💡Okay, so I have this problem where I need to determine if the numbers √(3v - 2), √(3v + 2), and √(6v) can be the side lengths of a triangle. If they can, I also need to find the measure of the largest angle. Hmm, let me think about how to approach this.First, I remember that for three numbers to be the sides of a triangle, they must satisfy the triangle inequality theorem. That means the sum of any two sides must be greater than the third side. So, I need to check all three combinations:1. √(3v - 2) + √(3v + 2) > √(6v)2. √(3v - 2) + √(6v) > √(3v + 2)3. √(3v + 2) + √(6v) > √(3v - 2)Before I dive into these inequalities, I should also make sure that all the expressions under the square roots are non-negative because you can't have a square root of a negative number in real numbers. So, let's set up those conditions:1. 3v - 2 ≥ 0 ⇒ v ≥ 2/32. 3v + 2 ≥ 0 ⇒ v ≥ -2/3 (but since v is under another square root, √(6v), v must be ≥ 0. So, combining these, v must be ≥ 2/3.Alright, so v has to be at least 2/3 for all these square roots to be real numbers. Now, let's tackle the triangle inequalities one by one.Starting with the first inequality: √(3v - 2) + √(3v + 2) > √(6v). Hmm, this looks a bit complicated. Maybe squaring both sides will help simplify it. Let me try that.Let’s denote A = √(3v - 2), B = √(3v + 2), and C = √(6v). So, the inequality becomes A + B > C. Squaring both sides:(A + B)^2 > C^2A² + 2AB + B² > C²Substituting back:(3v - 2) + 2√{(3v - 2)(3v + 2)} + (3v + 2) > 6vSimplify the left side:3v - 2 + 3v + 2 + 2√{(3v - 2)(3v + 2)} > 6v6v + 2√{(3v - 2)(3v + 2)} > 6vSubtract 6v from both sides:2√{(3v - 2)(3v + 2)} > 0Well, since v ≥ 2/3, both (3v - 2) and (3v + 2) are positive, so their product is positive, and the square root is positive. Therefore, 2 times a positive number is positive, which is always greater than 0. So, this inequality holds true for all v ≥ 2/3.Okay, that was the first inequality. Let's move on to the second one: √(3v - 2) + √(6v) > √(3v + 2). Again, let's denote A = √(3v - 2), C = √(6v), and B = √(3v + 2). So, the inequality is A + C > B.Squaring both sides:(A + C)^2 > B²A² + 2AC + C² > B²Substituting back:(3v - 2) + 2√{(3v - 2)(6v)} + 6v > 3v + 2Simplify the left side:3v - 2 + 6v + 2√{(3v - 2)(6v)} > 3v + 29v - 2 + 2√{(3v - 2)(6v)} > 3v + 2Subtract 3v and add 2 to both sides:6v + 2√{(3v - 2)(6v)} > 4Hmm, let's see. Since v ≥ 2/3, 6v is at least 4. So, 6v ≥ 4. Therefore, 6v + something positive is definitely greater than 4. So, this inequality also holds true for all v ≥ 2/3.Alright, moving on to the third inequality: √(3v + 2) + √(6v) > √(3v - 2). Again, let's denote B = √(3v + 2), C = √(6v), and A = √(3v - 2). So, the inequality is B + C > A.Squaring both sides:(B + C)^2 > A²B² + 2BC + C² > A²Substituting back:(3v + 2) + 2√{(3v + 2)(6v)} + 6v > 3v - 2Simplify the left side:3v + 2 + 6v + 2√{(3v + 2)(6v)} > 3v - 29v + 2 + 2√{(3v + 2)(6v)} > 3v - 2Subtract 3v and add 2 to both sides:6v + 2√{(3v + 2)(6v)} > -4Well, since v ≥ 2/3, 6v is at least 4, and the square root term is positive, so the left side is definitely greater than 4, which is way more than -4. So, this inequality also holds true for all v ≥ 2/3.Okay, so all three triangle inequalities hold true for v ≥ 2/3. That means these three expressions can indeed be the sides of a triangle for any v ≥ 2/3.Now, the next part is to find the measure of the largest angle. In a triangle, the largest angle is opposite the longest side. So, I need to figure out which of the three sides is the longest.Looking at the three sides: √(3v - 2), √(3v + 2), and √(6v). Let's compare them.First, √(3v + 2) is clearly larger than √(3v - 2) because 3v + 2 > 3v - 2 for all v.Now, let's compare √(3v + 2) and √(6v). Let's square both to make it easier:(√(3v + 2))² = 3v + 2(√(6v))² = 6vSo, we need to see whether 3v + 2 is greater than 6v or not.3v + 2 > 6vSubtract 3v from both sides:2 > 3vDivide both sides by 3:2/3 > vBut we know that v ≥ 2/3. So, when v = 2/3, 3v + 2 = 3*(2/3) + 2 = 2 + 2 = 4, and 6v = 6*(2/3) = 4. So, at v = 2/3, both are equal.For v > 2/3, 6v becomes larger than 3v + 2 because 6v - (3v + 2) = 3v - 2, which is positive when v > 2/3.Therefore, for v > 2/3, √(6v) is larger than √(3v + 2). So, the largest side is √(6v) when v > 2/3, and when v = 2/3, both √(6v) and √(3v + 2) are equal to 4, so it's a right triangle with two sides equal.Wait, at v = 2/3, let's check the sides:√(3*(2/3) - 2) = √(2 - 2) = 0. Wait, that can't be. Wait, hold on, 3v - 2 when v = 2/3 is 3*(2/3) - 2 = 2 - 2 = 0. So, √0 = 0. But a triangle can't have a side of length 0. So, actually, v must be strictly greater than 2/3, not equal to.So, v > 2/3. Therefore, √(6v) is the largest side for v > 2/3.Therefore, the largest angle is opposite √(6v). To find the measure of this angle, we can use the Law of Cosines.Law of Cosines states that for a triangle with sides a, b, c, opposite angles A, B, C respectively, c² = a² + b² - 2ab cos(C).In our case, the sides are:a = √(3v - 2)b = √(3v + 2)c = √(6v)So, we can write:(√(6v))² = (√(3v - 2))² + (√(3v + 2))² - 2 * √(3v - 2) * √(3v + 2) * cos(C)Simplify:6v = (3v - 2) + (3v + 2) - 2 * √{(3v - 2)(3v + 2)} * cos(C)Simplify the right side:6v = 3v - 2 + 3v + 2 - 2 * √{(3v - 2)(3v + 2)} * cos(C)6v = 6v + 0 - 2 * √{(3v - 2)(3v + 2)} * cos(C)So, subtract 6v from both sides:0 = -2 * √{(3v - 2)(3v + 2)} * cos(C)Divide both sides by -2 * √{(3v - 2)(3v + 2)} (which is non-zero since v > 2/3):0 = cos(C)So, cos(C) = 0. Therefore, angle C is 90 degrees.Wait, that's interesting. So, the angle opposite the largest side is 90 degrees. That means the triangle is a right triangle, and the largest angle is 90 degrees.But let me double-check this because sometimes when dealing with square roots and squaring, extraneous solutions can pop up, but in this case, since we've already established that v > 2/3, and all sides are positive, I think this holds.Alternatively, maybe I can check if the sides satisfy the Pythagorean theorem.Let's compute:(√(3v - 2))² + (√(3v + 2))² = (3v - 2) + (3v + 2) = 6vAnd (√(6v))² = 6vSo, indeed, (√(3v - 2))² + (√(3v + 2))² = (√(6v))²Therefore, by the Pythagorean theorem, the triangle is right-angled, and the largest angle is 90 degrees.So, putting it all together, yes, the numbers √(3v - 2), √(3v + 2), and √(6v) can be the sides of a triangle for v > 2/3, and the largest angle is 90 degrees.

