🧠:<think>Okay, so I have this problem where the numeral 58 in base c is equal to 85 in base d. I need to find the least possible value of c + d, and then express that sum as a Roman numeral. The options are XII, XIII, XIV, XV, and XVI. Hmm, okay, let's break this down step by step.First, I remember that when a number is in a certain base, each digit represents a power of that base. So, for example, the number 58 in base c would be equal to 5 times c to the power of 1 plus 8 times c to the power of 0. Similarly, 85 in base d would be 8 times d to the power of 1 plus 5 times d to the power of 0.So, mathematically, I can write this as:5c + 8 = 8d + 5Okay, so now I have an equation: 5c + 8 equals 8d + 5. I need to find positive integers c and d such that this equation holds true, and then find the smallest possible sum of c and d.Let me rearrange this equation to make it easier to solve for one variable in terms of the other. If I subtract 8d from both sides and subtract 5 from both sides, I get:5c - 8d = -3Hmm, so 5c - 8d equals -3. I need to find integer solutions for c and d here. This looks like a linear Diophantine equation. I remember that these types of equations can have multiple solutions, and I need to find the smallest positive integers c and d that satisfy this equation.Let me try to solve for one variable. Let's solve for d in terms of c. Starting with:5c - 8d = -3If I add 8d to both sides and add 3 to both sides, I get:5c + 3 = 8dThen, dividing both sides by 8:d = (5c + 3) / 8Okay, so d has to be an integer, which means (5c + 3) must be divisible by 8. So, 5c + 3 must be a multiple of 8. Let me write that as:5c + 3 ≡ 0 mod 8So, 5c ≡ -3 mod 8. But -3 mod 8 is the same as 5 mod 8 because 8 - 3 is 5. So, 5c ≡ 5 mod 8.Hmm, if 5c ≡ 5 mod 8, then dividing both sides by 5 (since 5 and 8 are coprime, I can do that), I get:c ≡ 1 mod 8That means c is congruent to 1 modulo 8. So, the possible values of c are 1, 9, 17, 25, and so on. But since we're dealing with bases, c has to be greater than the highest digit in the number 58. The highest digit is 8, so c has to be at least 9. Therefore, the smallest possible value for c is 9.Let me check that. If c is 9, then d would be:d = (5*9 + 3) / 8 = (45 + 3) / 8 = 48 / 8 = 6So, d is 6. Let me verify if this works. In base c=9, the number 58 is equal to 5*9 + 8 = 45 + 8 = 53. In base d=6, the number 85 is equal to 8*6 + 5 = 48 + 5 = 53. Yes, that works. So, c=9 and d=6 are valid bases.Now, the sum c + d is 9 + 6 = 15. The question asks for the least possible value of c + d written as a Roman numeral. The Roman numeral for 15 is XV.Wait, let me just make sure that there isn't a smaller sum. The next possible c would be 17, right? Because c has to be 1 mod 8, so 17 is the next one. Let's see what d would be then:d = (5*17 + 3)/8 = (85 + 3)/8 = 88/8 = 11So, c=17 and d=11, which sums to 28, which is much larger than 15. So, 15 is indeed the smallest possible sum.Therefore, the answer should be XV, which is option D.Final Answerboxed{D}</think>

🧠:<think>Okay, so I have this problem here where I need to figure out if a certain amount of material is enough to make a closed rectangular box that can hold at least 1995 unit cubes. The question has three parts: a) with 962 square units, b) with 960 square units, and c) with 958 square units. First, I need to understand what's being asked. A closed rectangular box means it's a rectangular prism with six faces. The material required would be the surface area of this box. So, if I can find a box that has a volume of at least 1995 unit cubes and a surface area less than or equal to the given amounts, then that amount of material is sufficient.Let's start by recalling the formulas. The volume ( V ) of a rectangular box is given by:[V = length times width times height]And