💡Okay, so I have this problem about two decks of playing cards. Each deck has a specific order: first the Big Joker, then the Little Joker, followed by the suits of Spades, Hearts, Diamonds, and Clubs. Each suit has cards numbered from 1 to 13, which includes numbers 1 through 10, and then Jack (J), Queen (Q), and King (K). The person stacks these two decks on top of each other, making a total of 108 cards (since each deck has 54 cards: 2 Jokers + 4 suits × 13 cards each). Then, they start discarding cards in a specific pattern: discard the first card, place the second card at the bottom of the deck, discard the third card, place the fourth card at the bottom, and so on. This process continues until only one card remains. The question is, what is that last remaining card?Alright, let me break this down. First, I need to understand the structure of the deck. Each deck has 54 cards, so two decks make 108. The order within each deck is Big Joker, Little Joker, followed by Spades (1-13), Hearts (1-13), Diamonds (1-13), and Clubs (1-13). So, when stacking two decks, we have:1. Big Joker (Deck 1)2. Little Joker (Deck 1)3. Spades 1 (Deck 1)4. Spades 2 (Deck 1)...54. Clubs 13 (Deck 1)55. Big Joker (Deck 2)56. Little Joker (Deck 2)57. Spades 1 (Deck 2)58. Spades 2 (Deck 2)...108. Clubs 13 (Deck 2)Now, the discarding process: starting from the top, discard the first card, place the second card at the bottom, discard the third, place the fourth at the bottom, etc. This is similar to the Josephus problem, where people are standing in a circle and every second person is eliminated until only one remains. However, in this case, instead of eliminating every second person, we're discarding every second card and moving the next one to the bottom.I remember that in the Josephus problem, the position of the last remaining person can be determined using a formula, especially when the number of people is a power of 2. The formula is J(n) = 2*(n - 2^m) + 1, where 2^m is the largest power of 2 less than n. But in our case, we have 108 cards, which is not a power of 2. The closest power of 2 less than 108 is 64 (2^6). So, if we consider the Josephus problem analogy, the last remaining card would be at position 2*(108 - 64) + 1 = 2*44 + 1 = 89. But wait, in the Josephus problem, we eliminate every second person, but in our case, we're discarding every second card and moving the next one to the bottom. I'm not sure if the same formula applies here.Let me think differently. Maybe I can simulate the process step by step, but with 108 cards, that would be too time-consuming. Instead, perhaps I can find a pattern or a mathematical approach to determine the last remaining card.I recall that when dealing with such discarding processes, the position of the last remaining card can be found by finding the largest power of 2 less than the total number of cards and then using that to calculate the position. So, for 108 cards, the largest power of 2 less than 108 is 64. Then, the position would be 2*(108 - 64) + 1 = 89, as I thought earlier.But wait, in our case, we're not eliminating every second card in a circle, but rather discarding and moving cards in a specific way. Maybe the formula still applies, but I need to verify it.Alternatively, I can think of the process as repeatedly removing every second card and moving the next one to the bottom. This is similar to a queue where we dequeue the first card, enqueue the second, dequeue the third, enqueue the fourth, and so on. This process continues until only one card remains.To model this, I can represent the deck as a queue and simulate the process:1. Initialize the queue with all 108 cards in order.2. While the queue has more than one card: a. Dequeue the first card (discard it). b. Dequeue the next card and enqueue it at the end of the queue.3. The last remaining card is the answer.However, simulating this for 108 cards manually would be tedious. Maybe I can find a pattern or a formula that can help me determine the last remaining card without simulating all steps.I found a resource that discusses this exact problem. It mentions that for a deck of size n, the last remaining card can be found using the formula:Last card = 2*(n - 2^floor(log2(n))) + 1But wait, this is similar to the Josephus problem formula. However, in our case, the process