💡Okay, so I have this problem where I have a sequence of numbers from 00 to 99, right? Each term is a two-digit number, and they're arranged in order: 00, 01, 02, ..., up to 99. Now, the task is to rearrange these terms such that each term is obtained from the previous one by either increasing or decreasing one of its digits by 1. For example, 29 can be followed by 19, 39, or 28, but not by 30 or 20 because that would require changing both digits or changing a digit by more than 1.The question is asking for the maximal number of terms that could remain on their original places after such a rearrangement. So, I need to figure out how many numbers can stay where they are without violating the rule that each subsequent number must differ by exactly 1 in one of its digits.First, let me try to understand the constraints better. Each term must be adjacent to the previous one by changing one digit up or down by 1. This reminds me of something like a knight's tour on a chessboard, where each move is restricted in a certain way. Maybe it's similar to a graph where each node is connected to its neighbors based on the digit change rule.But in this case, instead of moving on a chessboard, we're moving through two-digit numbers where each digit can be from 0 to 9. So, each number is a node, and edges connect numbers that can be reached by changing one digit by 1. For example, 00 is connected to 01 and 10, right? Similarly, 01 is connected to 00, 02, and 11.Now, the problem is about rearranging the sequence so that each consecutive term is connected by an edge in this graph. So, it's like finding a Hamiltonian path through this graph, but with the additional constraint that as many terms as possible remain in their original positions.Wait, Hamiltonian path is a path that visits every node exactly once. So, in this case, the rearranged sequence needs to be a Hamiltonian path in this graph. And we want as many nodes as possible to stay in their original positions.So, the question is, what's the maximum number of nodes that can remain fixed in a Hamiltonian path of this graph?Hmm, okay. Maybe I can model this as a graph and then analyze the properties of such a graph.First, let's think about the structure of the graph. Each node is a two-digit number, so there are 100 nodes in total. Each node is connected to up to four other nodes: increasing or decreasing the first digit or the second digit by 1, provided that the resulting number is still a valid two-digit number (i.e., digits don't go below 0 or above 9).For example, 00 is connected to 01 and 10. 01 is connected to 00, 02, and 11. 11 is connected to 01, 10, 12, and 21. So, the number of connections varies depending on the position. Corner nodes like 00, 09, 90, and 99 have only two connections each, edge nodes (but not corners) have three connections, and inner nodes have four connections.So, the graph is a 10x10 grid where each node is connected to its neighbors up, down, left, and right, except for the edges and corners which have fewer connections.Now, in such a grid graph, it's known that a Hamiltonian path exists. In fact, there are many Hamiltonian paths possible. But here, the twist is that we want as many nodes as possible to remain in their original positions.So, how can we maximize the number of fixed points in a Hamiltonian path? That is, how many nodes can stay in their original place while still allowing the rest of the nodes to form a path that visits every node exactly once.I think this relates to the concept of a derangement, but in this case, it's not exactly a derangement because we're not necessarily permuting all elements, but rather finding a permutation (the Hamiltonian path) with as many fixed points as possible.Wait, actually, in this case, the permutation is the Hamiltonian path, which is a specific kind of permutation where each element is visited exactly once in a sequence where consecutive elements are adjacent in the graph.So, the problem reduces to finding a Hamiltonian path in the 10x10 grid graph with the maximum number of nodes fixed in their original positions.I need to figure out what is the maximum number of fixed points possible in such a path.Let me think about the properties of the grid graph. It's a bipartite graph. In a bipartite graph, the nodes can be divided into two sets such that every edge connects a node from one set to the other.In the case of a grid graph, the bipartition can be based on the parity of the sum of the coordinates. For example, in a chessboard, the black and white squares form a bipartition. Similarly, here, each node (i,j) can be assigned to set A if i+j is even, and set B if i+j is odd.So, in this case, the grid graph is bipartite with two sets: even and odd parity nodes.In a Hamiltonian path in a bipartite graph, the path must alternate between the two sets. So, if the path starts in set A, the next node is in set B, then set A, and so on.Given that, the maximum length of a Hamiltonian path is 100 nodes, alternating between the two sets.Now, in our original sequence, the numbers are arranged from 00 to 99. Let's see how the original sequence relates to the bipartition.The original sequence is ordered by increasing the second digit first, then the first digit. So, 00, 01, 02, ..., 09, 10, 11, ..., 99.Let me check the parity of each term in the original sequence. For example:- 00: 0+0=0, even- 01: 0+1=1, odd- 02: 0+2=2, even- 03: 0+3=3, odd- ...- 09: 0+9=9, odd- 10: 1+0=1, odd- 11: 1+1=2, even- 12: 1+2=3, odd- ...