💡First, I recognize that there were initially 74 apples in the basket.By the end of the day, 32 apples remained, which means a total of 74 - 32 = 42 apples were removed.Let’s denote the number of apples Ricki removed as R. Since Samson removed twice as many apples as Ricki, Samson removed 2R apples.Together, Ricki and Samson removed R + 2R = 3R apples.Setting up the equation: 3R = 42.Solving for R: R = 42 / 3 = 14.Therefore, Ricki removed 14 apples.

💡First, I need to determine the unknown dimension of the room, which I'll call "x" feet.The room has four walls with the following dimensions:- Two walls are 25 feet long and 12 feet high.- The other two walls are "x" feet long and 12 feet high.Calculating the total area of the four walls:- Area of the two longer walls: 2 * (25 * 12) = 600 square feet.- Area of the two shorter walls: 2 * (x * 12) = 24x square feet.- Total area of the walls: 600 + 24x square feet.Next, I need to subtract the areas of the door and windows:- Area of the door: 6 * 3 = 18 square feet.- Area of one window: 4 * 3 = 12 square feet.- Total area of three windows: 3 * 12 = 36 square feet.- Total area to subtract: 18 + 36 = 54 square feet.Subtracting these areas from the total wall area:- Whitewashable area: 600 + 24x - 54 = 546 + 24x square feet.The cost of whitewashing is Rs. 2 per square foot:- Total cost: (546 + 24x) * 2 = 1092 + 48x.Given that the total cost is Rs. 1812, I can set up the equation:- 1092 + 48x = 1812.Solving for x:- 48x = 1812 - 1092 = 720.- x = 720 / 48 = 15 feet.Therefore, the unknown dimension of the room is 15 feet.

💡First, I need to understand the problem. There's an 8% error in measuring the side of a square, and I need to find out how this error affects the calculated area of the square.Let me denote the actual side length of the square as ( a ).If there's an 8% error in excess, the measured side length becomes ( a + 0.08a = 1.08a ).The actual area of the square is ( a^2 ).With the measured side length, the calculated area would be ( (1.08a)^2 = 1.1664a^2 ).The error in the area is the difference between the calculated area and the actual area: ( 1.1664a^2 - a^2 = 0.1664a^2 ).To find the percentage error, I divide the error in the area by the actual area and multiply by 100: ( left( frac{0.1664a^2}{a^2} right) times 100 = 16.64% ).So, the percentage of error in the calculated area of the square is 16.64%.

💡Alright, so I'm trying to find the smallest prime number that, when increased by 1, is exactly divisible by all these numbers: 823, 618, 3648, 60, 3917, 4203, 1543, and 2971. Hmm, that sounds a bit complicated, but let's break it down step by step.First, I need to understand what the problem is asking. It wants a prime number, let's call it 'p', such that when you add 1 to it, the result is divisible by all those numbers. So, mathematically, that means p + 1 is a multiple of each of those numbers. In other words, p + 1 is the least common multiple (LCM) of all those numbers. Therefore, p would be LCM - 1. But p also needs to be a prime number. So, my task is to find the smallest prime number that is one less than the LCM of these numbers.Okay, so the first step is to find the LCM of all these numbers. But calculating the LCM of so many large numbers manually seems really tough. Maybe I can simplify this by finding the prime factors of each number and then taking the highest powers of all the prime factors to compute the LCM.Let me try that. I'll list out the numbers and factorize them:1. **823**: I think 823 is a prime number. Let me check. It doesn't divide by 2, 3, 5, 7, 11, 13, 17, or 19. I'll assume it's prime for now.2. **618**: Let's see, 618 divided by 2 is 309. 