💡Okay, so I have this problem here where the sum of three different numbers is 80. The largest number is 9 more than the middle number, and the middle number is 4 more than the smallest number. Plus, the product of the two smaller numbers is 525. I need to find the value of the largest number. Hmm, let's break this down step by step.First, I'll assign variables to each of the numbers to make it easier. Let's say the smallest number is 'a'. Then, since the middle number is 4 more than the smallest, the middle number would be 'a + 4'. Similarly, the largest number is 9 more than the middle number, so that would be '(a + 4) + 9', which simplifies to 'a + 13'. So, now I have:- Smallest number: a- Middle number: a + 4- Largest number: a + 13Next, the problem says that the sum of these three numbers is 80. So, I can write that as an equation:a + (a + 4) + (a + 13) = 80Let me simplify this equation. Combining like terms:a + a + 4 + a + 13 = 80That's 3a + 17 = 80Now, I can solve for 'a' by subtracting 17 from both sides:3a = 80 - 173a = 63Then, divide both sides by 3:a = 63 / 3a = 21So, the smallest number is 21. Now, let's find the middle number, which is 'a + 4':Middle number = 21 + 4 = 25And the largest number is 'a + 13':Largest number = 21 + 13 = 34Wait a second, the problem also mentions that the product of the two smaller numbers is 525. Let me check if that's true. The two smaller numbers are 21 and 25. Multiplying them:21 * 25 = 525Yes, that's correct. So, all the conditions are satisfied.Therefore, the largest number is 34.

💡First, I recognize that Layla scored a total of 104 goals across four hockey games.Kristin scored fewer goals than Layla in the same four games.The average number of goals scored by both Layla and Kristin together is 92.To find out how many goals Kristin scored, I'll denote Kristin's total goals as ( K ).Using the average formula, the combined total goals scored by both Layla and Kristin is:[frac{104 + K}{2} = 92]Solving for ( K ):[104 + K = 184][K = 184 - 104][K = 80]Kristin scored 80 goals in total.To determine how many fewer goals Kristin scored compared to Layla, I'll subtract Kristin's total goals from Layla's total goals:[104 - 80 = 24]Therefore, Kristin scored 24 fewer goals than Layla.

💡Okay, so I have this problem where João starts making marks on his clock every 80 minutes, starting at noon. I need to figure out two things: a) After how much time will it no longer be necessary to make new marks on the clock?b) What is the sum of the interior angles of the polygon formed by the marks?Alright, let's tackle part (a) first. João makes a mark every 80 minutes. I know that the hour hand of a clock completes a full circle, which is 360 degrees, in 12 hours. Since 12 hours is 720 minutes, the hour hand takes 720 minutes to make a complete revolution.So, if João is making a mark every 80 minutes, I need to find out after how many marks the positions will start repeating. This sounds like a problem involving least common multiples (LCM). The LCM of 80 minutes and 720 minutes will give me the time after which the marks start repeating.Let me calculate the LCM of 80 and 720. First, I'll factor both numbers:- 80 = 2^4 * 5- 720 = 2^4 * 3^2 * 5The LCM is the product of the highest powers of all prime factors present in either number. So, that would be 2^4 * 3^2 * 5 = 16 * 9 * 5 = 720.Wait, so the LCM of 80 and 720 is 720 minutes. That means after 720 minutes, the marks will start repeating. So, João will have made all the necessary marks by then, and no new marks will be needed after that.But let me think again. If he makes a mark every 80 minutes, how many marks will he make in 720 minutes? 720 divided by 80 is 9. So, he'll make 9 marks. Since the clock has 12 hours, which is 720 minutes, and each hour mark is 60 minutes apart, but he's marking every 80 minutes. So, the positions of these marks will be spread out around the clock.But since 80 and 720 have an LCM of 720, it means that after 720 minutes, the marks will align again with the starting point, meaning no new marks are needed. So, the answer for part (a) is 720 minutes.Now, moving on to part (b). The sum of the interior angles of a polygon. I remember that the formula for the sum of interior angles of a polygon with n sides is (n-2)*180 degrees.From part (a), we found that João makes 9 marks before they start repeating. So, the polygon formed by these marks is a nonagon, which has 9 sides.Plugging into the formula: (9-2)*180 = 7*180 = 1260 degrees.Wait, let me verify that. If n=9, then yes, (9-2)=7, and 7*180=1260. That seems right.But just to be thorough, let me think about the polygon. Each mark is 80 minutes apart, which is 80/720 = 1/9 of the clock's circumference. So, the angle between each mark is 360/9 = 40 degrees. So, each interior angle of the polygon would be 40 degrees? Wait, no, that's the central angle, not the interior angle.Hmm, maybe I confused something here. The central angle is 40 degrees, but the interior angle of the polygon is different. Wait, no, actually, in a regular polygon, the central angle is equal to 360/n, and the interior angle is given by (n-2)*180/n. So, for a nonagon, the interior angle would be (9-2)*180/9 = 140 degrees. But the question asks for the sum of the interior angles, not each individual angle.So, the sum is indeed (9-2)*180 = 1260 degrees. So, my initial calculation was correct.Therefore, the answers are:a) 720 minutesb) 1260 degrees