- 19: 1+9=10, even- 20: 2+0=2, even- 21: 2+1=3, odd- ...So, the original sequence alternates between even and odd parity, except when moving from 09 to 10, which is from odd (9) to odd (1). Similarly, moving from 19 to 20 is from even (10) to even (2). So, in the original sequence, the parity alternates except at the transition from the end of one ten's place to the start of the next.Therefore, the original sequence is almost a Hamiltonian path in the bipartite graph, except that at the transition points (e.g., 09 to 10, 19 to 20, etc.), the parity doesn't alternate. Instead, it stays the same.So, in the original sequence, the parity alternates for the first ten numbers, then fails to alternate at the transition, then alternates again for the next ten, and so on.Therefore, the original sequence is not a valid Hamiltonian path in the bipartite graph because it doesn't alternate parity at the transition points.Hence, to make it a valid Hamiltonian path, we need to rearrange the sequence so that each consecutive term alternates parity.But the question is, how many terms can remain in their original positions?So, if we want to maximize the number of fixed points, we need as many terms as possible to stay in their original positions, but the rearranged sequence must alternate parity between consecutive terms.Given that, let's consider the original sequence's parity pattern.As observed, the original sequence alternates parity within each ten's place, but fails to alternate at the transition between ten's places.For example, from 09 (odd) to 10 (odd), the parity doesn't change. Similarly, from 19 (even) to 20 (even), it doesn't change, and so on.Therefore, in the original sequence, every 10th term (i.e., 09, 19, 29, ..., 99) is followed by a term with the same parity.So, in order to make the sequence alternate parity, we need to change the order around these transition points.Therefore, the terms at positions 10, 20, 30, ..., 100 (i.e., the 10th, 20th, etc.) cannot remain in their original positions because their original successor has the same parity, which violates the alternation requirement.Wait, actually, the term at position 10 is 09, which is followed by 10. But 09 has parity odd, and 10 has parity odd, so they are the same. So, to fix this, we need to rearrange the sequence so that after 09, we go to a term with even parity, not 10.Similarly, after 19, which is even, we need to go to an odd parity term, not 20.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 09, 19, ..., 89) cannot be followed by their original successors, which are 10, 20, ..., 90. Instead, they need to be followed by some other term with the opposite parity.Therefore, these transition points (positions 10, 20, ..., 90) cannot have their original successors, meaning that the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their predecessors would have to be different.Wait, actually, the term at position 10 is 09, which is followed by 10 in the original sequence. But in the rearranged sequence, 09 must be followed by a term with even parity, which 10 is odd, so 10 cannot follow 09. Therefore, 10 cannot be in position 11, but maybe it can be somewhere else.But the question is about how many terms can remain in their original positions. So, if 10 cannot be in position 11, but maybe it can be in another position. However, the term at position 10 is 09, which is odd, and in the rearranged sequence, it must be followed by an even term. So, 10 is odd, so it cannot follow 09. Therefore, 10 cannot be in position 11, but it can be in another position.Similarly, 20 cannot be in position 21, because 19 is even, and 20 is even, so 20 cannot follow 19. Therefore, 20 cannot be in position 21.Therefore, for each transition point, the term that was originally at position 10k cannot be in position 10k+1. So, these terms (10, 20, ..., 90) cannot be in their original positions.But wait, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid because they have the same parity. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.Wait, but there are 10 such terms: 10, 20, ..., 90, which is 9 terms, but actually, 10, 20, ..., 90 is 9 terms because 10*1=10, 10*2=20, ..., 10*9=90. So, 9 terms.But in the original sequence, the term at position 100 is 99, which is odd. So, the term after 99 would be... but since it's the last term, it doesn't need to be followed by anything. So, maybe 99 can stay.Wait, but in the original sequence, 99 is at position 100, and it's the last term. So, in the rearranged sequence, it's still the last term, so it doesn't need to follow any parity rule. Therefore, 99 can stay.But in the original sequence, the term before 99 is 98, which is even, and 99 is odd, so that alternation is correct. So, in the original sequence, the transition from 98 to 99 is valid because 98 is even and 99 is odd. Therefore, 99 can stay.Similarly, the term before 90 is 89, which is even, and 90 is even, so that's invalid. Therefore, 90 cannot stay.Wait, so in the original sequence, the transitions between 09-10, 19-20, ..., 89-90 are invalid because they have the same parity, but the transition from 98-99 is valid because they have different parities.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.But wait, actually, the term at position 10 is 09, which is odd, and the term at position 11 is 10, which is odd. So, in the original sequence, 09 is followed by 10, which is invalid. Therefore, in the rearranged sequence, 09 must be followed by an even term, so 10 cannot be in position 11. Therefore, 10 must be moved elsewhere.Similarly, 19 is followed by 20, which is invalid, so 20 must be moved.Therefore, the terms at positions 10, 20, ..., 90 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors would require them to be in a position where they have the same parity as the predecessor, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.Wait, but 10 is at position 11, right? Because the sequence starts at 00 as position 1, so 00 is 1, 01 is 2, ..., 09 is 10, 10 is 11, ..., 99 is 100.So, the term 10 is at position 11, not position 10. So, perhaps I made a mistake earlier.Let me clarify:- Position 1: 00- Position 2: 01- ...