💡Alright, so I've got this game between Alice and Bob on a strip of squares. Let me try to wrap my head around it step by step. First, the setup: There are ( n geq 3 ) squares in a strip. Alice's piece starts on the first square, and Bob's piece is on the last square. They take turns moving their pieces towards each other. Each move can be either one or two squares towards the opponent's piece, but they can't jump over each other. Alice goes first, and the person who can't move loses.Okay, so I need to figure out for which values of ( n ) Bob can always win, no matter how Alice plays, and for which ( n ) Alice can always win, regardless of Bob's moves.Let me start by looking at small values of ( n ) to see if I can spot a pattern.**Case 1: ( n = 3 )**- Alice is on square 1, Bob on square 3.- Alice moves first. She can move one or two squares towards Bob. - If she moves one square, she goes to square 2. - If she moves two squares, she goes to square 3, but Bob is already there, so she can't land on the same square. So, she can only move to square 2.- Now, Bob is on square 3, and Alice is on square 2. Bob can move one or two squares towards Alice. - If Bob moves one square, he goes to square 2, but Alice is already there. So he can't move there. - If Bob moves two squares, he would go to square 1, but that's behind Alice, which isn't allowed because he can't jump over her. So, Bob can't make a move.- Wait, that doesn't make sense. If Alice moves to square 2, Bob is on square 3. Bob can move one square to square 2, but Alice is already there, so he can't. If he moves two squares, he would go to square 1, which is behind Alice, which isn't allowed. So Bob can't move, meaning Alice wins? But that contradicts my initial thought.Wait, maybe I made a mistake. Let me think again.- Alice starts at 1, Bob at 3.- Alice moves to 2.- Now, Bob is at 3. He can move one square to 2, but Alice is there. So he can't. He can't move two squares because that would take him to 1, which is behind Alice. So Bob loses. So Alice wins when ( n = 3 ).But I thought Bob could win. Maybe I was wrong. Let me check ( n = 4 ).**Case 2: ( n = 4 )**- Alice at 1, Bob at 4.- Alice can move to 2 or 3. - If she moves to 2: - Bob can move to 3 or 2. If he moves to 3, then Alice can move to 3 (but Bob is there) or 4 (which is Bob's original spot). Wait, no, she can only move towards Bob. So from 2, she can move to 3 or 4. But 4 is Bob's original spot, which is now vacated. Wait, no, Bob is at 3 now. So Alice can move to 3 or 4. But 3 is occupied by Bob, so she can't. So she can only move to 4, but that's two squares away, which is allowed. So Alice moves to 4, but Bob is at 3. Wait, no, Bob is at 3, so Alice can move to 3 or 4. But 3 is occupied, so she can only move to 4, which is two squares away. So she moves to 4, but Bob is at 3. Now, Bob can move to 4, but Alice is there, so he can't. Or he can move to 2, but that's away from Alice, which isn't allowed. So Bob loses. So Alice wins.Wait, that's not right. If Alice moves to 2, Bob moves to 3, then Alice can move to 4, and Bob can't move. So Alice wins.Alternatively, if Alice moves to 3 on her first move:- Alice moves to 3, but Bob is at 4. So she can't move there because it's occupied. Wait, no, she can move to 2 or 3. If she moves to 3, Bob is at 4. So Bob can move to 3 or 2. If he moves to 3, Alice is there, so he can't. If he moves to 2, he moves two squares towards Alice, landing on 2. Now, Alice is at 3, Bob at 2. Now, Alice can move to 2 or 1, but 1 is behind, so she can only move to 2, but Bob is there. So Alice can't move, so Bob wins.Wait, so if Alice moves to 3 first, Bob can move to 2, and then Alice can't move, so Bob wins. But if Alice moves to 2 first, Bob moves to 3, and then Alice moves to 4, and Bob can't move, so Alice wins.So, depending on Alice's first move, the outcome changes. But since Alice is trying to win, she will choose the move that leads to her victory. So she will move to 2 first, forcing Bob into a losing position. Therefore, Alice can win when ( n = 4 ).Hmm, so for ( n = 3 ), Alice wins, and for ( n = 4 ), Alice can also win. Let me check ( n = 5 ).**Case 3: ( n = 5 )**- Alice at 1, Bob at 5.- Alice can move to 2 or 3. - If she moves to 2: - Bob can move to 4 or 3. - If Bob moves to 4: - Alice can move to 3 or 4. If she moves to 3, Bob can move to 3 or 2. If Bob moves to 3, Alice can't move. If Bob moves to 2, Alice can move to 2 or 1, but 1 is behind, so she can't. So Bob wins. - If Bob moves to 3: - Alice can move to 3 or 4. If she moves to 3, Bob can't move. If she moves to 4, Bob can move to 4 or 3. If Bob moves to 4, Alice can't move. So Bob wins. - If Alice moves to 3: - Bob can move to 4 or 3. If he moves to 4: - Alice can move to 4 or 5. If she moves to 4, Bob can't move. If she moves to 5, Bob can't move. So Alice wins. - If Bob moves to 3, he can't because Alice is there. So Bob must move to 4. - Then Alice can move to 4 or 5. If she moves to 4, Bob can't move. If she moves to 5, Bob can't move. So Alice wins.Wait, so if Alice moves to 3 first, she can win. But if she moves to 2 first, Bob can win. So Alice, being smart, will choose to move to 3 first, ensuring her win. Therefore, Alice can win when ( n = 5 ).Wait, that contradicts my initial thought. Maybe I'm missing something. Let me think again.If Alice moves to 3, Bob is at 5. Bob can move to 4 or 3. If he moves to 4, Alice can move to 4 or 5. If she moves to 4, Bob can't move. If she moves to 5, Bob can't move. So Alice wins.If Bob moves to 3, he can't because Alice is there. So Bob must move to 4. Then Alice can move to 4 or 5, and Bob can't move. So Alice wins.Alternatively, if Alice moves to 2 first, Bob can move to 4 or 3. If Bob moves to 4, Alice can move to 3 or 4. If she moves to 3, Bob can move to 3 or 2. If Bob moves to 3, Alice can't move. If Bob moves to 2, Alice can move to 2 or 1, but 1 is behind, so she can't. So Bob wins.If Bob moves to 3, Alice can move to 3 or 4. If she moves to 3, Bob can't move. If she moves to 4, Bob can move to 4 or 3. If Bob moves to 4, Alice can't move. So Bob wins.So, depending on Alice's first move, the outcome changes. But since Alice is trying to win, she will choose the move that leads to her victory. So she will move to 3 first, forcing Bob into a losing position. Therefore, Alice can win when ( n = 5 ).Wait, but earlier I thought Bob could win for ( n = 5 ). Maybe I was wrong. Let me check ( n = 6 ).