💡First, I need to determine how long it took Jordan to run 3 miles. Since Jordan ran 3 miles in a third of the time it took Steve to run 4 miles, and Steve took 32 minutes for 4 miles, I can calculate Jordan's time for 3 miles as ( frac{32}{3} ) minutes.Next, I'll find Jordan's running rate per mile by dividing the time it took Jordan to run 3 miles by 3. This gives me ( frac{32}{9} ) minutes per mile.Finally, to find out how long it would take Jordan to run 7 miles, I'll multiply the running rate per mile by 7. This results in ( frac{224}{9} ) minutes.

💡To solve the equation (log_{125}(5x + 3) = -frac{1}{3}), I start by converting the logarithmic equation to its exponential form. This gives me (125^{-1/3} = 5x + 3). Recognizing that (125) is (5^3), I simplify (125^{-1/3}) to (5^{-1}), which is (frac{1}{5}).Substituting this back into the equation, I have (frac{1}{5} = 5x + 3). To isolate (x), I subtract 3 from both sides, resulting in (frac{1}{5} - 3 = 5x). Simplifying the left side, I get (-frac{14}{5} = 5x). Finally, dividing both sides by 5, I find that (x = -frac{14}{25}).

💡Alright, so I came across this interesting problem: "Show that in any set of 10 consecutive integers there is one which is relatively prime to each of the others." At first glance, it seems a bit tricky, but I think I can break it down step by step.Okay, let's start by understanding what it means for two numbers to be relatively prime. Two numbers are relatively prime if their greatest common divisor (GCD) is 1. That means they don't share any common factors other than 1. So, in this problem, we need to show that within any 10 consecutive integers, there's at least one number that doesn't share any common factors with the other nine numbers in the set.Hmm, so how do I approach this? Maybe I can think about the properties of consecutive integers and their divisibility. Since we're dealing with 10 consecutive numbers, they cover a range of 10 numbers, which is exactly the number of residues modulo 10. That might be useful.Let me consider the possible prime factors that could divide these numbers. The primes less than 10 are 2, 3, 5, and 7. These are the only primes that could potentially divide any of the numbers in the set because if a number has a prime factor greater than 10, it won't affect the GCD with the other numbers in the set since they are consecutive.So, if I can find a number in the set that isn't divisible by 2, 3, 5, or 7, then that number would be relatively prime to all the others. Because if it doesn't share any of these prime factors, it can't share any common divisors greater than 1 with the other numbers.Wait, but how do I ensure that such a number exists in any set of 10 consecutive integers? Maybe I can use the pigeonhole principle here. Let's think about how many numbers in the set are divisible by each of these primes.For example, in any 10 consecutive numbers, exactly 5 will be even (divisible by 2), and exactly 2 will be divisible by 5. For 3, since 10 divided by 3 is approximately 3.33, there will be either 3 or 4 numbers divisible by 3 in the set. Similarly, for 7, since 10 divided by 7 is approximately 1.42, there will be either 1 or 2 numbers divisible by 7.So, adding these up, we have 5 (divisible by 2) + 2 (divisible by 5) + 3 or 4 (divisible by 3) + 1 or 2 (divisible by 7). That's potentially up to 5 + 2 + 4 + 2 = 13 numbers. But wait, we only have 10 numbers in the set. This means there must be some overlap—numbers that are divisible by more than one of these primes.Ah, so not all of these divisibility counts can be achieved simultaneously because some numbers are counted multiple times. For example, a number divisible by both 2 and 3 is counted in both the divisible by 2 and divisible by 3 categories. So, the actual number of distinct numbers divisible by 2, 3, 5, or 7 is less