is slightly different because we're not eliminating every second card but rather discarding and moving cards. So, I'm not sure if this formula applies directly.Let me test this formula with a smaller number of cards to see if it works. Suppose we have 4 cards:1. Card A2. Card B3. Card C4. Card DFollowing the process:- Discard A, move B to the bottom: Queue becomes [C, D, B]- Discard C, move D to the bottom: Queue becomes [B, D]- Discard B, move D to the bottom: Queue becomes [D]So, the last remaining card is D, which is card 4. According to the formula, n=4, 2^floor(log2(4))=4, so Last card = 2*(4 - 4) + 1 = 1. But the actual last card is 4, so the formula doesn't apply here.Hmm, that's confusing. Maybe the formula needs to be adjusted for this specific discarding process.Another approach is to recognize that each time we go through the deck, we're effectively halving the number of cards, but with some offset. Maybe I can use recursion to solve this.Let me define f(n) as the position of the last remaining card in a deck of size n. When we discard every second card and move the next one to the bottom, the position of the last card can be expressed recursively.After the first pass, we discard half of the cards, and the remaining cards are in a new order. The position of the last card in the new deck can be related to the position in the original deck.But I'm not sure about the exact recursion formula. Maybe I can look for a pattern with smaller n.Let's try n=1: Last card is 1.n=2:- Discard 1, move 2 to the bottom: Queue becomes [2]- Last card is 2.n=3:- Discard 1, move 2 to the bottom: Queue becomes [3, 2]- Discard 3, move 2 to the bottom: Queue becomes [2]- Last card is 2.n=4:- As before, last card is 4.n=5:- Discard 1, move 2 to the bottom: [3,4,5,2]- Discard 3, move 4 to the bottom: [5,2,4]- Discard 5, move 2 to the bottom: [4,2]- Discard 4, move 2 to the bottom: [2]- Last card is 2.n=6:- Discard 1, move 2 to the bottom: [3,4,5,6,2]- Discard 3, move 4 to the bottom: [5,6,2,4]- Discard 5, move 6 to the bottom: [2,4,6]- Discard 2, move 4 to the bottom: [6,4]- Discard 6, move 4 to the bottom: [4]- Last card is 4.n=7:- Discard 1, move 2 to the bottom: [3,4,5,6,7,2]- Discard 3, move 4 to the bottom: [5,6,7,2,4]- Discard 5, move 6 to the bottom: [7,2,4,6]- Discard 7, move 2 to the bottom: [4,6,2]- Discard 4, move 6 to the bottom: [2,6]- Discard 2, move 6 to the bottom: [6]- Last card is 6.n=8:- Following the same process, I think the last card would be 8.Looking at the results:n | Last Card---|---1 | 12 | 23 | 24 | 45 | 26 | 47 | 68 | 8Hmm, interesting. It seems that for powers of 2, the last card is n itself. For numbers between 2^k and 2^(k+1), the last card is 2*(n - 2^k). Let's test this:For n=5, 2^2=4, so 2*(5-4)=2. Correct.For n=6, 2*(6-4)=4. Correct.For n=7, 2*(7-4)=6. Correct.For n=9, let's predict:n=9, 2^3=8, so 2*(9-8)=2. Let's simulate:n=9:- Discard 1, move 2 to the bottom: [3,4,5,6,7,8,9,2]- Discard 3, move 4 to the bottom: [5,6,7,8,9,2,4]- Discard 5, move 6 to the bottom: [7,8,9,2,4,6]- Discard 7, move 8 to the bottom: [9,2,4,6,8]- Discard 9, move 2 to the bottom: [4,6,8,2]- Discard 4, move 6 to the bottom: [8,2,6]- Discard 8, move 2 to the bottom: [6,2]- Discard 6, move 2 to the bottom: [2]- Last card is 2. Correct.So, the formula seems to hold: Last card = 2*(n - 2^floor(log2(n))).But wait, in our earlier example with n=4, 2*(4 - 4)=0, but the last card was 4. So, maybe the formula is Last card = 2*(n - 2^floor(log2(n))) if n > 2^floor(log2(n)), else n.Yes, that makes sense. So, for n=4, since n=2^2, the last card is n itself.So, applying this formula to n=108:First, find the largest power of 2 less than 108. 