**Case 4: ( n = 6 )**- Alice at 1, Bob at 6.- Alice can move to 2 or 3. - If she moves to 2: - Bob can move to 5 or 4. - If Bob moves to 5: - Alice can move to 3 or 4. - If Alice moves to 3: - Bob can move to 4 or 3. If he moves to 4, Alice can move to 4 or 5. If she moves to 4, Bob can't move. If she moves to 5, Bob can't move. So Alice wins. - If Alice moves to 4: - Bob can move to 4 or 3. If he moves to 4, Alice can't move. If he moves to 3, Alice can move to 3 or 2. If she moves to 3, Bob can't move. If she moves to 2, Bob can move to 2 or 1, but 1 is behind, so he can't. So Alice wins. - If Bob moves to 4: - Alice can move to 3 or 4. If she moves to 3, Bob can move to 4 or 3. If Bob moves to 4, Alice can't move. If Bob moves to 3, Alice can move to 3 or 2. If she moves to 3, Bob can't move. If she moves to 2, Bob can move to 2 or 1, but 1 is behind, so he can't. So Alice wins. - If Alice moves to 3: - Bob can move to 5 or 4. - If Bob moves to 5: - Alice can move to 4 or 5. If she moves to 4, Bob can move to 4 or 3. If Bob moves to 4, Alice can't move. If Bob moves to 3, Alice can move to 3 or 2. If she moves to 3, Bob can't move. If she moves to 2, Bob can move to 2 or 1, but 1 is behind, so he can't. So Alice wins. - If Bob moves to 4: - Alice can move to 4 or 5. If she moves to 4, Bob can't move. If she moves to 5, Bob can't move. So Alice wins.So, regardless of Alice's first move, she can win when ( n = 6 ).Wait, so for ( n = 3 ), Alice wins; ( n = 4 ), Alice wins; ( n = 5 ), Alice wins; ( n = 6 ), Alice wins. Hmm, maybe I was wrong earlier about Bob winning for some ( n ). Let me check ( n = 7 ).**Case 5: ( n = 7 )**- Alice at 1, Bob at 7.- Alice can move to 2 or 3. - If she moves to 2: - Bob can move to 6 or 5. - If Bob moves to 6: - Alice can move to 3 or 4. - If Alice moves to 3: - Bob can move to 5 or 4. - If Bob moves to 5: - Alice can move to 4 or 5. If she moves to 4, Bob can move to 4 or 3. If Bob moves to 4, Alice can't move. If Bob moves to 3, Alice can move to 3 or 2. If she moves to 3, Bob can't move. If she moves to 2, Bob can move to 2 or 1, but 1 is behind, so he can't. So Alice wins. - If Bob moves to 4: - Alice can move to 4 or 5. If she moves to 4, Bob can't move. If she moves to 5, Bob can't move. So Alice wins. - If Alice moves to 4: - Bob can move to 5 or 4. If he moves to 5, Alice can move to 5 or 6. If she moves to 5, Bob can't move. If she moves to 6, Bob can't move. So Alice wins. - If Bob moves to 5: - Alice can move to 3 or 4. - If Alice moves to 3: - Bob can move to 5 or 4. If he moves to 5, Alice can move to 5 or 6. If she moves to 5, Bob can't move. If she moves to 6, Bob can't move. So Alice wins. - If Alice moves to 4: - Bob can move to 5 or 4. If he moves to 5, Alice can move to 5 or 6. If she moves to 5, Bob can't move. If she moves to 6, Bob can't move. So Alice wins. - If Alice moves to 3: - Bob can move to 6 or 5. - If Bob moves to 6: - Alice can move to 4 or 5. - If Alice moves to 4: - Bob can move to 5 or 4. If he moves to 5, Alice can move to 5 or 6. If she moves to 5, Bob can't move. If she moves to 6, Bob can't move. So Alice wins. - If Alice moves to 5: - Bob can't move because Alice is at 5 and Bob is at 6. So Alice wins. - If Bob moves to 5: - Alice can move to 4 or 5. - If Alice moves to 4: - Bob can move to 5 or 4. If he moves to 5, Alice can move to 5 or 6. If she moves to 5, Bob can't move. If she moves to 6, Bob can't move. So Alice wins. - If Alice moves to 5: - Bob can't move because Alice is at 5 and Bob is at 5. So Alice wins.So, again, regardless of Alice's first move, she can win when ( n = 7 ).Wait, this is getting confusing. Maybe I'm missing a pattern here. Let me try to generalize.It seems like for ( n = 3 ), Alice wins; ( n = 4 ), Alice wins; ( n = 5 ), Alice wins; ( n = 6 ), Alice wins; ( n = 7 ), Alice wins. So, is Alice always winning? But that contradicts the problem statement which suggests that Bob can win for some ( n ).Wait, maybe I'm not considering all possibilities correctly. Let me think differently.Perhaps the key is to look at the distance between Alice and Bob. Initially, the distance is ( n - 1 ) squares. Each move reduces this distance by 1 or 2, but without jumping over.So, if the distance is ( d ), each player can reduce it by 1 or 2, but they can't make the distance negative or jump over.The game ends when the distance is 0, meaning the players are on adjacent squares, and the next player can't move.Wait, no, the game ends when a player can't move. So, if the distance is 1, the player whose turn it is can't move because moving one or two squares would require jumping over or landing on the opponent's square, which is not allowed.Wait, actually, if the distance is 1, the player can't move because moving one square would land on the opponent's square, which is occupied, and