the surface area ( S ) is:[S = 2 times (length times width + length times height + width times height)]So, I need to find integer dimensions ( l, w, h ) such that ( l times w times h geq 1995 ) and ( 2(lw + lh + wh) ) is as small as possible.Since the problem is about sufficiency, if I can find such dimensions where the surface area is less than or equal to the given material, then it's sufficient. If not, it's not.I think the best approach is to find the box with the minimal surface area that can hold at least 1995 unit cubes. Then, compare that minimal surface area with the given amounts.So, how do I find such a box? I know that for a given volume, the shape that minimizes the surface area is a cube. But since we need a rectangular box, it's going to be as close to a cube as possible.Let me try to find the cube root of 1995 to get an idea of the dimensions.[sqrt[3]{1995} approx 12.6]So, dimensions around 13 might be good. But since the volume needs to be at least 1995, maybe 13x13x13 would be 2197, which is more than enough, but its surface area would be:[S = 2 times (13 times 13 + 13 times 13 + 13 times 13) = 2 times (507) = 1014]That's way higher than the given options. So, maybe I can find a more efficient box.I think I should look for three integers close to each other whose product is just above 1995 and whose sum of the products of their sides is minimal.Let me try 11, 13, and 14.Calculating the volume:[11 times 13 times 14 = 2002]That's just above 1995. Now, calculating the surface area:[S = 2 times (11 times 13 + 11 times 14 + 13 times 14)]Calculating each term:[11 times 13 = 143][11 times 14 = 154][13 times 14 = 182]Adding them up:[143 + 154 + 182 = 479]Then, multiplying by 2:[S = 2 times 479 = 958]Okay, so the surface area is 958. That means for part c), 958 is exactly sufficient. For parts a) and b), since 958 is less than both 962 and 960, those are also sufficient.Wait, but 958 is less than both 960 and 962, so that would mean all three are sufficient? But the question is asking if it's sufficient, so I guess yes, but the minimal one is 958.But let me double-check if 958 is indeed the minimal surface area for a box with volume at least 1995.Maybe I can find another set of dimensions with a smaller surface area.Let me try 10, 14, and 14.Volume:[10 times 14 times 14 = 1960]That's less than 1995, so not enough.How about 12, 12, and 13?Volume:[12 times 12 times 13 = 1872]Still less than 1995.12, 13, 13:[12 times 13 times 13 = 2028]That's more than 1995. Now, surface area:[S = 2 times (12 times 13 + 12 times 13 + 13 times 13)][= 2 times (156 + 156 + 169)][= 2 times 481 = 962]So, 962 is the surface area here. That's more than 958, which we got earlier.Another try: 11, 12, 15Volume:[11 times 12 times 15 = 1980]Close to 1995 but still less.11, 14, 14:[11 times 14 times 14 = 2156]Surface area:[S = 2 times (11 times 14 + 11 times 14 + 14 times 14)][= 2 times (154 + 154 + 196)][= 2 times 504 = 1008]That's even higher.How about 10, 15, 13:Volume:[10 times 15 times 13 = 1950]Too low.10, 16, 13:[10 times 16 times 13 = 2080]Surface area:[S = 2 times (10 times 16 + 10 times 13 + 16 times 13)][= 2 times (160 + 130 + 208)][= 2 times 498 = 996]Still higher than 958.Wait, what about 9, 14, 15:Volume:[9 times 14 times 15 = 1890]Too low.10, 12, 16:[10 times 12 times 16 = 1920]Still low.11, 12, 16:[11 times 12 times 16 = 2112]Surface area:[S = 2 times (11 times 12 + 11 times 16 + 12 times 16)][= 2 times (132 + 176 + 192)][= 2 times 400 = 800]Wait, that can't be right. Let me recalculate:11x12=132, 11x16=176, 12x16=192.Sum: 132+176=308+192=500.So, 2x500=1000.Oops, I messed up the addition earlier.So, surface area is 1000, which is higher than 958.Hmm, seems like 11x13x14 is giving me the minimal surface area so far.Let me try 12x12x14:Volume:[12 