than the sum of their individual counts.To get a better estimate, maybe I can use the inclusion-exclusion principle. But that might get complicated. Instead, perhaps I can think about the numbers that are not divisible by any of these primes. If I can show that there's at least one such number in the set, then that number would be relatively prime to all the others.So, how many numbers in the set are not divisible by 2, 3, 5, or 7? Let's see. In any 10 consecutive numbers, 5 are even, so 5 are odd. Out of these 5 odd numbers, how many are divisible by 3, 5, or 7?Well, in the 10 consecutive numbers, there are 2 numbers divisible by 5. One of them is even (ends with 0) and the other is odd (ends with 5). So, among the 5 odd numbers, one is divisible by 5.Similarly, for 3, in 10 consecutive numbers, there are either 3 or 4 numbers divisible by 3. Let's say 3 for simplicity. Among these, some are even and some are odd. Specifically, in 10 numbers, the number of odd multiples of 3 would be either 1 or 2, depending on where the set starts.Similarly, for 7, in 10 consecutive numbers, there's either 1 or 2 numbers divisible by 7. Again, one of them could be odd.So, putting this together, among the 5 odd numbers in the set:- 1 is divisible by 5.- 1 or 2 are divisible by 3.- 0 or 1 are divisible by 7.So, the maximum number of odd numbers divisible by 3, 5, or 7 is 1 (for 5) + 2 (for 3) + 1 (for 7) = 4. But we only have 5 odd numbers. So, at least one odd number is not divisible by 3, 5, or 7.Therefore, this number is not divisible by 2, 3, 5, or 7, meaning it's relatively prime to all the other numbers in the set.Wait, does this always hold? Let me test it with an example. Let's take the set from 1 to 10.Numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.Odd numbers: 1, 3, 5, 7, 9.Divisible by 5: 5.Divisible by 3: 3, 9.Divisible by 7: 7.So, among the odd numbers, 5 is divisible by 5, 3 and 9 are divisible by 3, and 7 is divisible by 7. That leaves 1, which is not divisible by any of these primes. So, 1 is relatively prime to all others.Another example: 11 to 20.Numbers: 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.Odd numbers: 11, 13, 15, 17, 19.Divisible by 5: 15.Divisible by 3: 15, 18 (but 18 is even, so among odd numbers, only 15).Divisible by 7: 14 (even) and 21 (not in the set). So, no odd numbers divisible by 7.So, among the odd numbers: 11, 13, 15, 17, 19.15 is divisible by 3 and 5, others are not. So, 11, 13, 17, 19 are not divisible by 2, 3, 5, or 7. So, multiple numbers are relatively prime to all others.Wait, but the problem says "there is one which is relatively prime to each of the others," not necessarily more than one. So, in this case, even if multiple numbers satisfy the condition, it's still true.Another example: 21 to 30.Numbers: 21, 22, 23, 24, 25, 26, 27, 28, 29, 30.Odd numbers: 21, 23, 25, 27, 29.Divisible by 5: 25.Divisible by 3: 21, 27.Divisible by 7: 21.So, among the odd numbers: 21 is divisible by 3 and 7, 25 is divisible by 5, 27 is divisible by 3, leaving 23 and 29. Both are primes, so they are relatively prime to all others.So, again, multiple numbers satisfy the condition.Wait, but what if the set starts at a number where more of the odd numbers are divisible by 3, 5, or 7? Let's try 100 to 109.Numbers: 100, 101, 102, 103, 104, 105, 106, 107, 108, 109.Odd numbers: 101, 103, 105, 107, 109.Divisible by 5: 105.Divisible by 3: 105, 108 (even), so only 105 among odd numbers.Divisible by 7: 105 (since 105 = 15*7).So, 105 is divisible by 3, 5, and 7. The other odd numbers: 101, 103, 107, 109. All of these are primes or not divisible by 2, 3, 5, or 7. So, again, multiple numbers satisfy the condition.Hmm, it seems consistent that in any set of 10 consecutive numbers, there's at least one number that's not divisible by 2, 3, 5, or 7, making it relatively prime