moving two squares would jump over, which is not allowed.So, the losing condition is when the distance is 1.Therefore, the game is about reducing the distance to 1 on your opponent's turn.So, the initial distance is ( d = n - 1 ).Players take turns reducing ( d ) by 1 or 2, but without making ( d ) negative or jumping over.The player who makes the distance 1 on their opponent's turn wins.Wait, no, actually, the player who cannot make a move loses. So, if the distance is 1 on your turn, you lose because you can't move.Therefore, the goal is to force the opponent into a position where the distance is 1.So, this is similar to a game where players reduce a number by 1 or 2, and the player who reduces it to 1 wins.Wait, but in this case, the distance starts at ( d = n - 1 ), and players alternately reduce it by 1 or 2, but the player who cannot move (i.e., when ( d = 1 )) loses.So, it's similar to a game where the target is to reduce the distance to 1, and the player who can't move loses.In such games, often the key is to leave the opponent with a multiple of 3.Wait, let me think about it.If the distance is a multiple of 3, say ( d = 3k ), then whatever the opponent does (reducing by 1 or 2), you can reduce by 2 or 1, respectively, to bring it back to a multiple of 3.So, if you can always leave the opponent with ( d = 3k ), you can force them into a losing position.But in our case, the losing condition is when ( d = 1 ). So, if you can force the opponent to have ( d = 1 ), they lose.Wait, but if ( d = 1 ), the player can't move, so they lose. So, the goal is to make the opponent face ( d = 1 ).So, if you can leave the opponent with ( d = 4 ), because then whatever they do (reduce by 1 or 2), you can reduce by 2 or 1, respectively, to make ( d = 1 ).Similarly, if you can leave the opponent with ( d = 7 ), because whatever they do, you can reduce it to 4, and so on.So, the key is to leave the opponent with ( d = 3k + 1 ), where ( k ) is a positive integer.Wait, let me test this.If ( d = 4 ):- If the opponent reduces by 1, making ( d = 3 ), then you can reduce by 1, making ( d = 2 ), and then the opponent can reduce by 1, making ( d = 1 ), and you lose.Wait, that doesn't make sense.Alternatively, if ( d = 4 ):- If the opponent reduces by 1, making ( d = 3 ), then you can reduce by 2, making ( d = 1 ), and the opponent loses.- If the opponent reduces by 2, making ( d = 2 ), then you can reduce by 1, making ( d = 1 ), and the opponent loses.So, yes, if you can leave the opponent with ( d = 4 ), you can win.Similarly, if ( d = 7 ):- If the opponent reduces by 1, making ( d = 6 ), you reduce by 2, making ( d = 4 ), and then you can win as above.- If the opponent reduces by 2, making ( d = 5 ), you reduce by 1, making ( d = 4 ), and then you can win.So, the pattern is that if you can leave the opponent with ( d = 3k + 1 ), you can force a win.Therefore, the winning strategy is to leave the opponent with ( d = 3k + 1 ).Now, the initial distance is ( d = n - 1 ).So, if ( n - 1 = 3k + 1 ), then ( n = 3k + 2 ).Therefore, if ( n equiv 2 mod 3 ), then Bob can win, because he can always respond to Alice's moves to leave her with ( d = 3k + 1 ).Wait, let me see.If ( n = 3k + 2 ), then ( d = n - 1 = 3k + 1 ).So, Bob can always respond to Alice's moves to reduce the distance by ( 3 - ) Alice's move, keeping the distance as ( 3(k - 1) + 1 ), and so on, until he leaves Alice with ( d = 1 ), making her lose.Therefore, Bob can win if ( n = 3k + 2 ).For other values of ( n ), Alice can win.Let me test this with the earlier cases.- ( n = 3 ): ( 3 = 3*0 + 3 ), not ( 3k + 2 ). So Alice wins. Correct.- ( n = 4 ): ( 4 = 3*1 + 1 ), not ( 3k + 2 ). So Alice wins. Correct.- ( n = 5 ): ( 5 = 3*1 + 2 ). So Bob wins. Wait, earlier I thought Alice could win, but according to this, Bob should win. Let me re-examine ( n = 5 ).**Re-examining ( n = 5 ):**- Alice at 1, Bob at 5. Distance ( d = 4 ).- Alice moves first. She can reduce ( d ) by 1 or 2, making ( d = 3 ) or ( d = 2 ). - If Alice reduces ( d ) by 1 to 3: - Bob can reduce ( d ) by 2 to 1, leaving Alice with ( d = 1 ), so she loses. - If Alice reduces ( d ) by 2 to 2: - Bob can reduce ( d ) by 1 to 1, leaving Alice with ( d = 1 ), so she loses.So, regardless of Alice's move, Bob can win when ( n = 5 ). So my earlier analysis was incorrect because I didn't consider that Bob can always respond to reduce the distance to 1.Therefore, the correct conclusion is that Bob can win when ( n = 3k + 2 ), and Alice can win otherwise.So, the answer is:- Bob can win for sure when ( n equiv 2 mod 3 ), i.e., ( n = 3k + 2 ) for some integer ( k geq 1 ).- Alice can win for sure otherwise.Let me check ( n = 5 ) again with this strategy.- ( n = 5 ), ( d = 4 ).- Alice moves first. She can reduce ( d ) to 3 or 2. - If she reduces to 3, Bob reduces to 1, Alice loses. - If she reduces to 2, Bob reduces to 1, Alice loses.So, Bob wins.Similarly, ( n = 8 ):- ( n = 8 ), ( d = 7 ).- Alice moves first. She can reduce ( d ) to 6 or 5. - If she reduces to 6, Bob reduces to 3, then Alice reduces to 2 or 1. If she reduces to 2, Bob reduces to 1, Alice loses. If she reduces to 1, Bob can't move, so Alice wins. Wait, that contradicts.Wait, no, if ( d = 7 ), Alice reduces to 6 or 5.- If Alice reduces to 6, Bob can reduce to 3 (since 6 - 3 = 3). Then Alice is faced with ( d = 3 ). She can reduce to 2 or 1. If she reduces to 2, Bob reduces to 1, Alice loses. If she reduces to 1, Bob can't move, so Alice wins. So Alice can choose to reduce to 1 and win.Wait, that contradicts the earlier conclusion. Maybe my strategy is flawed.Wait, no, if ( d = 7 ), which is ( 3*2 + 1 ), Bob can reduce it to ( 3*1 + 1 = 4 ) by reducing by 3, but he can only reduce by 1 or 2. So, if Alice reduces ( d ) to 6, Bob can reduce by 2 to make ( d = 4 ), which is ( 3*1 + 1 ). Then Alice is faced with ( d = 4 ), and whatever she does, Bob can reduce it to 1.Wait, let me go through it step by step.