times 12 times 14 = 2016]Surface area:[S = 2 times (12 times 12 + 12 times 14 + 12 times 14)][= 2 times (144 + 168 + 168)][= 2 times 480 = 960]Okay, so 960 surface area. That's less than 958? No, wait, 960 is more than 958.Wait, 960 is more than 958, so 958 is still better.Wait, but 960 is actually higher than 958, so 958 is minimal.Wait, let me confirm:11x13x14 gives 958.12x12x14 gives 960.So, 958 is better.Is there a box with dimensions closer to each other that gives a lower surface area?Let me see, 10x13x15:Volume:[10 times 13 times 15 = 1950]Too low.11x12x15:[11 times 12 times 15 = 1980]Still low.11x14x13 is 2002.Surface area 958.Is there a combination where the surface area is less than 958?Let me try 10x14x14:Volume 1960, surface area 958? Wait, no, earlier I calculated that 11x13x14 gives 958.Wait, 10x14x14 gives 1960, surface area:[2 times (10 times 14 + 10 times 14 + 14 times 14) = 2 times (140 + 140 + 196) = 2 times 476 = 952]Wait, that's 952. But wait, 10x14x14 is 1960, which is less than 1995, so it won't satisfy the volume requirement. So, even though the surface area is lower, the volume is insufficient.So, that's not acceptable.Therefore, the minimal surface area for a box with volume at least 1995 is 958, achieved by 11x13x14.So, going back to the original question:a) 962: Since 958 ≤ 962, yes, it's sufficient.b) 960: Since 958 ≤ 960, yes, it's sufficient.c) 958: Exactly sufficient.Therefore, all three are sufficient, but the minimal required is 958.I think that's the answer.Final AnswerThe minimal sufficient material is boxed{C}.</think>

🧠:<think>First, I need to understand the given function ( g(n) = log_{3003} n^3 ). This means that for any positive integer ( n ), the function ( g(n) ) calculates the logarithm of ( n^3 ) with base 3003.To find ( g(7) + g(11) + g(13) ), I can start by expressing each term individually:[g(7) = log_{3003} 7^3, quad g(11) = log_{3003} 11^3, quad g(13) = log_{3003} 13^3]Adding these together gives:[log_{3003} 7^3 + log_{3003} 11^3 + log_{3003} 13^3]Using the logarithmic identity that the sum of logs is the log of the product, this becomes:[log_{3003} (7^3 cdot 11^3 cdot 13^3)]Recognizing that ( 7 cdot 11 cdot 13 = 1001 ), I can rewrite the expression as:[log_{3003} (1001)^3]Since ( 3003 = 3 cdot 1001 ), I can further simplify:[log_{3003} (1001)^3 = 3 log_{3003} 1001 = 3 log_{3003} left( frac{3003}{3} right)]Applying the logarithmic identity ( log_b left( frac{a}{c} right) = log_b a - log_b c ), this becomes:[3 left( log_{3003} 3003 - log_{3003} 3 right)]Since ( log_{3003} 3003 = 1 ), the expression simplifies to:[3 times 1 - 3 log_{3003} 3]To evaluate ( log_{3003} 3 ), I use the change of base formula:[log_{3003} 3 = frac{1}{log_3 3003}]Given that ( 3003 = 3 cdot 1001 ), I find:[log_3 3003 = log_3 (3 cdot 1001) = 1 + log_3 1001]Assuming ( log_3 1001 approx 3 ), the expression becomes:[log_{3003} 3 approx frac{1}{1 + 3} = frac{1}{4}]Substituting back, the final result is:[3 times 1 - 3 times frac{1}{4} = 3 - frac{3}{4} = frac{9}{4}]</think>

🧠:<think>Alright, let's dive into this problem. So, we have four players: A, B, C, and D, who start with 16, 15, 14, and 13 tokens respectively. The rules are that in each round, the player with the most tokens gives one token to each of the other players and discards one token. The game ends when any player runs out of tokens. We need to figure out how many rounds the game will last.First, let me understand the mechanics of each round. The player with the most tokens is going to give away a total of four tokens: one to each of the other three players and one discarded. That means the player with the most tokens loses four tokens each round, while each of the other players gains one token. So, if I think about it, in each round, the total number of tokens