- Position 10: 09- Position 11: 10- Position 12: 11- ...- Position 20: 19- Position 21: 20- ...- Position 90: 89- Position 91: 90- ...- Position 100: 99So, the term 10 is at position 11, 20 is at position 21, ..., 90 is at position 91.Therefore, the terms at positions 11, 21, ..., 91 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors (09, 19, ..., 89) are of the same parity, which is invalid.Therefore, these 9 terms (10, 20, ..., 90) cannot remain in their original positions.So, in total, 9 terms cannot remain in their original positions. Therefore, the maximum number of terms that can remain is 100 - 9 = 91.But wait, that seems too high because in the original sequence, the parity alternates except at the transition points. So, if we fix the transitions, maybe we can keep more terms.Wait, perhaps my initial approach is flawed. Let me think differently.Since the graph is bipartite, with two sets A and B, where A contains numbers with even digit sum and B with odd digit sum. The original sequence alternates between A and B except at the transition points.In a Hamiltonian path, we must alternate between A and B. Therefore, the sequence must start in either A or B and alternate.In the original sequence, it starts with 00 (A), then 01 (B), 02 (A), ..., 09 (B), then 10 (B). So, at position 10, it's 09 (B), and position 11 is 10 (B), which is invalid because it's two B's in a row.Therefore, to fix this, we need to rearrange the sequence so that after 09 (B), we go to an A node, not 10 (B).Similarly, after 19 (A), we need to go to a B node, not 20 (A).Therefore, the terms at positions 11, 21, ..., 91 (i.e., 10, 20, ..., 90) cannot remain in their original positions because their original predecessors are of the same parity.Therefore, these 9 terms cannot remain. So, 9 terms must be moved.But, in addition, perhaps some other terms cannot remain because their positions require a certain parity.Wait, let's think about the entire sequence.In the original sequence, the parity alternates except at the transition points. So, in the first 10 terms, the parity alternates correctly: A, B, A, B, ..., A, B (positions 1-10: 00(A), 01(B), 02(A), ..., 09(B)).Then, at position 11, it's 10(B), which is the same as position 10 (B). So, invalid.Similarly, positions 21, 31, ..., 91 have the same issue.Therefore, to make the entire sequence alternate correctly, we need to adjust the transitions.One way to do this is to reverse the order of each ten's place after the first one.For example, keep the first ten terms as 00, 01, 02, ..., 09. Then, instead of going to 10, go to 19, 18, ..., 10. Then, go to 20, 21, ..., 29, then 39, 38, ..., 30, and so on.This way, the transitions between ten's places are fixed by reversing the order, ensuring that the parity alternates correctly.In this case, the terms 10, 20, ..., 90 are moved to the end of their respective reversed sequences.Therefore, in this rearrangement, the terms 10, 20, ..., 90 are moved, but the rest can remain.So, how many terms remain in their original positions?In the first ten's place (00-09), all terms remain.In the second ten's place (10-19), the terms are reversed: 19, 18, ..., 10. So, none of these terms remain in their original positions except possibly 10, but 10 was originally at position 11, which is now occupied by 19. So, none of the terms in the second ten's place remain.Similarly, in the third ten's place (20-29), we have 29, 28, ..., 20. None of these remain in their original positions.Wait, but actually, in the original sequence, the third ten's place starts at position 30: 20, 21, ..., 29. In the rearranged sequence, it's 29, 28, ..., 20, starting at position 21. So, none of these terms are in their original positions.Similarly, for the fourth ten's place (30-39), in the original sequence, it's 30, 31, ..., 39 at positions 40-49. In the rearranged sequence, it's 39, 38, ..., 30 at positions 30-39. So, none of these terms remain in their original positions.Wait, actually, in the rearranged sequence, the first ten's place (00-09) remains as is. Then, the second ten's place (10-19) is reversed and placed after 09, so positions 11-20 are 19, 18, ..., 10. Then, the third ten's place (20-29) is reversed and placed after 10, so positions 21-30 are 29, 28, ..., 20. And so on.Therefore, in this rearrangement, only the first ten's place (00-09) remains in their original positions. All other terms are moved.Therefore, in this case, only 10 terms remain in their original positions.But wait, that seems low. Maybe there's a better way to rearrange the sequence so that more terms remain.Alternatively, perhaps we can fix the transitions without reversing entire blocks.For example, at the transition from 09 to 10, instead of moving 10, maybe we can move another term to fix the parity.But if we do that, we might have to move multiple terms, which might not be efficient.Alternatively, perhaps we can pair the terms such that each pair consists of a term and its successor, and ensure that in each pair, only one can remain.Wait, considering the bipartition, since the graph is bipartite, any Hamiltonian path must alternate between the two sets. Therefore, the maximum number of fixed points is limited by the size of the smaller partition.Wait, in a bipartite graph, the maximum number of fixed points in a permutation is equal to the size of the smaller partition if the graph is balanced.But in our case, the graph is a 10x10 grid, which is bipartite with two sets of 50 nodes each (since 10x10=100, and each set has 50 nodes).Therefore, in a Hamiltonian path, which is a permutation of the nodes, the