💡Alright, so I have this problem here about 2016 students lining up and counting off from left to right according to numbers 1, 2, up to n, where n is at least 2. The key point is that if the 2016th student counts to n, then all students who counted to n in that round receive a New Year's gift. The question is asking how many students will definitely not receive a New Year's gift, regardless of what n is.Hmm, okay. Let me try to break this down. So, we have 2016 students in a row, each one counts off from 1 to n, and this repeats until the 2016th student. If the 2016th student counts to n, that means n has to be a number that divides evenly into 2016, right? Because if you're counting off in groups of n, the last person should land exactly on n for the last group.So, n must be a divisor of 2016. That makes sense. Therefore, n can be any number that divides 2016 without leaving a remainder. So, the possible values of n are the divisors of 2016.Now, the problem is asking about students who will definitely not receive a gift, no matter what n is. So, regardless of which divisor n we choose, certain students will never be in the position where they count to n. We need to find how many such students there are.Let me think about this. If a student is in position x, they will receive a gift if x is a multiple of n. So, for each n, the students who are multiples of n get a gift. But we need to find the students who are never multiples of any possible n, which are the divisors of 2016.Wait, so if a student's position is x, and x is not a multiple of any divisor of 2016, then they will never receive a gift. But actually, every number is a multiple of 1, but n is at least 2, so 1 is not considered here. So, we need to find numbers that are not multiples of any divisor of 2016 greater than or equal to 2.But actually, every number can be expressed as a multiple of some divisor. Hmm, maybe I'm approaching this incorrectly.Wait, perhaps it's the other way around. For a student to never receive a gift, their position x must not be congruent to 0 modulo n for any n that divides 2016. So, x should not be divisible by any divisor of 2016 except 1.But since n is at least 2, x must not be divisible by any n that divides 2016 and is at least 2. So, x must be such that it's not divisible by any of the divisors of 2016 except 1. But that would mean x is coprime with 2016.Ah, so the students who are in positions that are coprime with 2016 will never be counted as n, because n is a divisor of 2016, and if x is coprime with 2016, it can't be a multiple of any n that divides 2016.Therefore, the number of students who will definitely not receive a gift is equal to the number of integers between 1 and 2016 that are coprime with 2016. This is given by Euler's Totient function, φ(2016).Okay, so I need to calculate φ(2016). To do that, I first need the prime factorization of 2016.Let me factorize 2016:2016 ÷ 2 = 1008 1008 ÷ 2 = 504 504 ÷ 2 = 252 252 ÷ 2 = 126 126 ÷ 2 = 63 63 ÷ 3 = 21 21 ÷ 3 = 7 7 ÷ 7 = 1So, the prime factors are 2^5 * 3^2 * 7^1.Euler's Totient function is calculated as:φ(n) = n * (1 - 1/p1) * (1 - 1/p2) * ... * (1 - 1/pk)where p1, p2, ..., pk are the distinct prime factors of n.So, for 2016:φ(2016) = 2016 * (1 - 1/2) * (1 - 1/3) * (1 - 1/7)Let me compute each part:1 - 1/2 = 1/2 1 - 1/3 = 2/3 1 - 1/7 = 6/7So,φ(2016) = 2016 * (1/2) * (2/3) * (6/7)Let me compute this step by step:First, 2016 * (1/2) = 1008 1008 * (2/3) = 672 672 * (6/7) = 576So, φ(2016) = 576.Therefore, there are 576 students who are in positions coprime with 2016, meaning they will never be counted as n for any divisor n of 2016. Hence, these 576 students will definitely not receive a New Year's gift, regardless of what n is.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting with φ(2016):2016 = 2^5 * 3^2 * 7^1φ(2016) = 2016 * (1 - 1/2) * (1 - 1/3) * (1 - 1/7) = 2016 * 1/2 * 2/3 * 6/7Multiplying these fractions:1/2 * 2/3 = 1/3 1/3 * 6/7 = 2/7So, φ(2016) = 2016 * 2/7 = (2016 / 7) * 2 = 288 * 2 = 576Yes, that seems correct.So, the number of students who will definitely not receive a gift is 576.**Final Answer**boxed{576}

💡Alright, let's tackle this problem step by step. So, we have a number that, when multiplied by its first digit, gives 494; when multiplied by its second digit, gives 988; and when multiplied by its third digit, gives 1729. We need to find this number.First, let's denote the number as ( x ). Let's also denote its digits as ( a ) (first digit), ( b ) (second digit), and ( c ) (third digit). So, according to the problem:1. ( x times a = 494 )2. ( x times b = 988 )3. ( x times c = 1729 )Our goal is to find ( x ). To do this, we can observe that ( x ) must be a common factor of 494, 988, and 1729. So, the first step is to find the greatest common divisor (GCD) of these three numbers.Let's start by finding the GCD of 494 and 988. Since 988 is exactly twice 494, the GCD of 494 and 988 is 494.Next, we need to find the GCD of 494 and 1729. To do this, we can use the Euclidean algorithm:1. Divide 1729 by 494: - 494 goes into 1729 three times (3 × 494 = 1482) with a remainder of 247 (1729 - 1482 = 247).2. Now, find the GCD of 494 and 247: - 247 goes into 494 exactly twice (2 × 247 = 494) with no remainder.So, the GCD of 494 and 1729 is 247.Since 247 is the GCD of all three numbers (494, 988, and 1729), it must be the value of ( x ).To verify, let's check:1. ( 247 times a = 494 ) → ( a = 2 )2. ( 247 times b = 988 ) → ( b = 4 )3. ( 247 times c = 1729 ) → ( c = 7 )So, the number ( x ) is 247, and its digits are 2, 4, and 7, which satisfy all the given conditions.

💡First, I need to identify the largest two-digit number that can be formed using the digits 9, 4, 3, and 5 without repeating any digit. To do this, I'll consider all possible two-digit combinations and select the largest one.Next, I'll determine the smallest two-digit number that can be formed using the same set of digits. This involves examining all possible two-digit combinations and choosing the smallest one.Once I have both the largest and smallest two-digit numbers, I'll subtract the smallest number from the largest number to find the desired value.Finally, I'll present the result in a clear and concise manner within a box to indicate the final answer.