in the system decreases by one because one token is discarded each round. The initial total number of tokens is 16 + 15 + 14 + 13, which is 58 tokens. Since one token is discarded each round, the game will last at most 58 rounds because that's when all tokens would be discarded. However, the game could end earlier if a player runs out of tokens before all tokens are discarded.Let me consider how the token distribution changes each round. Let's denote the players' tokens as A, B, C, D.Starting with:- A: 16- B: 15- C: 14- D: 13In the first round, A has the most tokens, so A will give one token to B, C, D, and discard one. So, A loses 4 tokens, and B, C, D each gain 1 token.After Round 1:- A: 16 - 4 = 12- B: 15 + 1 = 16- C: 14 + 1 = 15- D: 13 + 1 = 14Now, the distribution is:- A: 12- B: 16- C: 15- D: 14In Round 2, B has the most tokens, so B will give one token to A, C, D, and discard one.After Round 2:- B: 16 - 4 = 12- A: 12 + 1 = 13- C: 15 + 1 = 16- D: 14 + 1 = 15Distribution:- A: 13- B: 12- C: 16- D: 15Round 3: C has the most tokens, so C gives one to each of the others and discards one.After Round 3:- C: 16 - 4 = 12- A: 13 + 1 = 14- B: 12 + 1 = 13- D: 15 + 1 = 16Distribution:- A: 14- B: 13- C: 12- D: 16Round 4: D has the most tokens, so D gives one to each of the others and discards one.After Round 4:- D: 16 - 4 = 12- A: 14 + 1 = 15- B: 13 + 1 = 14- C: 12 + 1 = 13Distribution:- A: 15- B: 14- C: 13- D: 12Hmm, interesting. After four rounds, the distribution is:- A: 15- B: 14- C: 13- D: 12Wait, this looks similar to the starting distribution but shifted. Initially, we had:- A: 16- B: 15- C: 14- D: 13After four rounds, the distribution is:- A: 15- B: 14- C: 13- D: 12So, each player has lost one token over four rounds. It seems like the distribution is decreasing by one for each player every four rounds. That suggests a pattern where the total number of tokens is decreasing by four each four rounds (since one token is discarded each round, so 4 rounds * 1 discard = 4 tokens discarded).If this pattern continues, then every four rounds, each player's token count decreases by one. So, starting from the initial distribution, each player's token count will decrease by one every four rounds.Let's see how this progresses.Starting from Round 4:- A: 15- B: 14- C: 13- D: 12After another four rounds (Round 8):- A: 14- B: 13- C: 12- D: 11After another four rounds (Round 12):- A: 13- B: 12- C: 11- D: 10Continuing this pattern, after 4k rounds, each player will have their initial token count minus k.Wait, let me check that. After 4 rounds, each player has lost 1 token. After 8 rounds, each has lost 2 tokens, and so on. So, after 4k rounds, each player has lost k tokens.Therefore, after 4k rounds:- A: 16 - k- B: 15 - k- C: 14 - k- D: 13 - kWe need to find when any player reaches zero tokens. So, we can set up the equation for each player and find the smallest k such that any of them reach zero.For A: 16 - k = 0 => k = 16For B: 15 - k = 0 => k = 15For C: 14 - k = 0 => k = 14For D: 13 - k = 0 => k = 13So, the smallest k is 13, which would mean after 4*13 = 52 rounds, player D would reach zero tokens.But wait, in the pattern I observed, each player loses one token every four rounds. So, after 52 rounds (which is 13 sets of four rounds), D would have 13 - 13 = 0 tokens.However, I need to verify this because the initial rounds don't follow this exact pattern perfectly. Let's check a few more rounds to ensure the pattern holds.After Round 12:- A: 13- B: 12- C: 11- D: 10Round 13: A has the most tokens (13). So, A gives one to each of B, C, D and discards one.After Round 13:- A: 13 - 4 = 9- B: 12 + 1 = 13- C: 11 + 1 = 12- D: 10 + 1 = 11Distribution:- A: 9- B: 13- C: 12- D: 11Hmm, this breaks the pattern I thought. After Round 12, A had 13, but after Round 13, A has 