309 divided by 3 is 103. 103 is a prime number. So, 618 = 2 × 3 × 103.3. **3648**: This is a larger number. Let's divide by 2: 3648 ÷ 2 = 1824. ÷2 again = 912. ÷2 = 456. ÷2 = 228. ÷2 = 114. ÷2 = 57. Now, 57 ÷ 3 = 19. So, 3648 = 2^6 × 3 × 19.4. **60**: Simple, 60 = 2^2 × 3 × 5.5. **3917**: Hmm, 3917. Let me check if it's prime. Divided by 2? No. 3? 3 × 1305.666, so no. 5? Doesn't end with 0 or 5. 7? 7 × 559.571, nope. 11? 11 × 356.09, no. 13? 13 × 301.307, no. I think 3917 is prime.6. **4203**: Let's see, 4203 ÷ 3 = 1401. 1401 ÷ 3 = 467. 467 is a prime number. So, 4203 = 3^2 × 467.7. **1543**: Checking if it's prime. Divided by 2? No. 3? 3 × 514.333, no. 5? Doesn't end with 0 or 5. 7? 7 × 220.428, no. 11? 11 × 140.27, no. I think 1543 is prime.8. **2971**: Checking for primality. Divided by 2? No. 3? 3 × 990.333, no. 5? No. 7? 7 × 424.428, no. 11? 11 × 270.09, no. I think 2971 is prime.Alright, so now I have the prime factors:- 823: 823- 618: 2 × 3 × 103- 3648: 2^6 × 3 × 19- 60: 2^2 × 3 × 5- 3917: 3917- 4203: 3^2 × 467- 1543: 1543- 2971: 2971To find the LCM, I need to take the highest power of each prime number present in the factorizations.So, let's list out all the primes:- 2: highest power is 2^6 (from 3648)- 3: highest power is 3^2 (from 4203)- 5: highest power is 5^1 (from 60)- 19: highest power is 19^1 (from 3648)- 103: highest power is 103^1 (from 618)- 467: highest power is 467^1 (from 4203)- 823: highest power is 823^1 (from 823)- 1543: highest power is 1543^1 (from 1543)- 2971: highest power is 2971^1 (from 2971)- 3917: highest power is 3917^1 (from 3917)So, the LCM would be:2^6 × 3^2 × 5 × 19 × 103 × 467 × 823 × 1543 × 2971 × 3917Wow, that's a massive number! Calculating this manually would be practically impossible. I think I need to use a calculator or some software to compute this. But since I don't have access to that right now, maybe I can think of another approach.Wait, the problem is asking for the smallest prime number p such that p + 1 is divisible by all these numbers. So, p + 1 is the LCM of these numbers, and p is prime. Therefore, p = LCM - 1.But if LCM is such a huge number, p would also be huge, and checking if it's prime would be challenging. Maybe there's a smarter way to approach this.Perhaps I can look for patterns or properties that can help me find p without computing the entire LCM. For example, if p + 1 is divisible by all these numbers, then p must be congruent to -1 modulo each of these numbers. So, p ≡ -1 mod n for each n in the list.This seems like a system of congruences, which can be solved using the Chinese Remainder Theorem (CRT). But CRT applies when the moduli are pairwise coprime. Are these numbers pairwise coprime?Looking back at the list:- 823: prime- 618: 2 × 3 × 103- 3648: 2^6 × 3 × 19- 60: 2^2 × 3 × 5- 3917: prime- 4203: 3^2 × 467- 1543: prime- 2971: primeClearly, many of these numbers share common factors, especially 2 and 3. So, they are not pairwise coprime. Therefore, CRT in its basic form doesn't apply directly.Hmm, maybe I need to find the least common multiple first, as I initially thought, and then subtract 1 to see if it's prime. But again, calculating that LCM is impractical manually.Alternatively, perhaps there's a way to find a smaller number that satisfies all these divisibility conditions. Maybe p + 1 doesn't have to be the absolute LCM, but just a common multiple. But the problem asks for the smallest prime p, so p + 1 should be the smallest such number, which would be the LCM.I'm stuck here. Maybe I can look for some properties or patterns in