2^6=64, 2^7=128. So, 2^6=64.Then, Last card = 2*(108 - 64) = 2*44=88.But wait, in our earlier small n examples, the last card was 2*(n - 2^k). But in the case where n=4, it was 4, which is 2*(4-4)=0, but we adjusted it to n itself when n is a power of 2.So, for n=108, since 108 > 64, Last card = 2*(108 - 64)=88.But let's verify this with n=8:n=8, 2*(8 - 8)=0, but the last card is 8. So, the formula should be:If n is a power of 2, Last card = n.Else, Last card = 2*(n - 2^floor(log2(n))).So, for n=108, it's 2*(108 - 64)=88.But wait, in our earlier example with n=6, 2*(6 - 4)=4, which was correct. For n=7, 2*(7 - 4)=6, correct. For n=9, 2*(9 - 8)=2, correct.So, applying this to n=108, the last card would be 88.But wait, in the initial problem, the deck has two decks of 54 cards each, so the total is 108. The order is:Deck 1:1. Big Joker2. Little Joker3. Spades 14. Spades 2...54. Clubs 13Deck 2:55. Big Joker56. Little Joker57. Spades 158. Spades 2...108. Clubs 13So, card number 88 is in Deck 2. Let's find out which card it is.Deck 2 starts at position 55. So, card 88 is 88 - 54 = 34th card in Deck 2.Now, Deck 2 has the same structure as Deck 1: Big Joker (55), Little Joker (56), Spades 1-13 (57-69), Hearts 1-13 (70-82), Diamonds 1-13 (83-95), Clubs 1-13 (96-108).So, card 34 in Deck 2 is:55: Big Joker56: Little Joker57-69: Spades 1-13 (13 cards)70-82: Hearts 1-13 (13 cards)83-95: Diamonds 1-13 (13 cards)96-108: Clubs 1-13 (13 cards)So, card 34 in Deck 2:First 2 cards: 55-56Next 13: 57-69 (Spades)Next 13: 70-82 (Hearts)Next 13: 83-95 (Diamonds)So, 34 - 2 (Jokers) -13 (Spades) -13 (Hearts) = 34 - 28 = 6.So, card 34 in Deck 2 is the 6th card in Diamonds, which is Diamonds 6.Therefore, the last remaining card is Diamonds 6.Wait, but according to our formula, the last card is 88, which is Diamonds 6 in Deck 2.But let me double-check the calculation:Deck 2 has 54 cards, starting at position 55.Card 88 is 88 - 54 = 34 in Deck 2.Deck 2 structure:55: Big Joker56: Little Joker57-69: Spades 1-1370-82: Hearts 1-1383-95: Diamonds 1-1396-108: Clubs 1-13So, card 34 in Deck 2:First 2: 55-56Next 13: 57-69 (Spades)Next 13: 70-82 (Hearts)Next 13: 83-95 (Diamonds)So, 34 - 2 (Jokers) = 3232 -13 (Spades) = 1919 -13 (Hearts) = 6So, the 6th card in Diamonds is Diamonds 6.Yes, that seems correct.But wait, in the initial problem, the user mentioned that the last remaining card is Diamonds 6, but in their thought process, they arrived at 88, which is Diamonds 6.However, in the initial problem, the user's thought process had a different calculation:"108 - 64 = 44.This means we need to discard 44 cards, which implies that when we follow the rule to discard 44 cards, the 88th card will be placed at the bottom of the remaining 64 cards. This card will also be the last remaining card.Therefore, this card is: 88 - 54 - 2 - 26 = 6 (54 for the first deck, 2 for the Big and Little Joker of the second deck, 26 for the Spades and Hearts of the second deck). Hence, the card is the Diamond 6."Wait, in their calculation, they subtracted 54 (Deck 1) + 2 (Jokers in Deck 2) + 26 (Spades and Hearts in Deck 2) from 88, getting 6, which is Diamonds 6.But in my calculation, I considered Deck 2 as starting at 55, so 88 is 34 in Deck 2, which is Diamonds 6.So, both methods lead to the same conclusion: Diamonds 6.But I want to make sure that the formula I used (Last card = 2*(n - 2^floor(log2(n)))) is correct for this specific discarding process.In the Josephus problem, the formula is J(n) = 2*l + 1, where l = n - 2^m, and m is the largest integer such that 2^m < n. But in our case, the process is slightly different because we're not eliminating every second card in a circle but rather discarding and moving cards in a specific way.However, through testing with smaller n, the formula seems to hold. For example:n=5: Last card=2, which is 2*(5-4)=2.n=6: Last card=4, which is 2*(6-4)=4.n=7: Last card=6, which is 2*(7-4)=6.n=9: Last card=2, which is 2*(9-8)=2.So, it seems consistent.Therefore, applying this formula to n=108:Largest power of 2 less than 108 is 64.Last card=2*(108-64)=88.Then, card 88 is Diamonds 6 in Deck 2.So, the last remaining card is Diamonds 6.But to be thorough, let me simulate the process for a smaller n where I know the result.Take n=4:Cards: 1,2,3,4.Process:1. Discard 1, move 2 to the bottom: [3,4,2]2. Discard 3, move 4 to the bottom: [2,4]3. Discard 2, move 4 to the bottom: [4]4. Last card is 4.According to the formula, Last card=2*(4-4)=0, but since 4 is a power of 2, the last card is 4 itself. So, the formula holds.Another test: n=6.Cards:1,2,3,4,5,6.Process:1. Discard 1, move 2 to the bottom: [3,4,5,6,2]2. Discard 3, move 4 to the bottom: [5,6,2,4]3. Discard 5, move 6 to the bottom: [2,4,6]4. Discard 2, move 4 to the bottom: [6,4]5. Discard 6, move 4 to the bottom: [4]6. Last card is 4.According to the formula, Last card=2*(6-4)=4. Correct.Another test: n=7.Cards:1,2,3,4,5,6,7.Process:1. Discard 1, move 2 to the bottom: [3,4,5,6,7,2]2. Discard 3, move 4 to the bottom: [5,6,7,2,4]3. Discard 5, move 6 to the bottom: [7,2,4,6]4. Discard 7, move 2 to the bottom: [4,6,2]5. Discard 4, move 6 to the bottom: [2,6]6. Discard 2, move 6 to the bottom: [6]7. Last card is 6.According to the formula, Last card=2*(7-4)=6. Correct.So, the formula seems reliable.Therefore, for n=108, the last card is 88, which is Diamonds 6.But wait, in the initial problem, the user's thought process arrived at the same conclusion, but they subtracted 54 (Deck 1) + 2 (Jokers in Deck 2) + 26 (Spades and Hearts in Deck 2) from 88, getting 6, which is Diamonds 6.Yes, that's consistent with my calculation.So, the last remaining card is Diamonds 6.

💡Okay, so I have this problem where I need to find all possible values of ( x_0 ) and ( x_1 ) such that the sequence defined by the recurrence relation[x_{n+1} = frac{x_{n-1} x_n}{3x_{n-1} - 2x_n}]for ( n ge 1 ) contains infinitely many natural numbers. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I should probably write out the first few terms of the sequence to get a sense of how it behaves. Let me denote ( x_0 = a ) and ( x_1 = b ), where ( a ) and ( b ) are some real numbers. Then,- ( x_2 = frac{x_0 x_1}{3x_0 - 2x_1} = frac{a b}{3a - 2b} )- ( x_3 = frac{x_1 x_2}{3x_1 - 2x_2} )Wait, this is getting complicated quickly. Maybe I can find a pattern or transform the recurrence into something more manageable.I notice that the recurrence involves both ( x_{n-1} ) and ( x_n ), and it's a rational function. Sometimes, taking reciprocals can simplify such expressions. Let me define ( y_n = frac{1}{x_n} ). Then, the recurrence becomes:[y_{n+1} = frac{3x_{n-1} - 2x_n}{x_{n-1} x_n} = frac{3}{x_n} - frac{2}{x_{n-1}} = 3y_n - 2y_{n-1}]Oh, that's much better! So, the transformed sequence ( y_n ) satisfies the linear recurrence:[y_{n+1} = 3y_n - 2y_{n-1}]This is a second-order linear homogeneous recurrence relation with constant coefficients. To solve this, I can find the characteristic equation. Let me set ( y_n = r^n ), then substituting into the recurrence gives:[r^{n+1} = 3r^n - 2r^{n-1}]Dividing both sides by ( r^{n-1} ) (assuming ( r neq 0 )):[r^2 = 3r - 2]So, the characteristic equation is:[r^2 - 3r + 2 = 0]Solving this quadratic equation:[r = frac{3 pm sqrt{9 - 8}}{2} = frac{3 pm 1}{2}]Thus, the roots are ( r = 2 ) and ( r = 1 ). Therefore, the general solution for ( y_n ) is:[y_n = A cdot 2^n + B cdot 1^n = A cdot 2^n + B]where ( A ) and ( B ) are constants determined by the initial conditions ( y_0 ) and ( y_1 ).Now, recalling that ( y_n = frac{1}{x_n} ), so:[x_n = frac{1}{A cdot 2^n + B}]For ( x_n ) to be a natural number, ( frac{1}{A cdot 2^n + B} ) must be a natural number. That means ( A cdot 2^n + B ) must be a positive integer that divides 1. The only positive integers that divide 1 are 1 itself. Therefore, we must have:[A cdot 2^n + B = 1]for infinitely many ( n ). But wait, ( A cdot 2^n + B ) is an exponential function in ( n ). The only way this can equal 1 for infinitely many ( n ) is if ( A = 0 ) and ( B = 1 ). Because if ( A neq 0 ), then as ( n ) increases, ( A cdot 2^n ) will dominate and the expression will grow without bound, never equaling 1 again after some point.So, if ( A = 0 ) and ( B = 1 ), then ( y_n = 1 ) for all ( n ), which implies ( x_n = 1 ) for all ( n ). Therefore, the sequence is constant and equal to 1, which is a natural number. But the problem states that the sequence should contain infinitely many natural numbers, not necessarily that all terms are natural numbers. However, in this case, since all terms are 1, it trivially contains infinitely many natural numbers.But wait, is this the only possibility? Let me think. If ( A neq 0 ), then ( x_n = frac{1}{A cdot 2^n + B} ) will tend to 0 as ( n ) increases because the denominator grows exponentially. So, for large ( n ), ( x_n ) becomes very small, approaching 0. Therefore, beyond some finite ( n ), ( x_n ) will be less than 1 and not a natural number. Hence, there can only be finitely many natural numbers in the sequence if ( A neq 0 ).Therefore, the only way for the sequence to contain infinitely many natural numbers is if ( A = 0 ) and ( B = 1 ), leading to ( x_n = 1 ) for all ( n ). This means that the initial terms ( x_0 ) and ( x_1 ) must both be 1.But let me check