maximum number of fixed points is 50, because you can't have more fixed points than the size of the smaller partition.Wait, but actually, in a permutation, the number of fixed points is not necessarily limited by the partition size. For example, the identity permutation has all 100 fixed points, but it's not a Hamiltonian path.But in our case, the permutation must be a Hamiltonian path, which alternates between the two partitions.Therefore, in a Hamiltonian path, the number of fixed points cannot exceed the size of the smaller partition because each fixed point must lie in one partition, and the path alternates between partitions.Wait, let me think carefully.In a bipartite graph with partitions A and B, a Hamiltonian path must start in A or B and alternate. Therefore, if the path starts in A, the sequence is A, B, A, B, ..., ending in A or B depending on the length.Since 100 is even, if we start in A, we end in B, having 50 nodes in A and 50 in B.Similarly, if we start in B, we end in A.Therefore, in a Hamiltonian path, exactly 50 nodes are in A and 50 in B.Now, in the original sequence, the number of nodes in A and B is 50 each, because the digit sums from 00 to 99 have equal numbers of even and odd sums.Therefore, in the original sequence, there are 50 nodes in A and 50 in B.Now, if we want to fix as many nodes as possible, we need to fix nodes in such a way that the fixed nodes alternate between A and B.But in the original sequence, the fixed nodes would have to alternate between A and B as well, because the Hamiltonian path must alternate.Therefore, the maximum number of fixed points is 50, because you can fix all nodes in one partition, but not both.Wait, but actually, you can fix nodes in both partitions, but they must alternate.Wait, no, because if you fix a node in A, the next fixed node must be in B, and so on.But in the original sequence, the nodes are already alternating between A and B, except at the transition points.Therefore, if we can fix the nodes in such a way that the fixed nodes follow the alternation, we can fix up to 50 nodes.But wait, in the original sequence, the alternation is almost correct except at the transitions.Therefore, if we fix the transitions, we can fix all the nodes except those at the transitions.But earlier, I thought that only 9 terms cannot remain, but that might not be the case.Wait, perhaps the maximum number of fixed points is 50.Because in the bipartite graph, the maximum number of fixed points in a Hamiltonian path is 50, as you can fix all nodes in one partition.But in our case, since the original sequence already has 50 nodes in A and 50 in B, and the Hamiltonian path must alternate between A and B, the maximum number of fixed points is 50.Therefore, the answer is 50.But wait, let me verify this.Suppose we fix all nodes in partition A. Then, the Hamiltonian path must go through all nodes in A and B, alternating. But if all nodes in A are fixed, then the path must go through them in order, but the nodes in B must be arranged around them.But in our case, the original sequence has nodes in A and B alternating, except at the transitions. Therefore, if we fix all nodes in A, we can have 50 fixed points.Similarly, if we fix all nodes in B, we can have 50 fixed points.But in the original sequence, the nodes in A and B are interleaved, so fixing all nodes in A would require that the nodes in B are arranged in between, which might not be possible without disturbing the original sequence.Wait, perhaps not all nodes in A can be fixed because their positions might require certain transitions.Alternatively, maybe the maximum number of fixed points is 50 because you can fix every other node.Wait, let me think of an example.Suppose we fix all even-positioned nodes. Since the original sequence alternates parity, fixing every other node would mean fixing 50 nodes, all in one partition.But in the original sequence, the parity alternates except at the transitions. So, if we fix the nodes in partition A (even parity), which are at positions 1, 3, 5, ..., 99, we can have 50 fixed points.But in reality, the original sequence has some nodes in A and some in B, but the transitions between ten's places break the alternation.Therefore, to fix 50 nodes, we need to ensure that the fixed nodes are in one partition and that the path alternates correctly.But perhaps it's not possible to fix 50 nodes because the transitions would require moving some nodes.Wait, maybe the maximum number of fixed points is 50 because that's the size of one partition, and you can't fix more than that without violating the alternation.Therefore, the answer is 50.But let me try to construct such a sequence.Suppose we fix all nodes in partition A (even parity). Then, the sequence would be:00 (A), 01 (B), 02 (A), 03 (B), ..., 09 (B), then instead of 10 (B), we go to 19 (A), then 18 (B), 17 (A), ..., 10 (B). Then, 20 (A), 21 (B), ..., 29 (A), then 39 (B), 38 (A), ..., 30 (B), and so on.In this case, the fixed nodes are 00, 02, 04, ..., 08, 20, 22, ..., 28, 40, 42, ..., 48, 60, 62, ..., 68, 80, 82, ..., 88, and 99 (if we fix it). Wait, but 99 is odd, so it's in partition B.Wait, actually, in this arrangement, the fixed nodes are all the even-positioned nodes in the original sequence, which are 50 nodes.But in reality, the original sequence has nodes in A and B alternating, except at the transitions. So, if we fix all nodes in A, which are 50 nodes, we can have them in their original positions, and arrange the nodes in B around them.But in the original sequence, the nodes in A are at positions 1, 3, 5, ..., 99, which are 50 nodes.Therefore, if we fix these 50 nodes, the rest can be arranged in between.Therefore, the maximum number of fixed points is 50.Hence, the answer is 50.