- ( n = 8 ), ( d = 7 ).- Alice moves first. She can reduce ( d ) to 6 or 5. - If she reduces to 6: - Bob can reduce by 2, making ( d = 4 ). - Alice now has ( d = 4 ). She can reduce to 3 or 2. - If she reduces to 3, Bob reduces to 1, Alice loses. - If she reduces to 2, Bob reduces to 1, Alice loses. - If she reduces to 5: - Bob can reduce by 1, making ( d = 4 ). - Alice now has ( d = 4 ). She can reduce to 3 or 2. - If she reduces to 3, Bob reduces to 1, Alice loses. - If she reduces to 2, Bob reduces to 1, Alice loses.So, regardless of Alice's first move, Bob can reduce the distance to 4, and then force Alice to lose. Therefore, Bob wins when ( n = 8 ).Similarly, for ( n = 11 ), which is ( 3*3 + 2 ), Bob can win.Therefore, the pattern holds: Bob wins when ( n = 3k + 2 ), and Alice wins otherwise.So, the final answer is:- Bob can win for sure when ( n equiv 2 mod 3 ), i.e., ( n = 3k + 2 ) for some integer ( k geq 1 ).- Alice can win for sure otherwise.**Final Answer**Bob can win for sure when ( n ) is of the form ( 3k + 2 ), and Alice can win for sure otherwise. Thus, the answer is:boxed{text{Bob wins if } n equiv 2 pmod{3} text{; otherwise, Alice wins.}}

💡Okay, so I have this problem here: "What is the sum of the eight smallest distinct positive integer multiples of 10?" Hmm, let's break this down step by step. First, I need to understand what it's asking for. It wants the sum of the first eight multiples of 10. Alright, so multiples of 10 are numbers like 10, 20, 30, and so on. Each of these is obtained by multiplying 10 by a positive integer. So, the first multiple is 10×1=10, the second is 10×2=20, the third is 10×3=30, and this pattern continues. Since the question is asking for the eight smallest distinct positive integer multiples, I need to list out the first eight of these. Let me write them down: 10, 20, 30, 40, 50, 60, 70, 80. That seems straightforward enough. Now, I need to find the sum of these numbers. So, I'll add them up one by one. Let's see:10 + 20 = 3030 + 30 = 6060 + 40 = 100100 + 50 = 150150 + 60 = 210210 + 70 = 280280 + 80 = 360Wait, so the total sum is 360? That seems right, but let me double-check to make sure I didn't make a mistake in my addition.Alternatively, I remember that when adding a series of numbers that form an arithmetic sequence, there's a formula that can be used. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 × (first term + last term). In this case, the first term is 10, the last term is 80, and the number of terms is 8. Plugging these into the formula:S_8 = 8/2 × (10 + 80) = 4 × 90 = 360Okay, that matches what I got when I added them one by one. So, that's reassuring. But just to be thorough, let me think about another way to approach this. Since all these numbers are multiples of 10, I can factor out the 10 and add the remaining coefficients. So, the sum becomes 10×(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8). Now, the sum inside the parentheses is the sum of the first 8 positive integers. I remember that the sum of the first n integers is given by n(n + 1)/2. So, for n=8:Sum = 8×9/2 = 72/2 = 36Therefore, the total sum is 10×36 = 360. Hmm, same result again. I think I'm confident now that 360 is the correct answer. Just to make sure I didn't overlook anything, let me revisit the problem statement. It specifies "eight smallest distinct positive integer multiples of 10." I've considered the first eight multiples, which are distinct and positive. So, I think I've covered all bases here.Another way to think about it is to list all multiples of 10 and pick the first eight. Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, and so on. The first eight are indeed 10 through 80, as I listed earlier. Adding them up gives 360.I could also use a calculator to add them individually to cross-verify:10 + 20 = 3030 + 30 = 6060 + 40 = 100100 + 50 = 150150 + 60 = 210210 + 70 = 280280 + 80 = 360Yep, same result. Alternatively, if I use the formula for the sum of an arithmetic series, which is S = n/2 × (2a + (n - 1)d), where n is the number of terms, a is the first term, and d is the common difference. In this case, a=10, d=10, n=8.Plugging in the values:S = 8/2 × (2×10 + (8 - 1)×10) = 4 × (20 + 70) = 4 × 90 = 360Again, the same answer. It seems consistent across different methods, so I'm pretty sure 360 is correct.Just to ensure I'm not missing any nuances in the problem statement, let's parse it again: "eight smallest distinct positive integer multiples of 10." - "Eight smallest": So, the first eight in the sequence of multiples of 10.- "Distinct": Each multiple is unique; no repeats.- "Positive integer multiples": Each multiple is obtained by multiplying 10 by a positive integer (1, 2, 3, etc.).Given that, my approach was correct. I listed the first eight multiples, added them, and verified using different methods. I don't think I made any calculation errors, as all methods led to the same result. It's always good to cross-verify, especially when dealing with sums and sequences, as it's easy to miscount or miscalculate.In summary, whether I add the numbers directly, use the arithmetic series formula, or factor out the common multiple and sum the coefficients, I consistently arrive at 360 as the sum of the eight smallest distinct positive integer multiples of 10. I feel confident that this is the correct answer.