9, which is a decrease of 4 tokens, not just 1. So, my initial assumption that each player loses one token every four rounds was incorrect because the player with the most tokens can lose more tokens in a single round.Therefore, I need to reconsider the approach.Let me try to simulate a few more rounds manually to see if a different pattern emerges.Round 14: B has the most tokens (13). So, B gives one to each of A, C, D and discards one.After Round 14:- B: 13 - 4 = 9- A: 9 + 1 = 10- C: 12 + 1 = 13- D: 11 + 1 = 12Distribution:- A: 10- B: 9- C: 13- D: 12Round 15: C has the most tokens (13). So, C gives one to each of A, B, D and discards one.After Round 15:- C: 13 - 4 = 9- A: 10 + 1 = 11- B: 9 + 1 = 10- D: 12 + 1 = 13Distribution:- A: 11- B: 10- C: 9- D: 13Round 16: D has the most tokens (13). So, D gives one to each of A, B, C and discards one.After Round 16:- D: 13 - 4 = 9- A: 11 + 1 = 12- B: 10 + 1 = 11- C: 9 + 1 = 10Distribution:- A: 12- B: 11- C: 10- D: 9So, after 16 rounds, the distribution is:- A: 12- B: 11- C: 10- D: 9Wait, this looks similar to the distribution after 4 rounds, which was:- A: 15- B: 14- C: 13- D: 12But now, after 16 rounds, it's:- A: 12- B: 11- C: 10- D: 9So, it seems like every 12 rounds, the distribution decreases by 3 tokens for each player. Wait, let me check.From Round 4 to Round 16 is 12 rounds, and each player has lost 3 tokens.From Round 4:- A: 15- B: 14- C: 13- D: 12To Round 16:- A: 12 (15 - 3)- B: 11 (14 - 3)- C: 10 (13 - 3)- D: 9 (12 - 3)So, every 12 rounds, each player loses 3 tokens. Therefore, the rate is 3 tokens lost per 12 rounds, which simplifies to 1 token lost every 4 rounds.Wait, that brings us back to the initial pattern I thought, but that conflicts with what happened after Round 13.Wait, let me think again. From Round 4 to Round 16, which is 12 rounds, each player loses 3 tokens. So, the rate is 1 token lost every 4 rounds. But in between, like after Round 13, A lost 4 tokens in one round, which disrupts the pattern.So, perhaps the overall trend is that every 12 rounds, each player loses 3 tokens, but within those 12 rounds, there are fluctuations where the player with the most tokens loses 4 tokens in a single round.Therefore, maybe the total number of rounds can be calculated by considering how many times each player can lose 4 tokens before reaching zero, but this seems complicated.Alternatively, perhaps we can model the total number of tokens and see when a player's tokens reach zero.The total number of tokens decreases by 1 each round, starting from 58. So, the game will last at most 58 rounds, but it will end earlier when a player reaches zero.We need to find when any of the players reaches zero.Let me consider the number of tokens each player has as a function of rounds.Let me denote the number of rounds as t.For each round, the player with the most tokens gives away 4 tokens (one to each of the other three and one discarded). So, the player with the most tokens decreases by 4, and the others increase by 1.This seems complex to model because the player with the most tokens can change each round.Alternatively, perhaps we can notice that the order of players having the most tokens cycles through A, B, C, D, and so on.From the initial rounds, we saw that A, B, C, D each took turns being the player with the most tokens.If this cycle continues, then every four rounds, each player has had a turn to be the one giving away tokens.In such a case, over four rounds, each player would have given away 4 tokens once, and received 1 token each from the other three players.Wait, no. Each player, when they are the one with the most tokens, gives away 4 tokens. The other players each receive 1 token.So, over four rounds, each player would have given away 4 tokens once and received 1 token three times (once from