answer：Alright, so I've got this problem where there are n points on a circle, and two players, Petya and Vasya, take turns drawing chords between non-adjacent points. The rule is that any two chords can only intersect at their endpoints, meaning they can't cross each other inside the circle. The player who can't make a move loses. I need to figure out for which values of n (where n is at least 5) Petya can always win, no matter how Vasya plays.Okay, let me start by understanding the game better. It's a two-player game with perfect information, meaning both players know the entire state of the game at all times. The players alternate turns, starting with Petya. Each move consists of drawing a chord between two non-adjacent points. The chords can't intersect except at their endpoints, so essentially, each chord divides the circle into two separate regions, and any subsequent chords must lie entirely within one of those regions or not cross existing chords.This reminds me a bit of the game of Nim or other impartial games where players take turns making moves that affect the game state, and the last player to make a move wins. But in this case, the losing condition is being unable to make a move, so it's similar but not exactly the same.Let me think about small values of n to see if I can spot a pattern.Starting with n=5. There are 5 points on a circle. The number of non-adjacent chords is limited. Each chord connects two points that aren't next to each other. In a pentagon, each point has two adjacent points, so each point can connect to two non-adjacent points. But wait, in a pentagon, the chords are actually the diagonals, right? So each chord is a diagonal.Wait, but in a pentagon, any two diagonals intersect unless they share a common endpoint. So if Petya draws a chord, then Vasya can draw another chord that doesn't intersect it only if they share a common endpoint or are non-crossing.Hmm, maybe I should visualize this. Let me draw a circle with 5 points labeled A, B, C, D, E. If Petya draws a chord from A to C, then Vasya can draw a chord from B to D or from C to E or from D to A, but wait, some of these might intersect.Wait, actually, in a pentagon, any two diagonals that don't share a common vertex will intersect inside the circle. So if Petya draws AC, then Vasya cannot draw BD or CE without intersecting AC. So Vasya's only options are chords that share a common vertex with AC, like CB or CD or AE or something?Wait, no. Let me think again. If Petya draws AC, then the circle is divided into two arcs: one from A to C and another from C back to A. Any chord drawn by Vasya must lie entirely within one of these arcs without crossing AC.So, in the arc from A to C, the points are A, B, C. So the only non-adjacent points in this arc are A and C, but that's already taken. So Vasya can't draw any chord in this arc. Similarly, in the arc from C back to A, the points are C, D, E, A. So the non-adjacent points here are C and D, D and E, E and A. But wait, C and D are adjacent, D and E are adjacent, E and A are adjacent. So actually, in this arc, there are no non-adjacent points to draw a chord.Wait, that can't be right. If Petya draws AC, then the remaining points are B, D, E. Wait, no, the points are still A, B, C, D, E. Drawing AC divides the circle into two arcs: one containing B and the other containing D and E.Wait, maybe I'm overcomplicating this. Let me think of it as a graph. Each chord is an edge connecting two non-adjacent vertices, and the game is about building a non-crossing graph where no two edges cross except at endpoints.So, in graph theory terms, this is equivalent to building a non-crossing chord diagram, and the game ends when no more chords can be added without crossing existing ones.So, the game is about building a non-crossing matching on the circle, with players taking turns adding chords.In this case, the game is similar to the game of Kayles or similar impartial games.I recall that in such games, the key is often to consider the Grundy numbers or Nimbers for each position, but I'm not sure if that's the right approach here.Alternatively, maybe I can think in terms of symmetry. If n is even, perhaps Petya can mirror Vasya's moves, ensuring that Petya always has a move. If n is odd, maybe Vasya can disrupt this symmetry.Wait, let's test this idea with n=5.If n=5, Petya draws a chord, say AC. Then Vasya can