to all others.But wait, what about a set where all the odd numbers are divisible by 3, 5, or 7? Is that possible?Let's see. Suppose we have 10 consecutive numbers where all 5 odd numbers are divisible by 3, 5, or 7. Is that possible?Well, let's try to construct such a set. Let's say we have numbers from n to n+9.We need all odd numbers in this range to be divisible by 3, 5, or 7.But considering the spacing, it's unlikely because the gaps between multiples of 3, 5, and 7 are larger than 10 in some cases.Wait, for example, multiples of 3 are every 3 numbers, multiples of 5 every 5, and multiples of 7 every 7. So, within 10 numbers, you can have at most 3 multiples of 3, 2 multiples of 5, and 1 multiple of 7.But since we're only considering odd numbers, which are 5 in total, and some of these multiples overlap (like 15 is a multiple of both 3 and 5), it's still not possible to cover all 5 odd numbers with just these multiples.Because even if we have 3 multiples of 3, 2 multiples of 5, and 1 multiple of 7, the total unique numbers covered would be 3 + 2 + 1 = 6, but considering overlaps, it's less. However, we only have 5 odd numbers, so even if all of them were covered, it's still possible, but in reality, the overlaps mean that not all can be covered.Wait, let's think about it. If we have 5 odd numbers, and we need to cover them with multiples of 3, 5, or 7.Each multiple of 3 can cover one odd number, each multiple of 5 can cover one odd number, and each multiple of 7 can cover one odd number.But in 10 consecutive numbers, there are at most 3 multiples of 3, but only some are odd. Similarly, at most 2 multiples of 5, one of which is odd, and at most 1 multiple of 7, which could be odd.So, the maximum number of odd numbers that can be covered is 3 (from 3) + 1 (from 5) + 1 (from 7) = 5. So, in theory, it's possible that all 5 odd numbers are divisible by 3, 5, or 7.But wait, is that actually possible? Let's try to find such a set.Let me try to find 10 consecutive numbers where all 5 odd numbers are divisible by 3, 5, or 7.Let's see. Let's start at n = 15.Numbers: 15, 16, 17, 18, 19, 20, 21, 22, 23, 24.Odd numbers: 15, 17, 19, 21, 23.15 is divisible by 3 and 5.17 is prime.19 is prime.21 is divisible by 3 and 7.23 is prime.So, 15, 21 are covered, but 17, 19, 23 are not. So, not all odd numbers are covered.Another try: n = 21.Numbers: 21, 22, 23, 24, 25, 26, 27, 28, 29, 30.Odd numbers: 21, 23, 25, 27, 29.21 divisible by 3 and 7.23 prime.25 divisible by 5.27 divisible by 3.29 prime.So, 21, 25, 27 are covered, but 23, 29 are not.Another try: n = 105.Numbers: 105, 106, 107, 108, 109, 110, 111, 112, 113, 114.Odd numbers: 105, 107, 109, 111, 113.105 divisible by 3, 5, 7.107 prime.109 prime.111 divisible by 3.113 prime.So, 105, 111 are covered, others are not.Hmm, seems challenging to find a set where all odd numbers are covered. Maybe it's not possible.Wait, let's think about the Chinese Remainder Theorem. If we try to set up a system where each odd number is congruent to 0 mod 3, 5, or 7, but given the spacing, it's impossible because the required congruences would overlap in a way that can't cover all 5 odd numbers without conflict.Therefore, it's impossible for all 5 odd numbers in a set of 10 consecutive integers to be divisible by 3, 5, or 7. Hence, there must be at least one odd number that is not divisible by any of these primes, making it relatively prime to all other numbers in the set.So, putting it all together, in any set of 10 consecutive integers, there are 5 odd numbers. Among these, at most 4 can be divisible by 3, 5, or 7 (due to overlaps and the limited number of multiples in 10 numbers). Therefore, at least one odd number remains that is not divisible by 2, 3, 5, or 7, ensuring it's relatively prime to all others.This seems to cover all cases, so I think this approach works.