each of the other rounds).Therefore, over four rounds, each player's token count would change as follows:- The player who gave away tokens: -4- The other three players: +1 eachBut since each player gets to give away tokens once every four rounds, over four rounds, each player's net change is:- For the player who gave away tokens: -4 + 3*(+1) = -4 + 3 = -1- For the other players: 3*(+1) = +3Wait, no. Let's clarify.Each player, when they give away tokens, loses 4 tokens, but in the other three rounds, they receive 1 token each time.So, over four rounds, each player will have one round where they lose 4 tokens, and three rounds where they gain 1 token each.Therefore, the net change for each player over four rounds is:-1 (because -4 + 3*1 = -1)So, each player's token count decreases by 1 every four rounds.Therefore, the number of tokens each player has decreases by 1 every four rounds.This seems to align with the initial pattern we saw.So, if this is the case, then the number of tokens for each player after t rounds can be approximated as:- A: 16 - (t / 4)- B: 15 - (t / 4)- C: 14 - (t / 4)- D: 13 - (t / 4)But since t must be an integer, we need to consider t as a multiple of 4.Wait, but in reality, the token changes happen every round, not every four rounds, so this approximation might not be exact.However, if the cycle of A, B, C, D giving away tokens continues, then over four rounds, each player's token count decreases by 1.Therefore, to find when a player reaches zero, we can set up the equation:For each player, 16 - (t / 4) = 0, 15 - (t / 4) = 0, 14 - (t / 4) = 0, 13 - (t / 4) = 0Solving for t:For A: t = 64For B: t = 60For C: t = 56For D: t = 52So, the smallest t is 52, which would mean after 52 rounds, player D would reach zero tokens.However, earlier, when we manually simulated up to Round 16, we saw that the distribution was:- A: 12- B: 11- C: 10- D: 9Which is consistent with each player losing 3 tokens over 12 rounds (since 12 / 4 = 3). So, after 52 rounds, which is 13 sets of four rounds, each player would have lost 13 tokens:- A: 16 - 13 = 3- B: 15 - 13 = 2- C: 14 - 13 = 1- D: 13 - 13 = 0Therefore, after 52 rounds, D would have 0 tokens, and the game would end.However, I need to verify this because in the earlier manual simulation, after Round 4, the distribution was:- A: 15- B: 14- C: 13- D: 12Which is each player having lost 1 token. After Round 8, it would be:- A: 14- B: 13- C: 12- D: 11And so on, until after Round 52, D would have 0.But wait, in the manual simulation up to Round 16, the distribution was:- A: 12- B: 11- C: 10- D: 9Which is consistent with losing 4 tokens each in four rounds, but actually, it's 12 rounds to lose 3 tokens, which is 1 token every four rounds.So, the calculation seems consistent.Therefore, the game would end after 52 rounds when D reaches zero tokens.But wait, let me think again. In the manual simulation, after Round 16, D had 9 tokens. So, after 16 rounds, D has 9 tokens. To reach zero, D needs to lose 9 more tokens. Since D loses 1 token every four rounds, it would take 9*4 = 36 more rounds, totaling 16 + 36 = 52 rounds.Yes, that makes sense.Therefore, the total number of rounds is 52.But wait, looking back at the options given: (A) 51, (B) 52, (C) 53, (D) 54, (E) 55So, 52 is an option, which is B.However, let me double-check because sometimes when counting rounds, the last round might be the one that causes the token to reach zero, so perhaps it's 53.Wait, let's think carefully. After 52 rounds, D would have 0 tokens. So, the game ends after the 52nd round because D cannot continue. So, the total number of rounds is 52.But let me simulate one more cycle to be sure.After Round 52, D would have 0 tokens. So, the game ends at Round 52.Therefore, the answer should be 52 rounds, which is option B.</think>