💡Alright, so I have this problem where I need to find all prime numbers ( p ) not exceeding 1000 such that ( 2p + 1 ) is a perfect power. A perfect power means that ( 2p + 1 = m^n ) where ( m ) and ( n ) are natural numbers, and ( n geq 2 ). First, I should understand what a perfect power is. It's a number that can be expressed as some integer raised to an exponent of at least 2. So examples include squares, cubes, etc. For instance, 4 is a perfect square because it's ( 2^2 ), and 8 is a perfect cube because it's ( 2^3 ).Given that ( p ) is a prime number, and ( 2p + 1 ) needs to be a perfect power, I can start by expressing this relationship as:[ 2p + 1 = m^n ]Where ( m ) and ( n ) are natural numbers, and ( n geq 2 ). My goal is to find all such primes ( p ) where this equation holds true, and ( p leq 1000 ).Let me rearrange the equation to solve for ( p ):[ 2p = m^n - 1 ][ p = frac{m^n - 1}{2} ]Since ( p ) must be a prime number, ( frac{m^n - 1}{2} ) must also be a prime. This gives me a condition to check for different values of ( m ) and ( n ).Now, let's think about the possible values of ( m ) and ( n ). Since ( m^n ) must be an odd number (because ( 2p + 1 ) is always odd), ( m ) itself must be odd. If ( m ) were even, then ( m^n ) would be even, and ( 2p + 1 ) would be odd, which is a contradiction. So, ( m ) must be odd.Therefore, ( m ) can be 3, 5, 7, etc. Let's start testing these values.**Case 1: ( m = 3 )**Let's see what values of ( n ) give us ( p ) as a prime number.- ( n = 2 ): [ p = frac{3^2 - 1}{2} = frac{9 - 1}{2} = frac{8}{2} = 4 ] 4 is not a prime number.- ( n = 3 ): [ p = frac{3^3 - 1}{2} = frac{27 - 1}{2} = frac{26}{2} = 13 ] 13 is a prime number.- ( n = 4 ): [ p = frac{3^4 - 1}{2} = frac{81 - 1}{2} = frac{80}{2} = 40 ] 40 is not a prime number.- ( n = 5 ): [ p = frac{3^5 - 1}{2} = frac{243 - 1}{2} = frac{242}{2} = 121 ] 121 is not a prime number (since ( 121 = 11^2 )).- ( n = 6 ): [ p = frac{3^6 - 1}{2} = frac{729 - 1}{2} = frac{728}{2} = 364 ] 364 is not a prime number.Continuing this, we can see that as ( n ) increases, ( p ) becomes larger, but it's not necessarily prime. Since we're limited to ( p leq 1000 ), let's check the next value:- ( n = 7 ): [ p = frac{3^7 - 1}{2} = frac{2187 - 1}{2} = frac{2186}{2} = 1093 ] 1093 is a prime number, but it exceeds our limit of 1000. So, we can stop here for ( m = 3 ).So, from ( m = 3 ), we get ( p = 13 ) as a valid prime number.**Case 2: ( m = 5 )**Let's try ( m = 5 ) and see if we can find any primes ( p ).- ( n = 2 ): [ p = frac{5^2 - 1}{2} = frac{25 - 1}{2} = frac{24}{2} = 12 ] 12 is not a prime number.- ( n = 3 ): [ p = frac{5^3 - 1}{2} = frac{125 - 1}{2} = frac{124}{2} = 62 ] 62 is not a prime number.- ( n = 4 ): [ p = frac{5^4 - 1}{2} = frac{625 - 1}{2} = frac{624}{2} = 312 ] 312 is not a prime number.- ( n = 5 ): [ p = frac{5^5 - 1}{2} = frac{3125 - 1}{2} = frac{3124}{2} = 1562 ] 1562 is way above 1000 and not a prime.So, ( m = 5 ) doesn't give us any valid primes ( p ) within our limit.**Case 3: ( m = 7 )**Let's try ( m = 7 ).- ( n = 2 ): [ p = frac{7^2 - 1}{2} = frac{49 - 1}{2} = frac{48}{2} = 24 ] 24 is not a prime number.- ( n = 3 ): [ p = frac{7^3 - 1}{2} = frac{343 - 1}{2} = frac{342}{2} = 171 ] 171 is not a prime number.- ( n = 4 ): [ p = frac{7^4 - 1}{2} = frac{2401 - 1}{2} = frac{2400}{2} = 1200 ] 1200 is above 1000 and not a prime.So, ( m = 7 ) also doesn't yield any valid primes ( p ).**Case 4: ( m = 9 )**Wait, ( m = 9 ) is the next odd number. Let's check.- ( n = 2 ): [ p = frac{9^2 - 1}{2} = frac{81 - 1}{2} = frac{80}{2} = 40 ] 40 is not a prime.- ( n = 3 ): [ p = frac{9^3 - 1}{2} = frac{729 - 1}{2} = frac{728}{2} = 364 ] 364 is not a prime.- ( n = 4 ): [ p = frac{9^4 - 1}{2} = frac{6561 - 1}{2} = frac{6560}{2} = 3280 ] 3280 is way above 1000.So, ( m = 9 ) doesn't help either.**Case 5: ( m = 11 )**Let's try ( m = 11 ).- ( n = 2 ): [ p = frac{11^2 - 1}{2} = frac{121 - 1}{2} = frac{120}{2} = 60 ] 60 is not a prime.