draw a chord that doesn't cross AC. But in a pentagon, any chord that doesn't cross AC must share a vertex with AC, right? Because otherwise, it would cross.Wait, no. If Petya draws AC, then Vasya can draw BD, but BD crosses AC. So Vasya can't do that. Alternatively, Vasya can draw CE, but CE would cross AC as well. Wait, is that true?Wait, in a pentagon, the chords AC and BD do cross each other inside the circle. Similarly, AC and CE also cross. So actually, after Petya draws AC, Vasya cannot draw any chord without crossing AC. So Vasya loses immediately. But that can't be right because n=5 is odd, and the problem states n >=5.Wait, maybe I'm making a mistake here. Let me think again.In a pentagon, if Petya draws AC, then the remaining points are B, D, E. Wait, no, the points are still all there, but the chords can't cross AC.So, in the arc from A to C, the points are A, B, C. The only non-adjacent points are A and C, which is already taken. In the arc from C to A, the points are C, D, E, A. The non-adjacent points here are C and E, D and A, but CD and DE and EA are adjacent.Wait, so can Vasya draw CE? CE doesn't cross AC because they share point C. Similarly, Vasya can draw CE. Wait, but CE is a chord from C to E, which doesn't cross AC because they share point C. So Vasya can draw CE.Similarly, Vasya can also draw BD, but BD would cross AC, so that's not allowed. So Vasya can only draw CE or maybe something else.Wait, let me think. After Petya draws AC, the circle is divided into two arcs: one containing B and the other containing D and E.In the arc containing B, the only points are A, B, C. So the only possible chord is AB or BC, but those are adjacent, so no non-adjacent chords.In the arc containing D and E, the points are C, D, E, A. So the non-adjacent pairs are C and E, D and A.So Vasya can draw CE or DA.Wait, DA is a chord from D to A, which doesn't cross AC because they share point A.So Vasya can choose to draw either CE or DA.Suppose Vasya draws CE. Then the circle is now divided into smaller arcs. Let me see.After AC and CE are drawn, the circle is divided into arcs: A to C, C to E, E to A.In the arc A to C, points are A, B, C. No non-adjacent chords.In the arc C to E, points are C, D, E. Non-adjacent chords: C and E is already taken, so no.In the arc E to A, points are E, A. Only two points, so no non-adjacent chords.Wait, so after Vasya draws CE, there are no more moves left. So Petya cannot move and loses. But that contradicts my earlier thought that Petya can win when n is even.Wait, but n=5 is odd, so maybe Petya cannot win. Hmm.Wait, maybe I'm not considering all possibilities. Let me try again.If Petya starts by drawing AC, Vasya can respond by drawing CE, leaving no moves for Petya. Alternatively, Vasya could draw DA instead of CE.If Vasya draws DA, then the circle is divided into arcs: A to D, D to C, C to A.In the arc A to D, points are A, B, C, D. Non-adjacent chords: A and C is already taken, so A and D is adjacent, B and D is non-adjacent.Wait, B and D are non-adjacent in the arc A to D. So Petya can draw BD.But BD would cross AC, right? Because AC is from A to C, and BD is from B to D, which crosses AC inside the circle.Wait, but the rule is that chords can only intersect at their endpoints. So BD cannot cross AC, meaning BD cannot be drawn if AC is already there.Wait, but in this case, after AC and DA are drawn, is BD crossing AC? Let me visualize.If AC is drawn, and then DA is drawn, then BD would start at B, go inside the circle, and end at D. But AC is from A to C, so BD would cross AC somewhere in the middle. Therefore, BD is not allowed because it crosses AC.So, after Vasya draws DA, Petya cannot draw BD because it would cross AC. Are there any other chords Petya can draw?In the arc A to D, points are A, B, C, D. The non-adjacent pairs are A and C (taken), B and D (crosses AC), so no.In the arc D to C, points are D, C. Only two points, so no.In the arc C to A, points are C, E, A. Non-adjacent pairs: C and E (if not taken), E and A (adjacent). So if Vasya drew DA, then CE is still available.Wait, no, CE is in the arc C to A, which is separate from the arc A to D.Wait, maybe I'm getting confused. Let me think step by step.After Petya draws AC, the circle is divided into two arcs: one containing B and the other containing D and E.Vasya can choose to draw either CE or DA.If Vasya draws CE, then the