💡Alright, so I have this problem where I need to figure out how many ways I can place four figures into a six-page homework assignment. The order of the figures is fixed, and each page can have at most two figures. The position of the figures on a page doesn't matter, so it's just about which figures go on which pages.Okay, let me break this down. First, I need to consider how the four figures can be distributed across the six pages. Since each page can have at most two figures, I have a few different scenarios to consider.**Case 1: Each figure is on a separate page.**In this case, I need to choose 4 pages out of the 6 to place each figure. Since the order of the figures is fixed, I don't need to worry about arranging them differently. The number of ways to choose 4 pages out of 6 is given by the combination formula:[binom{6}{4} = frac{6!}{4!(6-4)!} = frac{6!}{4! cdot 2!} = frac{720}{24 cdot 2} = 15]So, there are 15 ways to place each figure on a separate page.**Case 2: One page contains 2 figures, and the other two figures are on separate pages.**Here, I need to choose 3 pages out of the 6. One of these pages will hold two figures, and the other two pages will each hold one figure. First, I choose the 3 pages:[binom{6}{3} = frac{6!}{3!(6-3)!} = frac{6!}{3! cdot 3!} = frac{720}{6 cdot 6} = 20]Next, I need to choose which 2 of the 4 figures will be on the same page. This is another combination problem:[binom{4}{2} = frac{4!}{2! cdot 2!} = frac{24}{2 cdot 2} = 6]So, the total number of ways for this case is:[binom{6}{3} times binom{4}{2} = 20 times 6 = 120]**Case 3: Two pages each contain 2 figures.**In this scenario, I need to choose 2 pages out of the 6 to hold the figures. Each of these pages will have 2 figures. First, choose the 2 pages:[binom{6}{2} = frac{6!}{2!(6-2)!} = frac{6!}{2! cdot 4!} = frac{720}{2 cdot 24} = 15]Now, distribute the 4 figures across these 2 pages. I need to choose 2 figures out of 4 to go on the first page, and the remaining 2 will go on the second page:[binom{4}{2} = frac{4!}{2! cdot 2!} = frac{24}{2 cdot 2} = 6]So, the total number of ways for this case is:[binom{6}{2} times binom{4}{2} = 15 times 6 = 90]**Total Number of Ways**Now, I'll add up the number of ways from each case:[15 + 120 + 90 = 225]So, there are 225 ways to place the four figures on the six pages under the given constraints.

💡First, I recognize that A can complete the work in an unknown number of days, which I'll denote as ( x ). B can complete the same work in 12 days.Next, I calculate the work rates of A and B. A's work rate is ( frac{1}{x} ) and B's work rate is ( frac{1}{12} ).When A and B work together, their combined work rate is the sum of their individual work rates:[frac{1}{x} + frac{1}{12}]I know that together they can finish the work in 4 days, so their combined work rate is ( frac{1}{4} ).Setting up the equation:[frac{1}{x} + frac{1}{12} = frac{1}{4}]To solve for ( x ), I find a common denominator for the fractions, which is ( 12x ):[frac{12}{12x} + frac{x}{12x} = frac{3x}{12x}]Equating the numerators:[12 + x = 3x]Subtracting ( x ) from both sides:[12 = 2x]Dividing both sides by 2:[x = 6]Therefore, A can complete the work alone in 6 days.

💡Alright, so I have this problem here where I need to find the number of ways to draw 22 straight edges between 10 labeled points, with no three points being collinear. The key constraints are that no triangle should be formed, and there can be at most one edge between any two points. Then, I need to compute N divided by 9 factorial, where N is the number of such ways.First, let me try to understand the problem better. We have 10 labeled points, and we need to connect them with 22 edges. Since no three points are collinear, any three points can form a triangle unless we specifically avoid it. So, the main constraint is that the graph we form shouldn't contain any triangles.Hmm, so triangle-free graphs. I remember that bipartite graphs are triangle-free because they don't have any odd-length cycles, and triangles are cycles of length 3. So, maybe the graph we're looking at is bipartite.If the graph is bipartite, we can split the 10 points into two disjoint sets such that all edges go between the sets, and none within the sets. So, the first thing I need