- ( n = 3 ): [ p = frac{11^3 - 1}{2} = frac{1331 - 1}{2} = frac{1330}{2} = 665 ] 665 is not a prime (divisible by 5).- ( n = 4 ): [ p = frac{11^4 - 1}{2} = frac{14641 - 1}{2} = frac{14640}{2} = 7320 ] 7320 is above 1000.So, ( m = 11 ) doesn't give us a valid prime ( p ).**Case 6: ( m = 13 )**Trying ( m = 13 ).- ( n = 2 ): [ p = frac{13^2 - 1}{2} = frac{169 - 1}{2} = frac{168}{2} = 84 ] 84 is not a prime.- ( n = 3 ): [ p = frac{13^3 - 1}{2} = frac{2197 - 1}{2} = frac{2196}{2} = 1098 ] 1098 is above 1000.So, ( m = 13 ) doesn't help.At this point, I notice a pattern. For higher values of ( m ), even with ( n = 2 ), ( p ) becomes quite large, and often not prime. Also, as ( m ) increases, ( p ) increases rapidly, making it exceed 1000 quickly.Perhaps I should consider smaller exponents ( n ). Let's see if there are other values of ( m ) and ( n ) that could yield a prime ( p ).Wait, maybe I should also consider that ( m ) doesn't have to be prime. For example, ( m = 4 ) is even, but we already established ( m ) must be odd. So, ( m ) must be odd, but not necessarily prime.But earlier, I saw that ( m = 3 ) gave me ( p = 13 ), which is prime. Let me check if there are other ( m ) values that could give me primes.Wait, what if ( n = 2 )? Let's see if there are other ( m ) such that ( frac{m^2 - 1}{2} ) is prime.So, ( p = frac{m^2 - 1}{2} ). Let's see for different ( m ):- ( m = 3 ): ( p = 4 ) (not prime)- ( m = 5 ): ( p = 12 ) (not prime)- ( m = 7 ): ( p = 24 ) (not prime)- ( m = 9 ): ( p = 40 ) (not prime)- ( m = 11 ): ( p = 60 ) (not prime)- ( m = 13 ): ( p = 84 ) (not prime)- ( m = 15 ): ( p = 112 ) (not prime)- ( m = 17 ): ( p = 144 ) (not prime)- ( m = 19 ): ( p = 180 ) (not prime)- ( m = 21 ): ( p = 220 ) (not prime)- ( m = 23 ): ( p = 264 ) (not prime)- ( m = 25 ): ( p = 312 ) (not prime)- ( m = 27 ): ( p = 364 ) (not prime)- ( m = 29 ): ( p = 420 ) (not prime)- ( m = 31 ): ( p = 480 ) (not prime)- ( m = 33 ): ( p = 544 ) (not prime)- ( m = 35 ): ( p = 612 ) (not prime)- ( m = 37 ): ( p = 684 ) (not prime)- ( m = 39 ): ( p = 760 ) (not prime)- ( m = 41 ): ( p = 840 ) (not prime)- ( m = 43 ): ( p = 924 ) (not prime)- ( m = 45 ): ( p = 1012 ) (exceeds 1000)So, for ( n = 2 ), only ( m = 3 ) gives ( p = 4 ), which isn't prime. Wait, no, earlier I saw that for ( m = 3 ), ( n = 3 ) gives ( p = 13 ). So, maybe focusing on ( n = 3 ) could help.Let me check ( n = 3 ) for different ( m ):- ( m = 3 ): ( p = 13 ) (prime)- ( m = 5 ): ( p = 62 ) (not prime)- ( m = 7 ): ( p = 171 ) (not prime)- ( m = 9 ): ( p = 364 ) (not prime)- ( m = 11 ): ( p = 665 ) (not prime)- ( m = 13 ): ( p = 1098 ) (exceeds 1000)So, only ( m = 3 ) gives a prime ( p = 13 ) when ( n = 3 ).What about ( n = 4 )?- ( m = 3 ): ( p = 40 ) (not prime)- ( m = 5 ): ( p = 312 ) (not prime)- ( m = 7 ): ( p = 1200 ) (exceeds 1000)So, nothing useful here.What about ( n = 5 )?- ( m = 3 ): ( p = 121 ) (not prime)- ( m = 5 ): ( p = 1562 ) (exceeds 1000)Again, nothing useful.It seems that the only valid prime ( p ) under 1000 is 13 when ( m = 3 ) and ( n = 3 ).Wait, let me double-check if there are any other possibilities. Maybe ( m ) is a higher power, but ( n ) is small.For example, ( m = 15 ), ( n = 2 ):[ p = frac{15^2 - 1}{2} = frac{225 - 1}{2} = 112 ] (not prime)( m = 17 ), ( n = 2 ):[ p = frac{17^2 - 1}{2} = frac{289 - 1}{2} = 144 ] (not prime)( m = 19 ), ( n = 2 ):[ p = frac{19^2 - 1}{2} = frac{361 - 1}{2} = 180 ] (not prime)Hmm, still nothing.What if ( n = 6 )?