this. If ( x_0 = 1 ) and ( x_1 = 1 ), then:- ( x_2 = frac{1 cdot 1}{3 cdot 1 - 2 cdot 1} = frac{1}{1} = 1 )- ( x_3 = frac{1 cdot 1}{3 cdot 1 - 2 cdot 1} = 1 )- And so on.Yes, the sequence remains constant at 1. So, indeed, all terms are natural numbers, hence infinitely many.But wait, what if ( x_0 ) and ( x_1 ) are different? Suppose ( x_0 = k ) and ( x_1 = k ) for some natural number ( k ). Then, let's compute ( x_2 ):[x_2 = frac{k cdot k}{3k - 2k} = frac{k^2}{k} = k]Similarly, ( x_3 = frac{k cdot k}{3k - 2k} = k ), and so on. So, if ( x_0 = x_1 = k ), then the sequence remains constant at ( k ), which is a natural number. Therefore, in this case, the sequence contains infinitely many natural numbers.Wait a minute, so my earlier conclusion that ( x_n = 1 ) is the only possibility might be too restrictive. It seems that any constant sequence where ( x_0 = x_1 = k ) for some natural number ( k ) will satisfy the condition.But let me verify this with the transformed sequence. If ( x_n = k ) for all ( n ), then ( y_n = frac{1}{k} ) for all ( n ). Plugging into the recurrence for ( y_n ):[y_{n+1} = 3y_n - 2y_{n-1}]If ( y_n = frac{1}{k} ) for all ( n ), then:[frac{1}{k} = 3 cdot frac{1}{k} - 2 cdot frac{1}{k}][frac{1}{k} = frac{3}{k} - frac{2}{k} = frac{1}{k}]Which holds true. So, any constant sequence ( x_n = k ) satisfies the recurrence, and if ( k ) is a natural number, then all terms are natural numbers, hence infinitely many.But the problem asks for all possible values of ( x_0 ) and ( x_1 ) such that the sequence contains infinitely many natural numbers. So, not necessarily that all terms are natural numbers, but that infinitely many terms are natural numbers.Wait, so maybe there are other possibilities where the sequence isn't constant, but still has infinitely many natural numbers. Let me think about that.Suppose ( x_0 ) and ( x_1 ) are such that the sequence ( x_n ) is not constant, but still hits natural numbers infinitely often. Is that possible?Given the general solution for ( y_n ):[y_n = A cdot 2^n + B]Which implies:[x_n = frac{1}{A cdot 2^n + B}]For ( x_n ) to be a natural number, ( A cdot 2^n + B ) must equal 1 for infinitely many ( n ). But as I thought earlier, unless ( A = 0 ), ( A cdot 2^n + B ) will grow without bound, making ( x_n ) approach 0, which can't be a natural number beyond a certain point.Therefore, the only way for ( x_n ) to be a natural number infinitely often is if ( A = 0 ) and ( B = 1 ), leading to ( x_n = 1 ) for all ( n ). Hence, the sequence must be constant at 1.But wait, earlier I considered that if ( x_0 = x_1 = k ), then the sequence remains constant at ( k ). So, does that mean that for any natural number ( k ), setting ( x_0 = x_1 = k ) will result in a constant sequence with all terms equal to ( k ), hence containing infinitely many natural numbers?Yes, that seems to be the case. So, in fact, any pair ( x_0 = x_1 = k ) where ( k ) is a natural number will satisfy the condition.But let me check with ( k = 2 ). Let ( x_0 = 2 ) and ( x_1 = 2 ). Then,- ( x_2 = frac{2 cdot 2}{3 cdot 2 - 2 cdot 2} = frac{4}{6 - 4} = frac{4}{2} = 2 )- ( x_3 = frac{2 cdot 2}{3 cdot 2 - 2 cdot 2} = 2 )- And so on.Yes, it works. Similarly, for ( k = 3 ), it would also work.Therefore, the possible values of ( x_0 ) and ( x_1 ) are any pair where both are equal to the same natural number ( k ). So, ( x_0 = x_1 = k ) for some natural number ( k ).But wait, the problem says "contains infinitely many natural numbers," not necessarily that all terms are natural numbers. So, could there be non-constant sequences where infinitely many terms are natural numbers?Let me consider that. Suppose ( x_0 ) and ( x_1 ) are chosen such that ( A neq 0 ), but somehow ( A cdot 2^n + B = 1 ) for infinitely many ( n ). But as I thought earlier, ( A cdot 2^n + B ) is an exponential function, which can only equal 1 at most once unless ( A = 0 ). Because if ( A neq 0 ), ( 2^n ) grows exponentially, so ( A cdot 2^n + B ) will cross 1 at most once, meaning ( x_n ) can be 1 at most once. Therefore, beyond that point, ( x_n ) will be less than 1 and not a natural number.Hence, the only way to have infinitely many natural numbers in the sequence is if the sequence is constant, i.e., ( x_n = k ) for all ( n ), where ( k ) is a natural number. Therefore, the initial terms must satisfy ( x_0 = x_1 = k ).Therefore, the possible values of ( x_0 ) and ( x_1 ) are pairs where both are equal to the same natural number ( k ).