the numbers to simplify the problem.Let me see:- 60 is 2^2 × 3 × 5- 618 is 2 × 3 × 103- 3648 is 2^6 × 3 × 19- 4203 is 3^2 × 467- 823, 3917, 1543, 2971 are primesSo, the LCM needs to include:- 2^6 (from 3648)- 3^2 (from 4203)- 5 (from 60)- 19 (from 3648)- 103 (from 618)- 467 (from 4203)- 823, 3917, 1543, 2971 (primes)Therefore, LCM = 2^6 × 3^2 × 5 × 19 × 103 × 467 × 823 × 1543 × 2971 × 3917This is an enormous number. I think it's beyond practical computation without a computer.But the problem is asking for the smallest prime p such that p + 1 is divisible by all these numbers. So, p = LCM - 1. If LCM is this huge number, p would be just one less than that. But is p prime?Given how large LCM is, p would be an extremely large number, and checking its primality would require advanced algorithms and significant computational power.Maybe there's a mistake in my approach. Perhaps I don't need to compute the entire LCM. Maybe there's a way to find p without calculating such a massive number.Wait, let's think differently. If p + 1 is divisible by all these numbers, then p ≡ -1 mod n for each n. So, p ≡ -1 mod 823, p ≡ -1 mod 618, and so on.But since these moduli are not pairwise coprime, the system of congruences might not have a unique solution modulo the LCM. However, since p + 1 needs to be divisible by all of them, p must satisfy all these congruences simultaneously.This still leads me back to the LCM approach. I think I'm stuck because without computing the LCM, I can't find p.Maybe I can look for a pattern or see if any of these numbers have relationships that can simplify the problem.Looking at the numbers:- 60 and 618 share factors 2 and 3.- 60 and 3648 share factors 2 and 3.- 618 and 3648 share factors 2 and 3.- 60 and 4203 share factor 3.- 618 and 4203 share factor 3.- 3648 and 4203 share factor 3.So, the overlapping factors are mainly 2 and 3. The other numbers are primes or have unique prime factors.Given that, maybe I can separate the problem into parts:1. Find the LCM of the composite numbers, considering their prime factors.2. Then, include the primes in the LCM.But I'm not sure if that helps.Alternatively, perhaps I can find the LCM step by step:Start with 60 and 618:LCM(60, 618) = LCM(2^2 × 3 × 5, 2 × 3 × 103) = 2^2 × 3 × 5 × 103 = 4 × 3 × 5 × 103 = 6180Now, LCM(6180, 3648):Factorize 3648: 2^6 × 3 × 19So, LCM(6180, 3648) = LCM(2^2 × 3 × 5 × 103, 2^6 × 3 × 19) = 2^6 × 3 × 5 × 19 × 103 = 64 × 3 × 5 × 19 × 103Calculate that:64 × 3 = 192192 × 5 = 960960 × 19 = 18,24018,240 × 103 = 1,878,720So, LCM(60, 618, 3648) = 1,878,720Next, include 4203:Factorize 4203: 3^2 × 467So, LCM(1,878,720, 4203) = LCM(2^6 × 3 × 5 × 19 × 103, 3^2 × 467) = 2^6 × 3^2 × 5 × 19 × 103 × 467Calculate that:1,878,720 × 3 = 5,636,1605,636,160 × 467 = ?Hmm, that's a big multiplication. Let's see:5,636,160 × 400 = 2,254,464,0005,636,160 × 60 = 338,169,6005,636,160 × 7 = 39,453,120Adding them up:2,254,464,000 + 338,169,600 = 2,592,633,6002,592,633,600 + 39,453,120 = 2,632,086,720So, LCM(60, 618, 3648, 4203) = 2,632,086,720Now, include 823:Since 823 is prime and doesn't divide any of the previous LCM, we multiply:2,632,086,720 × 823 = ?Again, a huge multiplication. Let's approximate:2,632,086,720 × 800 = 2,105,669,376,0002,632,086,720 × 23 = ?2,632,086,720 × 20 = 52,641,734,4002,632,086,720 × 3 = 7,896,260,160Adding them up:52,641,734,400 + 7,896,260,160 = 60,537,994,560Now, total LCM:2,105,669,376,000 + 60,537,994,560 = 2,166,207,370,560So, LCM(60, 618, 3648, 4203, 823) = 2,166,207,370,560Next, include 3917:3917 is prime, so multiply:2,166,207,370,560 × 3917 = ?This is getting too large to compute