circle is divided into three arcs: A to C, C to E, and E to A.In each of these arcs, there are no non-adjacent points left, so Petya cannot move and loses.If Vasya draws DA instead, then the circle is divided into three arcs: A to D, D to C, and C to A.In the arc A to D, points are A, B, C, D. The non-adjacent pairs are A and C (taken), B and D (crosses AC, so not allowed). So no moves.In the arc D to C, points are D, C. No moves.In the arc C to A, points are C, E, A. The non-adjacent pair is C and E, which Vasya could have drawn instead.Wait, so if Vasya draws DA, then Petya can draw CE, right?Because CE is in the arc C to A, which doesn't interfere with DA or AC.So after Vasya draws DA, Petya can draw CE, leaving no moves for Vasya.Wait, so depending on Vasya's choice, the outcome changes.If Vasya draws CE, Petya loses.If Vasya draws DA, Petya can draw CE and win.But Vasya, being a good player, would choose the move that makes Petya lose, which is drawing CE.Therefore, in n=5, Vasya can force a win by responding to Petya's initial move with CE, leaving Petya without a move.So, for n=5, Vasya can win.Wait, but the problem states n >=5, so maybe n=5 is a losing position for Petya.Let me try n=6.n=6 is even. Let's see.Petya starts by drawing a chord, say AC.This divides the circle into two arcs: one containing B and the other containing D, E, F.In the arc containing B, points are A, B, C. Non-adjacent pairs: A and C is taken, so no moves.In the arc containing D, E, F, points are C, D, E, F, A.Wait, no, after drawing AC, the circle is divided into two arcs: one from A to C and the other from C back to A.In the arc from A to C, points are A, B, C. Non-adjacent pairs: A and C is taken, so no moves.In the arc from C back to A, points are C, D, E, F, A.The non-adjacent pairs here are C and E, D and F, E and A, etc.So Vasya can choose to draw a chord in this arc.Suppose Vasya draws CE.Now, the circle is divided into smaller arcs.After AC and CE are drawn, the circle is divided into arcs: A to C, C to E, E to A.In each of these arcs, we have:- A to C: A, B, C. No moves.- C to E: C, D, E. Non-adjacent pairs: C and E is taken, D and E is adjacent, so no moves.- E to A: E, F, A. Non-adjacent pairs: E and A is adjacent, F and A is non-adjacent.Wait, F and A are non-adjacent in the arc E to A.So Petya can draw FA.But FA would cross AC? Wait, FA is from F to A, which is in the arc E to A. AC is from A to C, so FA doesn't cross AC because they share point A.So Petya can draw FA.After FA is drawn, the circle is divided into arcs: A to F, F to E, E to A.In each of these:- A to F: A, B, C, D, E, F. Wait, no, after FA is drawn, the arc A to F contains points A, B, C, D, E, F, but FA is a chord, so it divides the arc into two: A to F and F to A.Wait, I'm getting confused again.Let me try to think differently. Maybe instead of focusing on specific moves, I should consider the parity of n.If n is even, Petya can always mirror Vasya's moves across the initial chord, ensuring that Petya always has a move.Wait, how?If n is even, say n=6, Petya can draw a diameter, dividing the circle into two equal halves. Then, whatever chord Vasya draws in one half, Petya can mirror it in the other half. This ensures that Petya always has a move as long as Vasya does.But wait, in n=6, if Petya draws a diameter, say AD, then the circle is divided into two triangles: A, B, C and A, D, E, F.Wait, no, n=6, so points are A, B, C, D, E, F.Drawing AD divides the circle into two arcs: A to D and D to A.In the arc A to D, points are A, B, C, D.In the arc D to A, points are D, E, F, A.So, if Vasya draws a chord in one arc, Petya can mirror it in the other arc.For example, if Vasya draws BE in the arc A to D, Petya can draw CF in the arc D to A.Wait, but BE and CF are both non-adjacent chords, and they don't cross each other because they're in separate arcs.This mirroring strategy ensures that Petya always has a move as long as Vasya does, meaning that Petya can always force a win when n is even.But in n=5, which is odd, this mirroring strategy doesn't work because after Petya draws a chord, the remaining arcs are asymmetric, and Vasya can disrupt the symmetry.So, perhaps the key is that when n is even, Petya can use a mirroring strategy to ensure victory, while when n is odd, Vasya can disrupt this strategy and force Petya to lose.Let me test this with n=6.Petya draws AD, dividing