💡Alright, so I'm trying to figure out if person A is guilty of the theft at the warehouse. Let's see what the information given is:1. The criminals took the stolen items in a car.2. The crime was committed by one or more of three individuals: A, B, or C.3. C never commits a crime without A.4. B knows how to drive.Okay, so first, the crime involved a car, which means someone had to drive it. B knows how to drive, so that's a possibility. But it could also be A or C driving. Now, it's possible that the crime was committed by one person, two people, or all three.Let's consider the possibilities step by step.**Case 1: Only one person is guilty.**- If only A is guilty, then A must have driven the car. There's no information that says A can't drive, so this is possible.- If only B is guilty, then B would have driven the car. Since B knows how to drive, this is also possible.- If only C is guilty, then C would have had to drive the car. But wait, point 3 says C never commits a crime without A. So, if C is guilty, A must also be guilty. Therefore, C cannot be the only guilty person.So, in the case of only one person being guilty, it could be either A or B.**Case 2: Two people are guilty.**- If A and B are guilty, then either A or B could have driven the car. Since B knows how to drive, this works.- If A and C are guilty, then C is involved, but according to point 3, C never commits a crime without A, which is already satisfied here. So, A and C together could have driven the car.- If B and C are guilty, then B could have driven the car, and C is involved. But wait, point 3 says C never commits a crime without A. So, if C is guilty, A must also be guilty. Therefore, B and C cannot be guilty without A.So, in the case of two people being guilty, the possibilities are A and B, or A and C.**Case 3: All three are guilty.**- If A, B, and C are all guilty, then any of them could have driven the car. Since B knows how to drive, this works. Also, C is involved, but since A is also involved, point 3 is satisfied.So, in the case of all three being guilty, it's possible.Now, let's summarize:- If only one person is guilty, it could be A or B.- If two people are guilty, it could be A and B, or A and C.- If all three are guilty, that's also possible.But the question is specifically asking if A is guilty. From the above analysis, in all scenarios except when only B is guilty, A is involved. However, if only B is guilty, then A is not involved. But wait, let's think about that.If only B is guilty, then B would have driven the car alone. But the crime was committed by one or more of A, B, or C. So, it's possible that only B did it. But let's revisit point 3: C never commits a crime without A. Since C is not involved in this case, it doesn't affect anything.But the question is about A's guilt. So, in the scenario where only B is guilty, A is not guilty. However, in all other scenarios, A is guilty.But the problem doesn't specify that multiple people were involved. It just says one or more. So, it's possible that only B is guilty, which would mean A is not guilty. But is there any information that contradicts this?Wait, point 3 says C never commits a crime without A. But if only B is guilty, C is not involved, so that point doesn't come into play. Therefore, it's possible that only B is guilty, and A is not involved.But let's think again. The crime was committed by one or more of A, B, or C. So, it's possible that only B did it, but is there any reason to believe that A must be involved?If only B is guilty, then B could have driven the car alone. There's no information that says that B couldn't have done it alone. So, it's possible that A is not guilty.But wait, let's consider the fact that the criminals took the stolen items in a car. If only B is guilty, then B took the items in a car. But if B is guilty alone, then A is not guilty. However, if B is guilty with someone else, then A might be involved.But the question is asking if A is guilty, not necessarily if A is the only guilty person. So, in some scenarios, A is guilty, and in others, A is not.But the question is phrased as "Is A guilty?" without specifying any additional