manually. Let's denote this as LCM1.Now, include 1543:LCM1 × 1543 = LCM2And finally, include 2971:LCM2 × 2971 = Final LCMAt this point, the LCM is astronomically large. Even if I could compute it, p = LCM - 1 would be an enormous number, and verifying its primality would be beyond my current capabilities.Maybe there's a different approach. Perhaps instead of finding the absolute LCM, I can find a smaller number that satisfies all the divisibility conditions. But I'm not sure how to do that without missing some constraints.Alternatively, maybe the problem has a trick or a pattern that I'm missing. For example, if all the numbers are pairwise coprime, then the LCM would just be their product, but they're not. So, that doesn't help.Wait, let's see if any of these numbers are multiples of others. For example, 60 divides 3648? 3648 ÷ 60 = 60.8, so no. 60 divides 618? 618 ÷ 60 = 10.3, no. 60 divides 4203? 4203 ÷ 60 = 70.05, no. So, no number is a multiple of another, except possibly larger numbers.But still, they share common factors like 2 and 3.I'm stuck. I think the only way to solve this is to compute the LCM, subtract 1, and check if it's prime. But without computational tools, this is impossible.Maybe the problem is designed to recognize that p + 1 must be the LCM, and p is prime. So, the answer is LCM - 1, and it's a prime number. But without computing it, I can't provide the exact value.Alternatively, perhaps the problem is testing the understanding that p + 1 is the LCM, and p is prime, so the answer is LCM - 1, which is prime. But again, without computing it, I can't confirm.I think I need to accept that without computational assistance, I can't find the exact value of p. However, I can outline the steps:1. Factorize all given numbers into their prime factors.2. Compute the LCM by taking the highest powers of all prime factors.3. Subtract 1 from the LCM to get p.4. Check if p is prime.Given the complexity, I suspect the problem expects recognizing that p = LCM - 1, and acknowledging that it's a prime number, even if we can't compute it exactly here.

💡Okay, so I have this problem here about a circle and a line, and I need to find the probability that they intersect. Let me try to break this down step by step.First, the circle is given by the equation ( x^2 + y^2 = 4 ). I remember that the general form of a circle is ( x^2 + y^2 = r^2 ), where ( r ) is the radius. So in this case, the radius ( r ) is 2 because ( 2^2 = 4 ). The center of the circle is at the origin, (0,0), since there are no shifts in the equation.Next, the line is given by ( y = x + b ), where ( b ) is a number between -5 and 5. So ( b ) can vary, and we need to find the probability that this line intersects the circle.I think the key here is to find out for which values of ( b ) the line ( y = x + b ) intersects the circle ( x^2 + y^2 = 4 ). Once I have that range of ( b ), I can figure out the probability by comparing the length of that range to the total possible range of ( b ), which is from -5 to 5.To find the intersection points, I can substitute ( y = x + b ) into the circle equation. Let me do that:Substitute ( y ) in ( x^2 + y^2 = 4 ):( x^2 + (x + b)^2 = 4 )Let me expand that:( x^2 + (x^2 + 2bx + b^2) = 4 )Combine like terms:( 2x^2 + 2bx + b^2 - 4 = 0 )So that's a quadratic equation in terms of ( x ). For the line to intersect the circle, this quadratic equation must have real solutions. That means the discriminant should be greater than or equal to zero.The discriminant ( D ) of a quadratic equation ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ). In our case, the quadratic