the expression for n is unique. So the expression for n must be unique and the induction is complete. It remains to show that f satisfies the required conditions. B3 a The process cannot terminate, because before the last move a single lamp would have been on. There are only finitely many states each lamp is on or off and the next move can be at one of finitely many lamps , hence the process must repeat.

The outcome of each step is uniquely determined by the state, so either the process moves around a single large loop, or there is an initial sequence of steps as far as state k and then the process goes around a loop back to k. However, the latter is not possible because then state k would have had two different precursors. But a state has only one possible precursor which can be found by toggling the lamp at the current position if the previous lamp is on and then moving the position back one.

Hence the process must move around a single large loop, and hence it must return to the initial state. Hence for each of its first n-1 moves lamp n-1 is not toggled and it retains its initial state. After each lamp has had n-1 moves, all of lamps 1 to n-2 are off. Finally over the next n-1 moves, lamps 1 to n-2 are turned on, so that all the lamps are on. For the first n-1 moves of each lamp the n left-hand and the n right-hand lamps are effectively insulated. Lamps n-1 and 2n-1 remain on.

Lamp 2n-1 being on means that lamps 0 to n-2 are in just the same situation that they would be with a set of only n lamps. Similarly, lamp n-1 being on means that lamps n to 2n-2 are in the same situation that they would be with a set of only n lamps. Hence after each lamp has had n-1 moves, all the lamps are off except for n-1 and 2n In the next n moves lamps 1 to n-2 are turned on, lamp n-1 is turned off, lamps n to 2n-2 remain off, and lamp 2n-1 remains on.

For the next n-1 moves for each lamp, lamp n-1 is not toggled, so it remains off. Hence all of n to 2n-2 also remain off and 2n-1 remains on. Lamps 0 to n-2 go through the same sequence as for a set of n lamps. Hence after these n-1 moves for each lamp, all the lamps are off, except for 2n Finally, over the next 2n-1 moves, lamps 0 to 2n-2 are turned on. This completes the induction. Counting moves, we see that there are n-1 sets of n moves, followed by n-1 moves, a total of n2 - 1.

Lamp 0 is off and lamp 1 is on. Suppose it is true for m. Lamp 2 has the same state as lamp 0 in the reduced case and both toggle since their predecessor lamps are on.

Hence lamps 3 to n - 1 behave the same as lamps 1 to n - 3 in the reduced case. That means that lamp n - 1 remains off. So after n - 2 moves for each lamp all lamps are off except lamp 1. In the next two moves nothing happens, then in the following n - 1 moves lamps 2 to n - 1 and lamp 0 are turned on.

So OQ is perpendicular to EF. A3 2, 4, So f k increases without limit, and since it only moves up in increments of 1, it never skips a number. We start by checking small values of n. Thanks to Gerhard Woeginger for pointing out the error in the original solution and supplying this solution.

Let A consists of all products of n distinct primes such that the smallest is greater than pn. For example: all primes except 2 are in A; 21 is not in A because it is a product of two distinct primes and the smallest is greater than 3.

Then pi1pi But pi2pi But both numbers are products of n distinct elements of S. Hence Q and Q' coincide. Let the expression given be E. Thanks to Gerhard Woeginger for pointing out that the original solution was wrong. Solution The first point to notice is that if no arrangement is possible for n, then no arrangement is possible for any higher integer.

We show that we cannot have two r's equal whether or not the 4 points form a convex pentagon. If they are on opposite sides, then the midpoint of A4A5 must lie on A1A2. The same argument can be applied to A1 and A3, and to A2 and A3. So we have established a contradiction. In particular, this shows that the 5 points cannot form a convex pentagon. Without loss of generality, we may take it to be A1A2A3A4. Again without loss of generality we may take it to be the latter, so that A1A2A5A4 is also a convex quadrilateral.

The final case is the convex hull a triangle, which we may suppose to be A1A2A3. B1 Answer For an even number of moves we can start with an arbitrary x0 and get back to it. Repeating brings you back to x0 after 2n moves. However, is odd! We show that this is the largest possible x0. Suppose we have a halvings followed by an inverse followed by b halvings followed by an inverse. All the a, b, The presence of equilateral triangles and quadrilaterals suggests using Ptolemy's inequality.

But we do not get quite what we want. We can repeat this process and get p sets in all. The sum of the elements in the set is increased by r each time. So, since p is prime, the sums must form a complete set of residues mod p.

In particular, they must all be distinct and hence all the subsets must be different. Suppose the elements in this intersection sum to k mod p. In other words, exactly one member of each group will have the sum of all its elements divisible by p. If r is divisible by 2, then a and b are both even or both odd. But that means that we can only access the black squares or the white squares assuming the rectangle is colored like a chessboard.

The two corners are of opposite color, so the task cannot be done. All squares are congruent to 0 or 1 mod 3, so if r is divisible by 3, then a and b must both be multiples of 3.

That means that if the starting square has coordinates 0,0 , we can only move to squares of the form 3m,3n. The required destination is 19,0 which is not of this form, so the task cannot be done. After some fiddling we find: 0,0 to 8,3 to 16,6 to 8,9 to 11,1 to 19,4 to 11,7 to 19,10 to 16,2 to 8,5 to 16,8 to 19,0.

As before, assume we start at 0,0. A good deal of fiddling around fails to find a solution, so we look for reasons why one is impossible. Call moves which change y by 4 "toggle" moves. Toggle moves must toggle us in and out of the strip.

Non-toggle moves cannot be made if we are in the strip and keep us out of it if we are out of it. Toggle moves also change the parity of the x-coordinate, whereas non-toggle moves do not. Now we start and finish out of the strip, so we need an even number of toggle moves. On the other hand, we start with even x and end with odd x, so we need an odd number of toggle moves. Hence the task is impossible.

A2 We need two general results: the angle bisector theorem; and the result about the feet of the perpendiculars from a general point inside a triangle. The second is not so well-known. So f n is a fixed point. Let k be the smallest non-zero fixed point. If k does not exist, then f n is zero for all n, which is a possible solution. Hence the fixed points are precisely the multiples of k.

But f n is a fixed point for any n, so f n is a multiple of k for any n. Let us take n1, n2, We can check that this satisfies the functional equation. So this is a solution and hence the most general solution. Hence m and n must both be divisible by This solution is straightforward, but something of a slog - all the residues have to be calculated. Hence m and n are both divisible by Advent Rising USA.

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