- ( m = 3 ): ( p = 364 ) (not prime)- ( m = 5 ): ( p = 3124 / 2 = 1562 ) (exceeds 1000)No luck.Wait, maybe I should consider ( m ) as a power of a smaller number. For example, ( m = 3^k ). Let's see:If ( m = 3^2 = 9 ), ( n = 3 ):[ p = frac{9^3 - 1}{2} = frac{729 - 1}{2} = 364 ] (not prime)If ( m = 3^3 = 27 ), ( n = 2 ):[ p = frac{27^2 - 1}{2} = frac{729 - 1}{2} = 364 ] (same as above)Not helpful.Alternatively, maybe ( m ) is a composite number, but I don't think that changes much because we're looking for ( p ) to be prime, so ( frac{m^n - 1}{2} ) must be prime regardless of ( m )'s primality.Another approach could be to consider the equation ( 2p + 1 = m^n ) and see if there are any known results or theorems about such equations. For example, Catalan's conjecture (now Mihăilescu's theorem) states that the only solution in the natural numbers of ( x^a - y^b = 1 ) for ( x, y > 1 ) and ( a, b > 1 ) is ( 3^2 - 2^3 = 1 ). But in our case, it's ( m^n - 2p = 1 ). Not sure if directly applicable, but perhaps similar reasoning.Wait, if we rearrange ( m^n = 2p + 1 ), then ( m^n - 2p = 1 ). If ( m ) and ( p ) are both primes, but ( m ) is odd, as we saw earlier. However, ( p ) is a prime, but ( m ) doesn't have to be. So, perhaps not directly applicable.Alternatively, maybe looking for primes ( p ) such that ( 2p + 1 ) is a perfect power. So, perhaps checking known perfect powers and seeing if ( (m^n - 1)/2 ) is prime.Let me list some perfect powers and see:Perfect squares:- ( 3^2 = 9 ): ( p = (9 - 1)/2 = 4 ) (not prime)- ( 5^2 = 25 ): ( p = 12 ) (not prime)- ( 7^2 = 49 ): ( p = 24 ) (not prime)- ( 9^2 = 81 ): ( p = 40 ) (not prime)- ( 11^2 = 121 ): ( p = 60 ) (not prime)- ( 13^2 = 169 ): ( p = 84 ) (not prime)- ( 15^2 = 225 ): ( p = 112 ) (not prime)- ( 17^2 = 289 ): ( p = 144 ) (not prime)- ( 19^2 = 361 ): ( p = 180 ) (not prime)- ( 21^2 = 441 ): ( p = 220 ) (not prime)- ( 23^2 = 529 ): ( p = 264 ) (not prime)- ( 25^2 = 625 ): ( p = 312 ) (not prime)- ( 27^2 = 729 ): ( p = 364 ) (not prime)- ( 29^2 = 841 ): ( p = 420 ) (not prime)- ( 31^2 = 961 ): ( p = 480 ) (not prime)- ( 33^2 = 1089 ): ( p = 544 ) (not prime)- ( 35^2 = 1225 ): ( p = 612 ) (not prime)- ( 37^2 = 1369 ): ( p = 684 ) (not prime)- ( 39^2 = 1521 ): ( p = 760 ) (not prime)- ( 41^2 = 1681 ): ( p = 840 ) (not prime)- ( 43^2 = 1849 ): ( p = 924 ) (not prime)- ( 45^2 = 2025 ): ( p = 1012 ) (exceeds 1000)So, no primes from perfect squares.Perfect cubes:- ( 3^3 = 27 ): ( p = 13 ) (prime)- ( 5^3 = 125 ): ( p = 62 ) (not prime)- ( 7^3 = 343 ): ( p = 171 ) (not prime)- ( 9^3 = 729 ): ( p = 364 ) (not prime)- ( 11^3 = 1331 ): ( p = 665 ) (not prime)- ( 13^3 = 2197 ): ( p = 1098 ) (exceeds 1000)So, only ( m = 3 ), ( n = 3 ) gives a prime ( p = 13 ).Perfect fourth powers:- ( 3^4 = 81 ): ( p = 40 ) (not prime)- ( 5^4 = 625 ): ( p = 312 ) (not prime)- ( 7^4 = 2401 ): ( p = 1200 ) (exceeds 1000)No primes here.Perfect fifth powers:- ( 3^5 = 243 ): ( p = 121 ) (not prime)- ( 5^5 = 3125 ): ( p = 1562 ) (exceeds 1000)No primes.Perfect sixth powers:- ( 3^6 = 729 ): ( p = 364 ) (not prime)- ( 5^6 = 15625 ): ( p = 7812 ) (exceeds 1000)No primes.So, from this exhaustive check, it seems that the only prime ( p ) under 1000 satisfying the condition is 13.Wait, just to be thorough, let me check if there are any other perfect powers with higher exponents that might yield a prime ( p ). For example, ( m = 3 ), ( n = 7 ):[ p = frac{3^7 - 1}{2} = frac{2187 - 1}{2} = 1093 ]1093 is a prime number, but it exceeds our limit of 1000. So, it's out of consideration.Similarly, ( m = 3 ), ( n = 4 ):[ p = frac{81 - 1}{2} = 40 ] (not prime)So, nothing else comes up.Therefore, after checking all possible cases, the only prime number ( p ) not exceeding 1000 such that ( 2p + 1 ) is a perfect power is 13.

💡First, I need to determine how many more years Norine needs to work until her age and the number of years she has worked add up to 85. Currently, she is 50 years old and has worked for 19 years. I'll set up an equation where her future age plus her future years of work equals 85. Let ( x ) represent the number of years until she retires. Her future age will be ( 50 + x ) and her future years of work will be ( 19 + x ). So, the equation becomes:[(50 + x) + (19 + x) = 85]Simplifying the equation:[69 + 2x = 85]Subtracting 69 from both sides:[2x = 16]Dividing both sides by 2:[x = 8]Therefore, Norine will need to work for 8 more years. Adding this to her current age:[50 + 8 = 58]So, Norine will be 58 years old when she retires.