💡Okay, so I have this problem here: What is the largest digit N for which 5217N is divisible by 6? Hmm, let me think about how to approach this. I remember that a number is divisible by 6 if and only if it's divisible by both 2 and 3. So, I need to check the conditions for divisibility by 2 and 3 separately and then find the largest digit N that satisfies both.First, let's tackle divisibility by 2. I know that a number is divisible by 2 if its last digit is even. In this case, the last digit is N. So, N must be an even number. The possible even digits are 0, 2, 4, 6, and 8. So, N can be one of these.Next, let's think about divisibility by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. So, I need to calculate the sum of the digits of 5217N and see what N makes that sum divisible by 3.Let's add up the known digits first: 5 + 2 + 1 + 7. That equals 15. So, the sum of the digits is 15 plus N. Therefore, 15 + N must be divisible by 3.Now, I need to find the largest even digit N such that 15 + N is divisible by 3. Let's list the possible even digits and check which ones satisfy the condition.Starting with the largest even digit, which is 8:15 + 8 = 23. Is 23 divisible by 3? Let's see, 3 times 7 is 21, and 23 minus 21 is 2, so no, 23 is not divisible by 3. So, N cannot be 8.Next, let's try N = 6:15 + 6 = 21. Is 21 divisible by 3? Yes, because 3 times 7 is 21. So, N = 6 works.Just to be thorough, let's check the other even digits in case I made a mistake:N = 4:15 + 4 = 19. 19 divided by 3 is 6 with a remainder of 1, so not divisible by 3.N = 2:15 + 2 = 17. 17 divided by 3 is 5 with a remainder of 2, so not divisible by 3.N = 0:15 + 0 = 15. 15 is divisible by 3, but since we're looking for the largest digit, 6 is larger than 0.So, the largest digit N that makes 5217N divisible by 6 is 6. I think that's it. Let me double-check my calculations to make sure I didn't make any errors.Sum of digits: 5 + 2 + 1 + 7 = 15. Adding N = 6 gives 21, which is divisible by 3. Since 6 is even, it also satisfies the divisibility by 2. Yep, that seems correct. I don't think I missed anything here.

💡Alright, so I have this equation to solve: (x^3 + x^2 + x + frac{1}{3} = 0). Hmm, it's a cubic equation. I remember that solving cubic equations can be a bit tricky, but maybe I can figure it out step by step.First, let me write down the equation again to make sure I have it right:(x^3 + x^2 + x + frac{1}{3} = 0)Okay, so it's a cubic equation, which means it has a degree of 3. I recall that cubic equations can have up to three real roots or one real root and two complex roots. I wonder which case this equation falls into.I also remember that sometimes, cubic equations can be factored, especially if they have rational roots. Maybe I can try the Rational Root Theorem to see if there are any rational solutions. The Rational Root Theorem says that any possible rational root, expressed in lowest terms as (frac{p}{q}), has (p) as a factor of the constant term and (q) as a factor of the leading coefficient.In this equation, the constant term is (frac{1}{3}) and the leading coefficient is 1. So, the possible rational roots could be (pm1), (pmfrac{1}{3}). Let me test these values.First, let's try (x = 1):(1^3 + 1^2 + 1 + frac{1}{3} = 1 + 1 + 1 + frac{1}{3} = 3 + frac{1}{3} = frac{10}{3}), which is not zero.Next, (x = -1):((-1)^3 + (-1)^2 + (-1) + frac{1}{3} = -1 + 1 - 1 + frac{1}{3} = -1 + frac{1}{3} = -frac{2}{3}), which is also not zero.Now, (x = frac{1}{3}):(left(frac{1}{3}right)^3 + left(frac{1}{3}right)^2 + frac{1}{3} + frac{1}{3} = frac{1}{27} + frac{1}{9} + frac{1}{3} + frac{1}{3})Let me calculate that:(frac{1}{27} + frac{3}{27} + frac{9}{27} + frac{9}{27} = frac{1 + 3 + 9 + 9}{27} = frac{22}{27}), which is not zero.Lastly, (x = -frac{1}{3}):(left(-frac{1}{3}right)^3 + left(-frac{1}{3}right)^2 + left(-frac{1}{3}right) + frac{1}{3} = -frac{1}{27} + frac{1}{9} - frac{1}{3} + frac{1}{3})Simplifying:(-frac{1}{27} + frac{3}{27} - frac{9}{27} + frac{9}{27} = frac{-1 + 3 - 9 + 9}{27} = frac{2}{27}), which is still not zero.So, none of the possible rational roots are actual roots of the equation. That means either the equation has irrational roots or complex roots, or perhaps it can be factored in some other way.Since factoring doesn't seem straightforward, maybe I can try to use the method of depressed cubic or Cardano's formula. I remember that for a general cubic equation of the form (x^3 + ax^2 + bx + c = 0), we can make a substitution to eliminate the (x^2) term. The substitution is (x = y - frac{a}{3}). In this case, (a = 1), so the substitution would be (x = y - frac{1}{3}).Let me try that substitution. Let (x = y - frac{1}{3}). Then, I need to express the original equation in terms of (y).First, compute (x = y - frac{1}{3}), so:(x + frac{1}{3} = y)Now, let's compute each term of the original equation in terms of (y):1. (x^3 = left(y - frac{1}{3}right)^3 = y^3 - y^2 cdot frac{1}{3} cdot 3 + y cdot left(frac{1}{3}right)^2 cdot 3 - left(frac{1}{3}right)^3)Wait, that seems complicated. Maybe it's better to expand it step by step.