the circle into two arcs: A to D and D to A.Vasya draws BE in the arc A to D.Petya mirrors by drawing CF in the arc D to A.Now, the circle has chords AD, BE, CF.Vasya's next move could be to draw, say, BF in the arc D to A.But wait, BF would cross CF, so that's not allowed.Alternatively, Vasya could draw CD in the arc A to D.But CD is adjacent to C and D, so it's not a non-adjacent chord.Wait, in the arc A to D, points are A, B, C, D.Non-adjacent pairs are A and C, B and D.A and C is non-adjacent, but if Vasya draws AC, that's allowed.Wait, but AC is in the arc A to D.Wait, no, AC is from A to C, which is in the arc A to D.But AC is a chord that doesn't cross AD.So Vasya can draw AC.Then Petya can mirror by drawing DF in the arc D to A.Wait, DF is from D to F, which is non-adjacent.So Petya draws DF.Now, the circle has chords AD, BE, CF, AC, DF.Vasya's next move could be to draw BD in the arc A to D.But BD is non-adjacent, so Vasya draws BD.Then Petya mirrors by drawing EF in the arc D to A.Now, the circle has chords AD, BE, CF, AC, DF, BD, EF.At this point, are there any moves left?In the arc A to D, points are A, B, C, D.Chords drawn: AC, BD, BE.Wait, BE is from B to E, which is in the arc D to A.Wait, no, BE is in the arc A to D.Wait, I'm getting confused again.Let me try to list all possible non-adjacent chords in n=6.In a hexagon, the non-adjacent chords are the ones that skip one or more points.So, for n=6, the possible non-adjacent chords are:- AC, AD, AE- BD, BE, BF- CE, CF, CA- DF, DA, DBWait, no, in a hexagon, each point has three non-adjacent points: skipping one, two, or three points.But chords that skip three points are diameters.So, in n=6, the non-adjacent chords are:From A: AC, AD, AEFrom B: BD, BE, BFFrom C: CE, CF, CAFrom D: DF, DA, DBFrom E: EA, EB, ECFrom F: FC, FD, FEBut many of these are duplicates.So, the unique non-adjacent chords are:AC, AD, AE, BD, BE, BF, CE, CF, DF, EA, EB, EC, FC, FD, FE.Wait, but actually, in a hexagon, the non-adjacent chords are those that are not edges of the hexagon. So, each vertex connects to three non-adjacent vertices: the ones two steps away, three steps away, etc.But in a hexagon, the chords can be of length 2, 3, or 4, but since it's a circle, length 4 is the same as length 2 in the other direction.So, the non-adjacent chords are those of length 2 and 3.So, from each vertex, there are two non-adjacent chords of length 2 and one of length 3 (the diameter).So, in total, for n=6, there are 6 diameters and 12 chords of length 2, but since each chord is counted twice, the total number of non-adjacent chords is 6 + 6 = 12.Wait, no, for n=6, the number of non-adjacent chords is C(n,2) - n = 15 - 6 = 9. Wait, that's not right.Wait, C(n,2) is the total number of chords, including adjacent ones. So, for n=6, C(6,2)=15. The number of adjacent chords is 6 (the edges of the hexagon). So, the number of non-adjacent chords is 15 - 6 = 9.So, there are 9 non-adjacent chords in a hexagon.Let me list them:From A: AC, AD, AEFrom B: BD, BE, BFFrom C: CE, CFFrom D: DFFrom E: EAFrom F: FBWait, that's 3 + 3 + 2 + 1 + 1 + 1 = 11, which is more than 9. I must be double-counting.Actually, each chord is counted twice, once from each endpoint. So, the unique non-adjacent chords are:AC, AD, AE, BD, BE, BF, CE, CF, DF.That's 9 unique chords.So, in n=6, there are 9 non-adjacent chords.Now, back to the game.Petya starts by drawing AD, a diameter.This divides the circle into two arcs: A to D and D to A.In the arc A to D, points are A, B, C, D.Non-adjacent chords here are AC, BD.In the arc D to A, points are D, E, F, A.Non-adjacent chords here are DF, EA.So, Vasya can choose to draw either AC, BD, DF, or EA.Suppose Vasya draws AC.Then, Petya can mirror by drawing DF.Now, the circle has chords AD, AC, DF.Vasya's next move could be to draw BD in the arc A to D.Petya mirrors by drawing EA in the arc D to A.Now, the circle has chords AD, AC, DF, BD, EA.Vasya's next move could be to draw CE in the arc A to D.But CE is from C to E, which is in the arc D to A, right?Wait, no, CE is from C to E, which spans across the diameter AD.Wait, does CE cross AD? CE goes from C to E, which is on the opposite side of the circle from C.Wait, in a hexagon, CE would cross AD if drawn.But the rule is that chords cannot cross except at endpoints.So, CE cannot be drawn because it would cross AD.Therefore, Vasya cannot draw