information. So, based on the given information, it's possible that A is guilty, but it's also possible that A is not guilty if only B is guilty.However, let's think about the fact that C never commits a crime without A. So, if C is involved, A must be involved. But if C is not involved, then A might not be involved. But the problem doesn't say anything about A's driving skills or any other constraints.Wait, but the crime was committed by one or more of A, B, or C. So, it's possible that only B is guilty, and A is not involved. Therefore, A is not necessarily guilty.But let's think again. If only B is guilty, then B could have driven the car alone. There's no information that says that B couldn't have done it alone. So, it's possible that A is not guilty.But wait, the problem says "one or more of three individuals: A, B, or C." So, it's possible that only B is guilty, and A is not involved. Therefore, A is not necessarily guilty.But wait, the problem is asking "Is A guilty?" not "Is A necessarily guilty?" So, based on the information, it's possible that A is guilty, but it's also possible that A is not guilty.But let's think again. If only B is guilty, then A is not guilty. If A and B are guilty, then A is guilty. If A and C are guilty, then A is guilty. If all three are guilty, then A is guilty.So, in some scenarios, A is guilty, and in one scenario, A is not guilty. Therefore, based on the given information, we cannot definitively say that A is guilty, because it's possible that only B is guilty.But wait, let's think about the fact that the crime was committed by one or more of A, B, or C. So, it's possible that only B is guilty, but is there any information that contradicts this?Point 3 says C never commits a crime without A. But if only B is guilty, C is not involved, so that point doesn't come into play. Therefore, it's possible that only B is guilty, and A is not involved.Therefore, based on the given information, we cannot definitively conclude that A is guilty, because it's possible that only B is guilty.But wait, the problem is asking "Is A guilty?" not "Is A necessarily guilty?" So, based on the information, it's possible that A is guilty, but it's also possible that A is not guilty.But the problem is asking "Is A guilty?" which implies whether A is definitely guilty or not. Given that it's possible that only B is guilty, A is not necessarily guilty.Therefore, the answer is that we cannot definitively conclude that A is guilty based on the given information.

💡Okay, so I have this equation to solve: |2y - 44| + |y - 24| = |3y - 66|. Hmm, absolute value equations can be tricky because they involve multiple cases depending on the value inside the absolute value. Let me try to break this down step by step.First, I notice that all the terms inside the absolute values are linear in y. Maybe I can simplify them or express them in terms of a common variable. Let me see:Looking at |2y - 44|, I can factor out a 2, so that becomes 2|y - 22|. Similarly, |3y - 66| can be factored as 3|y - 22|. So, substituting these back into the equation, it becomes:2|y - 22| + |y - 24| = 3|y - 22|Hmm, that's interesting. Now, I can subtract 2|y - 22| from both sides to simplify:|y - 24| = |y - 22|So, now the equation is |y - 24| = |y - 22|. This means that the distance of y from 24 is equal to the distance of y from 22. On the number line, this happens when y is exactly halfway between 22 and 24. The midpoint between 22 and 24 is (22 + 24)/2 = 23. So, y = 23.Wait, but let me make sure I didn't skip any steps or make any mistakes. Maybe I should check if y = 23 satisfies the original equation.Plugging y = 23 into the original equation:Left side: |2*23 - 44| + |23 - 24| = |46 - 44| + | -1 | = |2| + | -1 | = 2 + 1 = 3Right side: |3*23 - 66| = |69 - 66| = |3| = 3Yes, both sides equal 3, so y = 23 is indeed a solution.But wait, are there any other solutions? Absolute value equations can sometimes have more than one solution. Let me think about this.The equation |y - 24| = |y - 22| implies that y is equidistant from 24 and 22. On the number line, the only point equidistant from 22 and 24 is 23. So, there are no other solutions. Therefore, y = 23 is the only solution.I think that's it. I don't see any other possibilities, and checking the solution confirms it works. So, y must be 23.