to figure out is how to partition these 10 points into two sets.Since 10 is an even number, the most straightforward partitions are 5 and 5, or 6 and 4. I wonder if both of these partitions are possible or if one is more suitable than the other.Let me think: for a bipartite graph with partitions of size m and n, the maximum number of edges is m*n. So, for a 5-5 partition, the maximum number of edges is 25. For a 6-4 partition, it's 24. But we need 22 edges, which is less than both maximums. So, both partitions are possible.Wait, but 22 is less than 24 and 25, so maybe both partitions can give us the required number of edges. So, perhaps I need to consider both cases.So, the total number of ways N will be the sum of the number of ways for the 5-5 partition and the 6-4 partition.Let me formalize this:Case 1: Partition into sets of size 5 and 5.Case 2: Partition into sets of size 6 and 4.For each case, I need to compute the number of ways to choose the partition and then the number of ways to choose 22 edges without forming a triangle.Starting with Case 1: 5-5 partition.First, how many ways are there to partition 10 labeled points into two sets of 5? That's the combination C(10,5). But wait, since the two sets are indistinct (i.e., swapping the two sets doesn't create a new partition), I need to divide by 2 to avoid double-counting. So, the number of distinct partitions is C(10,5)/2.Once the partition is fixed, the maximum number of edges is 25. We need 22 edges, so we need to remove 3 edges. The number of ways to remove 3 edges from 25 is C(25,3).Therefore, the number of ways for Case 1 is [C(10,5)/2] * C(25,3).Moving on to Case 2: 6-4 partition.Here, how many ways are there to partition 10 labeled points into sets of 6 and 4? That's simply C(10,4), since choosing 4 points automatically defines the other 6.Once the partition is fixed, the maximum number of edges is 6*4=24. We need 22 edges, so we need to remove 2 edges. The number of ways to remove 2 edges from 24 is C(24,2).Therefore, the number of ways for Case 2 is C(10,4) * C(24,2).So, the total number of ways N is the sum of the two cases:N = [C(10,5)/2 * C(25,3)] + [C(10,4) * C(24,2)].Now, I need to compute this value and then divide it by 9!.Let me compute each term step by step.First, compute C(10,5):C(10,5) = 10! / (5! * 5!) = (10*9*8*7*6)/(5*4*3*2*1) = 252.So, C(10,5)/2 = 252 / 2 = 126.Next, compute C(25,3):C(25,3) = 25! / (3! * 22!) = (25*24*23)/(3*2*1) = (25*24*23)/6.Let me compute that:25*24 = 600,600*23 = 13,800,13,800 / 6 = 2,300.So, C(25,3) = 2,300.Therefore, the first term is 126 * 2,300.Let me compute that:126 * 2,300 = (100 + 26) * 2,300 = 100*2,300 + 26*2,300 = 230,000 + 59,800 = 289,800.So, the first term is 289,800.Now, compute the second term: C(10,4) * C(24,2).First, compute C(10,4):C(10,4) = 10! / (4! * 6!) = (10*9*8*7)/(4*3*2*1) = (5040)/(24) = 210.Next, compute C(24,2):C(24,2) = 24! / (2! * 22!) = (24*23)/2 = 276.So, the second term is 210 * 276.Let me compute that:210 * 276 = (200 + 10) * 276 = 200*276 + 10*276 = 55,200 + 2,760 = 57,960.Therefore, the second term is 57,960.Now, summing both terms:N = 289,800 + 57,960 = 347,760.So, N = 347,760.Now, I need to compute N / 9!.First, compute 9!:9! = 362,880.So, N / 9! = 347,760 / 362,880.Let me simplify this fraction.First, let's see if both numerator and denominator can be divided by 10: 347,760 / 10 = 34,776; 362,880 /10 = 36,288.So, now we have 34,776 / 36,288.Let me see if both can be divided by 24:34,776 ÷ 24 = 1,449.36,288 ÷ 24 = 1,512.So, now we have 1,449 / 1,512.Let me see if both can be divided by 3:1,449 ÷ 3 = 483.1,512 ÷ 3 = 504.So, now we have 483 / 504.Again, divide by 3:483 ÷ 3 = 161.504 ÷ 3 = 168.So, now we have 161 / 168.Check if both can be divided by 7:161 ÷ 7 = 23.168 ÷ 7 = 24.So, now we have 23 / 24.Therefore, N / 9! = 23/24.Wait, that seems too clean. Let me verify my calculations.Starting from N = 347,760 and 9! = 362,880.Compute 347,760 / 362,880.Divide numerator and denominator by 10: 34,776 / 36,288.Divide by 24: 34,776 ÷24=1,449; 36,288 ÷24=1,512.Divide by 3: 1,449 ÷3=483; 1,512 ÷3=504.Divide by 3: 483 ÷3=161; 504 ÷3=168.Divide by 7: 161 ÷7=23; 168 ÷7=24.Yes, so 23/24 is correct.Therefore, the final answer is 23/24.**Final Answer**boxed{dfrac{23}{24}}