CE.Alternatively, Vasya could draw CF in the arc D to A.But CF is from C to F, which is in the arc D to A.Wait, C is in the arc A to D, and F is in the arc D to A.So, CF would cross AD, right? Because it goes from C to F, passing through the diameter AD.Therefore, CF cannot be drawn.Similarly, BF is from B to F, which would cross AD.So, Vasya cannot draw BF.Wait, so after Petya draws EA, the remaining non-adjacent chords are CE, CF, BF, but all of these cross AD.Therefore, Vasya cannot make a move and loses.Wait, but that contradicts my earlier thought that Petya can mirror Vasya's moves.Wait, maybe I made a mistake in the mirroring strategy.Alternatively, maybe the mirroring strategy works differently.Wait, perhaps Petya should not draw a diameter initially but instead draw a chord that doesn't split the circle into two equal halves.Wait, but in n=6, drawing a diameter is the most symmetric move.Alternatively, maybe Petya should draw a chord that splits the circle into two regions with an even number of points each, allowing for mirroring.Wait, but in n=6, drawing a diameter splits it into two regions of 3 points each, which is odd.Wait, maybe that's the problem.Wait, if n is even, but the regions after the first move are odd, then mirroring might not work.Wait, maybe I need to reconsider.Perhaps the key is not just the parity of n, but also the structure of the game.I recall that in such games, the first player can win if the total number of available moves is odd, and the second player can win if it's even.But in this case, the number of non-adjacent chords is C(n,2) - n = n(n-1)/2 - n = n(n-3)/2.For n=5, that's 5(2)/2=5.For n=6, that's 6(3)/2=9.So, for n=5, there are 5 non-adjacent chords.For n=6, there are 9.So, the total number of moves possible is n(n-3)/2.If this number is odd, the first player can win; if even, the second player can win.Wait, for n=5, 5 is odd, so Petya would win.But earlier, I thought that in n=5, Vasya can win.Hmm, maybe this approach is incorrect.Alternatively, perhaps the game is similar to the game of Nim with heaps, where each region is a heap, and players can make moves in one heap at a time.In this case, each chord divides the circle into regions, and each region can be considered a separate heap.So, the game is equivalent to a disjunctive game where each region is a heap, and players can make moves in one heap at a time.In such games, the Grundy number of the entire game is the XOR of the Grundy numbers of each heap.So, if we can compute the Grundy numbers for regions of size k, we can determine the winning positions.But this might be complicated.Alternatively, perhaps the game is equivalent to the game of Kayles, where players take turns removing edges from a graph, and the last to remove an edge wins.But in this case, the graph is a circle, and the edges are non-adjacent chords.Wait, maybe not.Alternatively, perhaps the game is equivalent to the game of cutting a cake, where each cut divides the cake into smaller pieces, and players take turns making cuts.In this case, the first player can win if the number of regions is odd, and the second player can win if it's even.But I'm not sure.Wait, let me think about the game in terms of moves.Each chord drawn reduces the problem into smaller independent subproblems (the regions divided by the chord).So, the game is a disjunctive game, and the Grundy number of the whole game is the XOR of the Grundy numbers of the subgames.Therefore, to determine if the starting position is a winning or losing position, we need to compute the Grundy number for a circle with n points.But computing Grundy numbers for such games can be complex.Alternatively, perhaps there's a pattern based on parity.From my earlier analysis, in n=5, Vasya can win, and in n=6, Petya can win.Let me test n=7.n=7 is odd.Petya draws a chord, say AC.This divides the circle into two arcs: A to C and C to A.In the arc A to C, points are A, B, C.Non-adjacent pairs: A and C is taken, so no moves.In the arc C to A, points are C, D, E, F, G, A.Non-adjacent pairs: C and E, D and F, E and G, F and A, etc.Vasya can choose to draw a chord in this arc.Suppose Vasya draws CE.Now, the circle is divided into three arcs: A to C, C to E, E to A.In each of these:- A to C: A, B, C. No moves.- C to E: C, D, E. Non-adjacent pairs: C and E is taken, D and E is adjacent, so no moves.