is ( 2x^2 + 2bx + (b^2 - 4) = 0 ), so ( a = 2 ), ( b = 2b ), and ( c = b^2 - 4 ).Wait, hold on, the coefficients are a bit confusing because the coefficient of ( x ) is also called ( b ). Maybe I should use different letters to avoid confusion. Let me rewrite the quadratic equation:( 2x^2 + 2bx + (b^2 - 4) = 0 )So here, the coefficient ( a = 2 ), the coefficient of ( x ) is ( 2b ), and the constant term is ( b^2 - 4 ).So the discriminant ( D ) is:( D = (2b)^2 - 4 * 2 * (b^2 - 4) )Let me compute that:( D = 4b^2 - 8(b^2 - 4) )Expand the second term:( D = 4b^2 - 8b^2 + 32 )Combine like terms:( D = -4b^2 + 32 )For the quadratic to have real solutions, ( D geq 0 ):( -4b^2 + 32 geq 0 )Let me solve for ( b ):First, subtract 32 from both sides:( -4b^2 geq -32 )Divide both sides by -4. But wait, when I divide by a negative number, the inequality sign flips:( b^2 leq 8 )So ( b^2 leq 8 ) implies that ( |b| leq sqrt{8} ). Simplify ( sqrt{8} ):( sqrt{8} = 2sqrt{2} approx 2.828 )So ( b ) must be between ( -2sqrt{2} ) and ( 2sqrt{2} ) for the line to intersect the circle.Now, the total possible range of ( b ) is from -5 to 5, which is a length of 10 units. The range where the line intersects the circle is from ( -2sqrt{2} ) to ( 2sqrt{2} ), which is a length of ( 4sqrt{2} ) units.Therefore, the probability ( P ) that the line intersects the circle is the ratio of the favorable range to the total range:( P = frac{4sqrt{2}}{10} )Simplify this fraction:( P = frac{2sqrt{2}}{5} )So the probability is ( frac{2sqrt{2}}{5} ).Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I substituted ( y = x + b ) into the circle equation correctly.2. Expanded and combined like terms correctly to get the quadratic equation.3. Calculated the discriminant correctly, making sure to handle the coefficients properly.4. Solved the inequality correctly, remembering to flip the inequality when dividing by a negative.5. Converted ( sqrt{8} ) to ( 2sqrt{2} ) correctly.6. Calculated the lengths of the intervals correctly.7. Simplified the fraction correctly.Everything seems to check out. So I think my final answer is correct.**Final Answer**The probability that the circle ( O ) and the line ( l ) have a common point is boxed{dfrac{2sqrt{2}}{5}}.

💡First, I need to determine the time David takes to swim each of the four events, considering both his natural swimming speeds and the external factors like wind resistance and water current.Starting with the freestyle, David swims 100 meters in 48 seconds. This is our baseline time.Next, for the backstroke, David swims 4 seconds slower than his freestyle time without any external factors. This gives me a base time of 52 seconds for the backstroke. However, when wind resistance is added, it increases the time by an additional 2 seconds, resulting in a total of 54 seconds for the backstroke.Moving on to the butterfly, David swims 3 seconds slower than his backstroke time without external factors. This means the butterfly has a base time of 55 seconds. With the added effect of a water current, the time increases by another 3 seconds, making the total butterfly time 58 seconds.Finally, for the breaststroke, David swims 2 seconds slower than his butterfly time without external factors, giving a base time of 57 seconds. However, the combination of wind resistance and water current reduces this time by 1 second, resulting in a final breaststroke time of 56 seconds.To find the combined total time for all four events, I add up the times for each event: 48 seconds (freestyle) + 54 seconds (backstroke) + 58 seconds (butterfly) + 56 seconds (breaststroke) equals 216 seconds.