💡First, I need to determine the total capacity of bucket B. The problem states that pouring 20% of the water from bucket A fills ( frac{2}{5} ) of bucket B, and an additional 6 kg is needed to fill bucket B completely.Let’s denote the total capacity of bucket B as ( B ). According to the problem, ( frac{2}{5}B ) is filled by 20% of the water from bucket A, and the remaining ( frac{3}{5}B ) corresponds to the additional 6 kg needed to fill bucket B.To find ( B ), I can set up the equation:[frac{3}{5}B = 6 text{ kg}]Solving for ( B ):[B = 6 text{ kg} times frac{5}{3} = 10 text{ kg}]Now that I know the total capacity of bucket B is 10 kg, I can determine how much water was originally in bucket A. Since 20% of the water from bucket A fills ( frac{2}{5} ) of bucket B, which is 4 kg, I can set up the equation:[0.20 times A = 4 text{ kg}]Solving for ( A ):[A = frac{4 text{ kg}}{0.20} = 20 text{ kg}]Therefore, the original amount of water in bucket A was 20 kg.

💡Alright, so I've got this problem about a truck delivering four bags of cement, and a worker moving them around. The worker can carry one bag at a time, either from the truck to the gate or from the gate to the shed. The worker chooses randomly between these two options each time with a probability of 0.5. Eventually, all the bags end up in the shed. Part (a) asks for the probability that the bags end up in the shed in the reverse order compared to how they were placed in the truck. Hmm, okay. So, if the bags are stacked in the truck as Bag 1 at the bottom, Bag 2, Bag 3, and Bag 4 at the top, we want the probability that they end up in the shed as Bag 4 at the bottom, Bag 3, Bag 2, and Bag 1 at the top.To reverse the order, the worker must carry each bag directly from the truck to the shed without any intermediate stops at the gate. That means every time the worker has a choice, they must choose to carry the bag to the shed rather than the gate. Since the worker has a 50% chance each time, and there are three opportunities to make the wrong choice (since the first bag doesn't have a choice, it's just carried to the shed), the probability would be (1/2)^3 = 1/8.Wait, let me think again. There are four bags, so the worker has to make three decisions after the first bag is moved. Each decision has a 50% chance of being correct (i.e., moving the bag to the shed instead of the gate). So, the probability of making all three correct choices is indeed (1/2)^3 = 1/8.Okay, that seems right.Now, part (b) asks for the probability that the bag that was second from the bottom in the truck ends up as the bottom bag in the shed. So, Bag 2 needs to end up at the bottom of the shed.For Bag 2 to be at the bottom of the shed, all the bags above it in the truck (Bags 3 and 4) must be moved to the gate first. Then, when Bag 2 is reached, it must be moved directly to the shed. After that, Bags 3 and 4 can be moved from the gate to the shed on top of Bag 2.So, the worker must first move Bags 3 and 4 to the gate. Each time, the worker has a 50% chance to move a bag to the gate. Since there are two bags to move, the probability is (1/2)^2 = 1/4.Once Bags 3 and 4 are at the gate, the worker must then move Bag 2 directly to the shed. This is a single choice with a 50% probability, so we multiply by 1/2.Therefore, the total probability is (1/4) * (1/2) = 1/8.Wait, but is there another way this could happen? What if the worker moves some bags back and forth between the gate and the shed? Could that still result in Bag 2 being at the bottom?Hmm, actually, no. Because once a bag is moved to the shed, it's placed on top of the existing stack. So, if Bag 2 is to be at the bottom, it must be the first bag moved to the shed. But since it's the second bag from the bottom, it's underneath Bags 3 and 4. So, the only way for Bag 2 to be moved to the shed first is if Bags 3 and 4 are moved to the gate first.Therefore, the initial reasoning holds, and the probability is indeed 1/8.Yeah, I think that's correct.