- E to A: E, F, G, A. Non-adjacent pairs: E and G, F and A.So, Petya can draw EG or FA.Suppose Petya draws EG.Now, the circle has chords AC, CE, EG.Vasya's next move could be to draw FA in the arc E to A.But FA is from F to A, which is non-adjacent.So Vasya draws FA.Now, the circle has chords AC, CE, EG, FA.Petya's next move could be to draw DF in the arc C to E.Wait, DF is from D to F, which is non-adjacent.But DF would cross CE, right? Because CE is from C to E, and DF is from D to F, which is in the arc E to A.Wait, no, DF is in the arc C to E, which is separate from the arc E to A.Wait, no, DF is from D to F, which spans across the arc C to E and E to A.Wait, actually, DF would cross CE because CE is from C to E, and DF is from D to F, which is on the other side.Wait, no, in the arc C to E, points are C, D, E.So, DF is from D to F, but F is in the arc E to A, so DF would cross the chord CE.Therefore, DF cannot be drawn.Alternatively, Petya could draw DG in the arc E to A.But DG is from D to G, which is non-adjacent.Wait, D is in the arc C to E, and G is in the arc E to A.So, DG would cross CE.Therefore, DG cannot be drawn.Alternatively, Petya could draw FC in the arc E to A.But FC is from F to C, which is non-adjacent.But FC would cross CE, so it's not allowed.Wait, so after Vasya draws FA, Petya cannot make a move.Therefore, Petya loses.Wait, but n=7 is odd, and Petya loses.Similarly, in n=5, Petya loses.In n=6, Petya can win.So, perhaps the pattern is that when n is even, Petya can win, and when n is odd, Vasya can win.But let me test n=4, even though the problem states n >=5.n=4, a square.Petya draws a chord, say AC.This divides the circle into two arcs: A to C and C to A.In each arc, points are A, B, C and C, D, A.Non-adjacent pairs: in each arc, only AC is taken, so no moves.Wait, but in n=4, the non-adjacent chords are the two diagonals, AC and BD.So, Petya draws AC, Vasya draws BD, and then no moves left. So Petya loses.Wait, but n=4 is even, but Petya loses.Hmm, so my earlier conclusion might be incorrect.Wait, but n=4 is a special case because it's the smallest even n >=4.Wait, the problem states n >=5, so maybe n=4 is not considered.But still, in n=6, Petya can win, while in n=4, Petya cannot.So, perhaps the pattern is that for even n >=6, Petya can win, but for n=4, Petya cannot.But the problem states n >=5, so maybe n=5 and above.Wait, but n=5 is odd, and Petya cannot win.n=6 is even, Petya can win.n=7 is odd, Petya cannot win.n=8 is even, Petya can win.So, the pattern seems to be that for even n >=6, Petya can win, and for odd n >=5, Vasya can win.But wait, in n=6, Petya can win by mirroring, but in n=4, Petya cannot.So, perhaps the rule is that for even n >=6, Petya can win, and for odd n >=5, Vasya can win.But the problem states n >=5, so the answer would be that Petya can win when n is even and n >=6.But let me think again.In n=6, Petya can win by mirroring.In n=5, Vasya can win.In n=7, Vasya can win.In n=8, Petya can win.So, the general rule is that Petya can win when n is even, and Vasya can win when n is odd, for n >=5.But wait, in n=4, which is even, Petya cannot win, but n=4 is below the problem's constraint.So, considering n >=5, Petya can win when n is even, and Vasya can win when n is odd.Therefore, the answer is that Petya can win when n is even and n >=6, but since the problem states n >=5, we can say that Petya can win when n is even.Wait, but in n=6, Petya can win, and in n=5, Vasya can win.So, the conclusion is that Petya can win when n is even, and Vasya can win when n is odd, for n >=5.Therefore, the answer is that Petya will win if and only if n is even.

answer：First, I need to convert the decimal number 253 to its binary equivalent. I'll start by identifying the largest power of 2 that is less than or equal to 253, which is 128 (2^7). Subtracting 128 from 253 leaves me with 125.Next, I'll find the largest power of 2 less than or equal to 125, which is 64 (2^6). Subtracting 64 from 125 gives me 61.Continuing this process, the largest power of 2 less than or equal to 61 is 32 (2^5). Subtracting 32 from 61 leaves me with 29.The largest power of 2 less than or equal to 29 is 16 (2^4). Subtracting 16 from 29 gives me 13.Next, the largest power of 2 less than or equal to 13 is 8 (2^3). Subtracting 8 from 13 leaves me with 5.The largest power of 2 less than or equal to 5 is 4 (2^2). Subtracting 4 from 5 gives me 1.Finally, the largest power of 2 less than or equal to 1 is 1 (2^0). Subtracting 1 from 1 leaves me with 0, completing the conversion.So, the binary representation of 253 is 11111101.Now, I'll count the number of zeros (x) and ones (y) in the binary number:- Number of 1s (y) = 7- Number of 0s (x) = 1Finally, I'll calculate the difference between the number of ones and zeros:y - x = 7 - 1 = 6