💡First, I need to determine which set of line segments can form a right triangle by applying the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.**For set A: 4cm, 5cm, 6cm**I check if (4^2 + 5^2 = 6^2):- (4^2 = 16)- (5^2 = 25)- (6^2 = 36)So, (16 + 25 = 41 neq 36). Therefore, set A cannot form a right triangle.**For set B: 1cm, (sqrt{2})cm, 2.5cm**I check if (1^2 + (sqrt{2})^2 = 2.5^2):- (1^2 = 1)- ((sqrt{2})^2 = 2)- (2.5^2 = 6.25)So, (1 + 2 = 3 neq 6.25). Therefore, set B cannot form a right triangle.**For set C: 2cm, 3cm, 4cm**I check if (2^2 + 3^2 = 4^2):- (2^2 = 4)- (3^2 = 9)- (4^2 = 16)So, (4 + 9 = 13 neq 16). Therefore, set C cannot form a right triangle.**For set D: 1.5cm, 2cm, 2.5cm**I check if (1.5^2 + 2^2 = 2.5^2):- (1.5^2 = 2.25)- (2^2 = 4)- (2.5^2 = 6.25)So, (2.25 + 4 = 6.25 = 6.25). Therefore, set D can form a right triangle.Hence, the correct answer is (boxed{D}).

💡Okay, so I have this problem where I need to arrange three identical red balls and three identical black balls in a row, and then number them from left to right as 1, 2, 3, 4, 5, 6. The question is asking for the number of arrangements where the sum of the numbers of the red balls is less than the sum of the numbers of the black balls. The options given are A: 10, B: 11, C: 15, D: 18.Hmm, let me try to understand this step by step. First, arranging three red and three black balls in a row. Since the balls of the same color are identical, the number of distinct arrangements is given by the combination formula. Specifically, it's the number of ways to choose 3 positions out of 6 for the red balls (the rest will automatically be black). So, the total number of arrangements is C(6,3), which is 20. That makes sense because 6 choose 3 is 20.Now, the problem is about comparing the sums of the numbers assigned to the red and black balls. Each position in the row is numbered from 1 to 6, so depending on where the red and black balls are placed, their sums will vary.The key point here is that for each arrangement, either the sum of the red balls will be less than the sum of the black balls, or it will be greater. There's no possibility of them being equal because the total sum of all positions is 1+2+3+4+5+6 = 21. If the sum of red balls is S, then the sum of black balls is 21 - S. For S to equal 21 - S, S would have to be 10.5, which isn't possible since all the numbers are integers. So, there are no cases where the sums are equal. Therefore, the number of arrangements where red sum is less than black sum should be exactly half of the total arrangements, right?Since the total number of arrangements is 20, dividing by 2 gives 10. So, the answer should be 10, which is option A.Wait, but let me double-check this reasoning. Sometimes in combinatorics, symmetry can be tricky. Is there a reason why the number of arrangements where red sum is less than black sum might not be exactly half?Let me think about it. Each arrangement corresponds to a unique subset of three numbers (positions) for the red balls. The sum of these three numbers can vary. For each arrangement where the red sum is S, there is a complementary arrangement where the red sum is 21 - S (by swapping red and black balls). So, for every arrangement where red sum is less than 10.5, there's a corresponding arrangement where red sum is greater than 10.5. Since 10.5 is the midpoint, and there are no sums equal to 10.5, the number of arrangements where red sum is less than black sum should indeed be equal to the number where it's greater. Therefore, each should be half of 20, which is 10.So, I think my initial reasoning was correct. The answer is 10, which is option A.But just to be thorough, maybe I should consider some specific examples to see if this holds.For instance, consider the arrangement where red balls are in positions 1, 2, 3. The sum is 1+2+3=6. The black balls are in positions 4,5,6, sum is 4+5+6=15. So, red sum is less than black sum.Now, the complementary arrangement would have red balls in positions 4,5,6, sum=15, and black balls in positions 1,2,3, sum=6. So, red sum is greater than black sum.Similarly, take another arrangement: red balls in positions 1,2,4. Sum is 1+2+4=7. Black balls are in positions 3,5,6. Sum is 3+5+6=14. Again, red sum is less.The complementary arrangement would have red balls in 3,5,6, sum=14, and black balls in 1,2,4, sum=7. So, red sum is greater.This seems consistent. Every arrangement where red sum is less than 10.5 has a complementary arrangement where red sum is greater. Therefore, the counts should be equal.Therefore, the number of arrangements where red sum is less than black sum is 10.I think that solidifies the reasoning. So, I'm confident the answer is A: 10.**Final Answer**boxed{A}

💡First, calculate the total earnings from selling ducks and chickens.The farmer sells 5 chickens at 8 each:5 chickens * 8/chicken = 40The farmer sells 2 ducks at 10 each:2 ducks * 10/duck = 20Total earnings from selling ducks and chickens:40 (from chickens) + 20 (from ducks) = 60Next, determine how much the farmer spent on the wheelbarrow.Someone paid the farmer double the price he paid for the wheelbarrow, resulting in an additional 60:60 / 2 = 30The farmer spent 30 on the wheelbarrow.Finally, find the ratio of the amount spent on the wheelbarrow to the total earnings:30 (spent on wheelbarrow) / 60 (total earnings) = 1/2The ratio is 1:2.