(left(y - frac{1}{3}right)^3 = y^3 - 3y^2 cdot frac{1}{3} + 3y cdot left(frac{1}{3}right)^2 - left(frac{1}{3}right)^3)Simplify each term:- (y^3)- (-3y^2 cdot frac{1}{3} = -y^2)- (3y cdot frac{1}{9} = frac{1}{3}y)- (-frac{1}{27})So, (x^3 = y^3 - y^2 + frac{1}{3}y - frac{1}{27})Next, (x^2 = left(y - frac{1}{3}right)^2 = y^2 - frac{2}{3}y + frac{1}{9})Then, (x = y - frac{1}{3})Now, substitute all these back into the original equation:(x^3 + x^2 + x + frac{1}{3} = 0)Becomes:(left(y^3 - y^2 + frac{1}{3}y - frac{1}{27}right) + left(y^2 - frac{2}{3}y + frac{1}{9}right) + left(y - frac{1}{3}right) + frac{1}{3} = 0)Now, let's combine like terms:1. (y^3): Only one term, so (y^3)2. (y^2): (-y^2 + y^2 = 0)3. (y): (frac{1}{3}y - frac{2}{3}y + y = left(frac{1}{3} - frac{2}{3} + 1right)y = left(-frac{1}{3} + 1right)y = frac{2}{3}y)4. Constants: (-frac{1}{27} + frac{1}{9} - frac{1}{3} + frac{1}{3})Let's compute the constants:- (-frac{1}{27} + frac{1}{9} = -frac{1}{27} + frac{3}{27} = frac{2}{27})- (-frac{1}{3} + frac{1}{3} = 0)- So, total constants: (frac{2}{27})Putting it all together:(y^3 + frac{2}{3}y + frac{2}{27} = 0)Hmm, so now we have a depressed cubic equation in terms of (y):(y^3 + frac{2}{3}y + frac{2}{27} = 0)I think this is the depressed cubic form (t^3 + pt + q = 0), where (p = frac{2}{3}) and (q = frac{2}{27}).Now, I recall that for a depressed cubic equation (t^3 + pt + q = 0), we can use Cardano's formula to find the roots. The formula involves finding two numbers (u) and (v) such that:(u^3 + v^3 = -q)and(u^3 v^3 = -frac{p^3}{27})Wait, let me recall the exact steps. I think it's something like:We set (y = u + v), then substitute into the equation:((u + v)^3 + p(u + v) + q = 0)Expanding ((u + v)^3):(u^3 + 3u^2 v + 3u v^2 + v^3 + p(u + v) + q = 0)Then, grouping terms:(u^3 + v^3 + 3uv(u + v) + p(u + v) + q = 0)Now, if we set (3uv = -p), then the terms involving (u + v) will cancel out. So, let's set:(3uv = -p)Which gives:(uv = -frac{p}{3})In our case, (p = frac{2}{3}), so:(uv = -frac{2}{9})Now, we also have:(u^3 + v^3 = -q)Since (q = frac{2}{27}), we have:(u^3 + v^3 = -frac{2}{27})So now, we have a system of equations:1. (u^3 + v^3 = -frac{2}{27})2. (u^3 v^3 = left(-frac{p}{3}right)^3 = left(-frac{2}{9}right)^3 = -frac{8}{729})Wait, actually, I think I might have made a mistake here. Let me double-check.From the substitution, we have:(u^3 + v^3 = -q = -frac{2}{27})and(u^3 v^3 = left(frac{p}{3}right)^3 = left(frac{2}{9}right)^3 = frac{8}{729})Wait, no, because (uv = -frac{p}{3}), so (u^3 v^3 = left(-frac{p}{3}right)^3 = -frac{p^3}{27}). So, in our case:(u^3 v^3 = -frac{(2/3)^3}{27} = -frac{8/27}{27} = -frac{8}{729})So, the two equations are:1. (u^3 + v^3 = -frac{2}{27})2. (u^3 v^3 = -frac{8}{729})Now, let me denote (u^3 = A) and (v^3 = B). Then, we have:1. (A + B = -frac{2}{27})2. (AB = -frac{8}{729})So, we can write a quadratic equation in terms of (A) and (B):(t^2 - (A + B)t + AB = 0)Which becomes:(t^2 + frac{2}{27}t - frac{8}{729} = 0)Let me solve this quadratic equation for (t):Using the quadratic formula:(t = frac{-frac{2}{27} pm sqrt{left(frac{2}{27}right)^2 - 4 cdot 1 cdot left(-frac{8}{729}right)}}{2})First, compute the discriminant:(left(frac{2}{27}right)^2 - 4 cdot 1 cdot left(-frac{8}{729}right) = frac{4}{729} + frac{32}{729} = frac{36}{729} = frac{4}{81})So, the square root of the discriminant is:(sqrt{frac{4}{81}} = frac{2}{9})Now, plug back into the quadratic formula:(t = frac{-frac{2}{27} pm frac{2}{9}}{2})Let's compute both possibilities:1. (t = frac{-frac{2}{27} + frac{2}{9}}{2} = frac{-frac{2}{27} + frac{6}{27}}{2} = frac{frac{4}{27}}{2} = frac{2}{27})2. (t = frac{-frac{2}{27} - frac{2}{9}}{2} = frac{-frac{2}{27} - frac{6}{27}}{2} = frac{-frac{8}{27}}{2} = -frac{4}{27})So, the solutions for (A) and (B) are (A = frac{2}{27}) and (B = -frac{4}{27}), or vice versa.Now, we need to find (u) and (v) such that (u^3 = A) and (v^3 = B). So, let's compute (u) and (v):1. If (A = frac{2}{27}), then (u = sqrt[3]{frac{2}{27}} = frac{sqrt[3]{2}}{3})2. If (B = -frac{4}{27}), then (v = sqrt[3]{-frac{4}{27}} = -frac{sqrt[3]{4}}{3})Alternatively, if (A = -frac{4}{27}) and (B = frac{2}{27}), then:1. (u = sqrt[3]{-frac{4}{27}} = -frac{sqrt[3]{4}}{3})2. (v = sqrt[3]{frac{2}{27}} = frac{sqrt[3]{2}}{3})In either case, (u) and (v) are real numbers, so we can take the real cube roots.Now, recall that (y = u + v). So, let's compute (y):Case 1:(y = frac{sqrt[3]{2}}{3} + left(-frac{sqrt[3]{4}}{3}right) = frac{sqrt[3]{2} - sqrt[3]{4}}{3})Case 2:(y = -frac{sqrt[3]{4}}{3} + frac{sqrt[3]{2}}{3} = frac{-sqrt[3]{4} + sqrt[3]{2}}{3})Wait, both cases give the same expression for (y), just written in a different order. So, (y = frac{sqrt[3]{2} - sqrt[3]{4}}{3})But I think I might have made a mistake here because the depressed cubic equation (y^3 + frac{2}{3}y + frac{2}{27} = 0) might have only one real root and two complex roots. Let me check.Alternatively, maybe I should consider that the cubic equation has one real root and two complex conjugate roots. So, perhaps only one real solution for (y), and the other two are complex.But for now, let's proceed with the real root.So, (y = frac{sqrt[3]{2} - sqrt[3]{4}}{3})But I can simplify (sqrt[3]{4}) as (sqrt[3]{2^2} = (sqrt[3]{2})^2). Let me denote (a = sqrt[3]{2}), then (sqrt[3]{4} = a^2).So, (y = frac{a - a^2}{3})Now, recall that (x = y - frac{1}{3}), so:(x = frac{a - a^2}{3} - frac{1}{3} = frac{a - a^2 - 1}{3})But I think this might not be the simplest form. Maybe I can factor out something.Alternatively, perhaps I can write (x = -frac{1 + a^2 - a}{3}), but I'm not sure if that helps.Wait, let me think differently. Maybe I can express (x) in terms of (a):Since (a = sqrt[3]{2}), then (a^3 = 2).Let me see if I can express (x) in terms of (a):(x = frac{a - a^2 - 1}{3})Hmm, not sure if that's helpful. Maybe I can rationalize the denominator or find a better expression.Alternatively, perhaps I can write (x = -frac{1}{1 + a}), but let me check:Let me assume (x = -frac{1}{1 + a}), then:(x = -frac{1}{1 + sqrt[3]{2}})Let me see if this satisfies the original equation.Compute (x^3 + x^2 + x + frac{1}{3}):First, (x = -frac{1}{1 + a}), where (a = sqrt[3]{2}).Compute (x^3):(left(-frac{1}{1 + a}right)^3 = -frac{1}{(1 + a)^3})Compute (x^2):(left(-frac{1}{1 + a}right)^2 = frac{1}{(1 + a)^2})Compute (x):(-frac{1}{1 + a})So, the equation becomes:(-frac{1}{(1 + a)^3} + frac{1}{(1 + a)^2} - frac{1}{1 + a} + frac{1}{3} = 0)Let me find a common denominator, which is ((1 + a)^3):(-frac{1}{(1 + a)^3} + frac{1 + a}{(1 + a)^3} - frac{(1 + a)^2}{(1 + a)^3} + frac{(1 + a)^3}{3(1 + a)^3} = 0)Simplify each term:1. (-frac{1}{(1 + a)^3})2. (frac{1 + a}{(1 + a)^3})3. (-frac{(1 + a)^2}{(1 + a)^3} = -frac{1 + 2a + a^2}{(1 + a)^3})4. (frac{(1 + a)^3}{3(1 + a)^3} = frac{1}{3})Now, combine all terms over the common denominator:(-frac{1}{(1 + a)^3} + frac{1 + a}{(1 + a)^3} - frac{1 + 2a + a^2}{(1 + a)^3} + frac{1}{3} = 0)Combine the first three terms:(left(-1 + (1 + a) - (1 + 2a + a^2)right) / (1 + a)^3 + frac{1}{3} = 0)Simplify the numerator:-1 + 1 + a -1 - 2a - a^2 = (-1 + 1 -1) + (a - 2a) + (-a^2) = (-1) + (-a) + (-a^2) = -1 - a - a^2So, the equation becomes:(frac{-1 - a - a^2}{(1 + a)^3} + frac{1}{3} = 0)Now, let's compute ((1 + a)^3):((1 + a)^3 = 1 + 3a + 3a^2 + a^3)But since (a^3 = 2), this becomes:(1 + 3a + 3a^2 + 2 = 3 + 3a + 3a^2 = 3(1 + a + a^2))So, ((1 + a)^3 = 3(1 + a + a^2))Now, substitute back into the equation:(frac{-1 - a - a^2}{3(1 + a + a^2)} + frac{1}{3} = 0)Simplify the first term:(frac{-(1 + a + a^2)}{3(1 + a + a^2)} = -frac{1}{3})So, the equation becomes:(-frac{1}{3} + frac{1}{3} = 0)Which is indeed true. Therefore, (x = -frac{1}{1 + sqrt[3]{2}}) is a solution to the original equation.Now, since it's a cubic equation, there should be two more roots. But since the original equation has real coefficients, the other roots could be real or complex conjugates. However, given the form of the equation, it's likely that there is one real root and two complex roots.But for the purpose of this problem, I think the real root is sufficient, and the other roots might be complex, which are not required here.So, the solution to the equation (x^3 + x^2 + x + frac{1}{3} = 0) is:(x = -frac{1}{1 + sqrt[3]{2}})I think that's the final answer.

💡First, I need to determine the initial number of pigs in the barn.I know that after 86 pigs joined, the total number of pigs became 150.To find the initial number of pigs, I can subtract the number of pigs that joined from the total number of pigs now.So, I'll subtract 86 from 150.This calculation will give me the number of pigs that were originally in the barn.