answer：First, I'll denote the total number of workers in the company as ( W ).A third of the workers do not have a retirement plan, so the number of workers without a retirement plan is ( frac{1}{3}W ).60% of the workers without a retirement plan are women, which means the number of women without a retirement plan is ( 0.6 times frac{1}{3}W = frac{1}{5}W ).40% of the workers with a retirement plan are men. Since two-thirds of the workers have a retirement plan, the number of men with a retirement plan is ( 0.4 times frac{2}{3}W = frac{8}{30}W ) or ( frac{4}{15}W ).Given that there are 120 men in total, this includes men with and without a retirement plan. Therefore, the total number of men can be expressed as:[frac{4}{15}W + text{(Number of men without a retirement plan)} = 120]Since 40% of the workers without a retirement plan are men, the number of men without a retirement plan is ( 0.4 times frac{1}{3}W = frac{4}{30}W ) or ( frac{2}{15}W ).Now, the total number of men is:[frac{4}{15}W + frac{2}{15}W = 120]Combining the fractions:[frac{6}{15}W = 120]Simplifying:[frac{2}{5}W = 120]To find ( W ), multiply both sides by the reciprocal of ( frac{2}{5} ):[W = 120 times frac{5}{2} = 60 times 5 = 300]So, there are 300 workers in total. Since 120 of them are men, the number of women in the company is:[text{Number of women} = text{Total number of workers} - text{Number of men} = 300 - 120 = 180]

answer：Alright, so I have this equation: (2^m - 3^n = 1), and I need to find all the solutions where (m) and (n) are natural numbers. Hmm, okay. Let me think about how to approach this.First off, natural numbers usually mean positive integers, right? So (m) and (n) are both 1, 2, 3, and so on. I guess I should start by plugging in small values for (m) and (n) and see if the equation holds.Let's try (m = 1). Then the equation becomes (2^1 - 3^n = 1), which simplifies to (2 - 3^n = 1). Subtracting 2 from both sides gives (-3^n = -1), so (3^n = 1). The only way this is true is if (n = 0), because (3^0 = 1). But wait, is 0 considered a natural number here? Sometimes natural numbers start at 1, sometimes at 0. I should probably check both cases. If (n = 0) is allowed, then ((1, 0)) is a solution. If not, then maybe not.Moving on, let's try (m = 2). Then the equation becomes (2^2 - 3^n = 1), which is (4 - 3^n = 1). Subtracting 4 from both sides gives (-3^n = -3), so (3^n = 3). That means (n = 1). So ((2, 1)) is definitely a solution.What about (m = 3)? Plugging that in, we get (2^3 - 3^n = 1), which is (8 - 3^n = 1). Subtracting 8 from both sides gives (-3^n = -7), so (3^n = 7). Hmm, 7 isn't a power of 3, so there's no natural number (n) that satisfies this. So no solution here.Next, (m = 4): (2^4 - 3^n = 1) becomes (16 - 3^n = 1). Subtracting 16 gives (-3^n = -15), so (3^n = 15). Again, 15 isn't a power of 3, so no solution.Trying (m = 5): (32 - 3^n = 1), which simplifies to (3^n = 31). Not a power of 3, so no solution.(m = 6): (64 - 3^n = 1), so (3^n = 63). 63 isn't a power of 3 either.Hmm, seems like as (m) increases, (2^m) grows much faster than (3^n), so maybe there are no more solutions beyond (m = 2). But I should check a few more to be sure.(m = 7): (128 - 3^n = 1), so (3^n = 127). Nope, not a power of 3.(m = 8): (256 - 3^n = 1), so (3^n = 255). Still not a power of 3.Okay, maybe I can approach this differently instead of just plugging in numbers. Let's think about the equation modulo some number to see if we can find constraints on (m) and (n).Let's consider the equation modulo 3. So, (2^m - 3^n equiv 1 pmod{3}). Since (3^n equiv 0 pmod{3}), the equation simplifies to (2^m equiv 1 pmod{3}).What is (2^m) modulo 3? Let's see:- (2^1 equiv 2 pmod{3})- (2^2 equiv 4 equiv 1 pmod{3})- (2^3 equiv 8 equiv 2 pmod{3})- (2^4 equiv 16 equiv 1 pmod{3})So it alternates between 2 and 1 modulo 3. Therefore, (2^m equiv 1 pmod{3}) when (m) is even. So (m) must be even.Alright, that's a useful constraint. So (m) has to be even. Let's denote (m = 2k) for some natural number (k).So the equation becomes (2^{2k} - 3^n = 1), which is (4^k - 3^n = 1).Hmm, maybe I can analyze this further. Let's look at small values of (k):- (k = 1): (4 - 3^n = 1) → (3^n = 3) → (n = 1). So that's the solution we already found: ((2, 1)).- (k = 2): (16 - 3^n = 1) → (3^n = 15). Not a power of 3.- (k = 3): (64 - 3^n = 1) → (3^n = 63). Not a power of 3.- (k = 4): (256 - 3^n = 1) → (3^n = 255). Not a power of 3.Same as before, no solutions beyond (k = 1). Maybe I can try another modulus.How about modulo 4? Let's see:Original equation: (2^m - 3^n = 1).Modulo 4:- (2^m) modulo 4: If (m = 1), (2); if (m geq 2), (0).- (3^n) modulo 4: (3^1 equiv 3), (3^2 equiv 1), (3^3 equiv 3), (3^4 equiv 1), etc. So it alternates between 3 and 1.So the equation modulo 4 is:If (m = 1): (2 - 3^n equiv 1 pmod{4}). So (2 - 3^n equiv 1) → (-3^n equiv -1) → (3^n equiv 1 pmod{4}). Since (3^n) is 3 when (n) is odd and 1 when (n) is even, this implies (n) is even.If (m geq 2): (0 - 3^n equiv 1 pmod{4}) → (-3^n equiv 1) → (3^n equiv -1 equiv 3 pmod{4}). So (3^n equiv 3 pmod{4}), which happens when (n) is odd.So, combining with our earlier result that (m) must be even, let's see:If (m) is even and (m geq 2), then (n) must be odd.But from our earlier trials, even when (m) is even, we didn't find any solutions beyond (m = 2). Maybe I can explore this further.Let me try another modulus, perhaps modulo 8.Original equation: (2^m - 3^n = 1).Modulo 8:- (2^m) modulo 8: For (m = 1), 2; (m = 2), 4; (m geq 3), 0.- (3^n) modulo 8: Let's compute powers of 3 modulo 8: - (3^1 = 3) - (3^2 = 9 equiv 1) - (3^3 = 27 equiv 3) - (3^4 = 81 equiv 1) So it alternates between 3 and 1 modulo 8.So, the equation modulo 8 is:If (m = 1): (2 - 3^n equiv 1 pmod{8}) → (2 - 3^n equiv 1) → (-3^n equiv -1) → (3^n equiv 1 pmod{8}). So (n) must be even.If (m = 2): (4 - 3^n equiv 1 pmod{8}) → (4 - 3^n equiv 1) → (-3^n equiv -3) → (3^n equiv 3 pmod{8}). So (n) must be odd.If (m geq 3): (0 - 3^n equiv 1 pmod{8}) → (-3^n equiv 1) → (3^n equiv -1 equiv 7 pmod{8}). But looking at (3^n) modulo 8, it's either 1 or 3. So (3^n equiv 7 pmod{8}) is impossible. Therefore, there are no solutions for (m geq 3).So, combining all these results:- When (m = 1), (n) must be even. From earlier, (n = 0) is a solution if allowed.- When (m = 2), (n) must be odd, and we found (n = 1).- For (m geq 3), no solutions exist.Therefore, the only solutions are ((1, 0)) and ((2, 1)), assuming (n = 0) is considered a natural number. If not, then only ((2, 1)) is the solution.But in the initial problem statement, it says ((m, n) in mathbf{N}^2). Depending on the definition, (mathbf{N}) sometimes includes 0 and sometimes doesn't. If it does, then both solutions are valid. If not, only ((2, 1)) is valid.To be thorough, let's check if (n = 0) is acceptable. If (n = 0), then (3^0 = 1), so the equation becomes (2^m - 1 = 1) → (2^m = 2) → (m = 1). So yes, that works. If (n = 0) is allowed, then ((1, 0)) is a solution.In conclusion, the solutions are ((1, 0)) and ((2, 1)).

answer：First, I need to determine the Greatest Common Divisor (GCD) of 8 and 12. To do this, I'll list the factors of each number:- Factors of 8: 1, 2, 4, 8- Factors of 12: 1, 2, 3, 4, 6, 12The common factors are 1, 2, and 4. The largest of these is 4, so the GCD of 8 and 12 is 4.Next, I'll find the Least Common Multiple (LCM) of 8 and 12. One way to calculate the LCM is by using the formula:[text{LCM}(a, b) = frac{a times b}{text{GCD}(a, b)}]Plugging in the values:[text{LCM}(8, 12) = frac{8 times 12}{4} = frac{96}{4} = 24]Finally, I'll add the GCD and LCM together:[4 + 24 = 28]

answer：To find the approximate average price John paid per bottle, I need to calculate the total cost of the large and small bottles separately and then determine the overall average.First, I'll calculate the total cost for the large bottles by multiplying the number of large bottles by the price per large bottle.Next, I'll calculate the total cost for the small bottles by multiplying the number of small bottles by the price per small bottle.After obtaining both total costs, I'll add them together to get the combined total cost for all the bottles.Then, I'll find the total number of bottles by adding the number of large bottles to the number of small bottles.Finally, I'll divide the combined total cost by the total number of bottles to determine the approximate average price paid per bottle.