💡Alright, so I have this problem where I need to find all positive integers ( n ) such that the numbers ( n+1, n+3, n+7, n+9, n+13, n+15 ) are all prime. Hmm, okay, let me try to figure this out step by step.First, I know that prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, I need to make sure that each of these expressions ( n+1, n+3, ) etc., results in a prime number.Maybe I should start by testing small values of ( n ) and see if they work. Let's try ( n = 1 ) first.For ( n = 1 ):- ( n+1 = 2 ) (prime)- ( n+3 = 4 ) (not prime) Okay, so ( n = 1 ) doesn't work because 4 is not prime.Next, let's try ( n = 2 ):- ( n+1 = 3 ) (prime)- ( n+3 = 5 ) (prime)- ( n+7 = 9 ) (not prime) Again, 9 isn't prime, so ( n = 2 ) doesn't work either.How about ( n = 3 ):- ( n+1 = 4 ) (not prime) Right away, 4 isn't prime, so ( n = 3 ) is out.Moving on to ( n = 4 ):- ( n+1 = 5 ) (prime)- ( n+3 = 7 ) (prime)- ( n+7 = 11 ) (prime)- ( n+9 = 13 ) (prime)- ( n+13 = 17 ) (prime)- ( n+15 = 19 ) (prime) Wow, all of these are prime numbers! So, ( n = 4 ) works. That's promising.But I shouldn't stop here. Maybe there are other values of ( n ) that also satisfy this condition. Let's check ( n = 5 ):- ( n+1 = 6 ) (not prime) 6 isn't prime, so ( n = 5 ) doesn't work.What about ( n = 6 ):- ( n+1 = 7 ) (prime)- ( n+3 = 9 ) (not prime) 9 isn't prime, so ( n = 6 ) is out.Trying ( n = 7 ):- ( n+1 = 8 ) (not prime) 8 isn't prime, so ( n = 7 ) doesn't work.Let's try ( n = 8 ):- ( n+1 = 9 ) (not prime) 9 isn't prime, so ( n = 8 ) is out.How about ( n = 9 ):- ( n+1 = 10 ) (not prime) 10 isn't prime, so ( n = 9 ) doesn't work.Hmm, seems like after ( n = 4 ), the next few values don't work. Maybe I need to check higher values, but this could take a while. Perhaps there's a pattern or a mathematical way to determine if there are more solutions.I remember that primes greater than 5 are of the form ( 6k pm 1 ), but I'm not sure if that helps here. Alternatively, maybe considering modulo operations could help identify if any of these expressions must be composite for certain ( n ).Let's consider the numbers modulo 5, since 5 is a small prime and might help identify patterns.For ( n ) modulo 5, the possible residues are 0, 1, 2, 3, 4.Let's compute each expression modulo 5:- ( n+1 equiv (n mod 5) + 1 )- ( n+3 equiv (n mod 5) + 3 )- ( n+7 equiv (n mod 5) + 2 ) (since 7 mod 5 is 2)- ( n+9 equiv (n mod 5) + 4 ) (since 9 mod 5 is 4)- ( n+13 equiv (n mod 5) + 3 ) (since 13 mod 5 is 3)- ( n+15 equiv (n mod 5) + 0 ) (since 15 mod 5 is 0)So, depending on ( n mod 5 ), we can see which expressions might be divisible by 5, hence composite unless they equal 5.Let's analyze each case:1. **If ( n equiv 0 mod 5 ):** - ( n+15 equiv 0 + 0 = 0 mod 5 ) - So, ( n+15 ) is divisible by 5. Since ( n ) is positive, ( n+15 geq 16 ), which is greater than 5, hence composite.2. **If ( n equiv 1 mod 5 ):** - ( n+4 equiv 1 + 4 = 5 equiv 0 mod 5 ) - Wait, but ( n+4 ) isn't one of our expressions. Let me check: - ( n+1 equiv 2 mod 5 ) - ( n+3 equiv 4 mod 5 ) - ( n+7 equiv 3 mod 5 ) - ( n+9 equiv 5 equiv 0 mod 5 ) - So, ( n+9 equiv 0 mod 5 ). Since ( n geq 1 ), ( n+9 geq 10 ), which is composite.3. **If ( n equiv 2 mod 5 ):** - ( n+3 equiv 2 + 3 = 5 equiv 0 mod 5 ) - So, ( n+3 equiv 0 mod 5 ). ( n+3 geq 5 ), but if ( n+3 = 5 ), then ( n = 2 ). Let's check ( n = 2 ): - ( n+7 = 9 ) which is not prime. So, even if ( n+3 = 5 ), other expressions fail.4. **If ( n equiv 3 mod 5 ):** - ( n+7 equiv 3 + 2 = 5 equiv 0 mod 5 ) - So, ( n+7 equiv 0 mod 5 ). ( n+7 geq 10 ), which is composite.5. **If ( n equiv 4 mod 5 ):** - ( n+1 equiv 4 + 1 = 5 equiv 0 mod 5 ) - So, ( n+1 equiv 0 mod 5 ). ( n+1 geq 5 ), but if ( n+1 = 5 ), then ( n = 4 ). Let's check ( n = 4 ): - All expressions ( 5, 7, 11, 13, 17, 19 ) are prime. So, ( n = 4 ) works.From this analysis, it seems that for any ( n > 4 ), at least one of the expressions ( n+1, n+3, n+7, n+9, n+13, n+15 ) will be divisible by 5 and greater than 5, hence composite. Therefore, the only solution is ( n = 4 ).I should double-check this reasoning. For ( n = 4 ), all expressions are prime as we saw earlier. For any other ( n ), at least one expression is composite due to the modulo 5 analysis. So, I think ( n = 4 ) is indeed the only solution.

💡Okay, so I have this problem where a biologist is trying to estimate the number of fish in a lake. Let me try to understand what's going on here step by step.First, on May 1st, the biologist catches 60 fish, marks them, and releases them back into the lake. That part is straightforward—marking the fish so that they can be identified later. Then, on September 1st, he catches another 70 fish, and out of those, 3 are marked. Now, there are some additional assumptions given. It says that 25% of the fish present on May 1st were no longer in the lake by September 1st. That could be due to death or migration. So, that means 75% of the fish from May 1st are still in the lake by September 1st. Also, it's mentioned that 40% of the fish present in the lake on September 1st were not there on May 1st. These could be new births or migration into the lake. So, that implies that 60% of the fish on September 1st were there on May 1st.The problem also states that the fish caught on September 1st represent the entire fish population of the lake. So, the sample caught on September 1st is considered to be a representative sample of the entire population at that time.Alright, so I need to find out how many fish were in the lake on May 1st. Let's denote the number of fish on May 1st as N.First, on May 1st, 60 fish are marked. So, the proportion of marked fish in the lake on May 1st is 60/N.By September 1st, 25% of the fish from May 1st are no longer in the lake. So, the number of fish remaining from May 1st is 75% of N, which is 0.75N.Additionally, 40% of the fish on September 1st were not there on May 1st. So, the total number of fish on September 1st can be considered as the remaining fish from May 1st plus the new fish. Let's denote the total number of fish on September 1st as M.So, M = 0.75N + 0.40M.Wait, that doesn't seem right. Let me think again. If 40% of M are new fish, then 60% of M are the remaining fish from May 1st. So, 0.60M = 0.75N.Yes, that makes sense. So, 0.60M = 0.75N.From this, we can solve for M in terms of N:M = (0.75 / 0.60) * NM = (75/60) * NM = (5/4) * NM = 1.25NSo, the total number of fish on September 1st is 1.25 times the number of fish on May 1st.Now, on September 1st, the biologist catches 70 fish, and 3 of them are marked. So, the proportion of marked fish in the sample is 3/70.Assuming that this proportion is representative of the entire population on September 1st, the proportion of marked fish in the lake on September 1st is also 3/70.But how many marked fish are there in the lake on September 1st? On May 1st, 60 fish were marked, and 25% of them are no longer in the lake by September 1st. So, the number of marked fish remaining is 75% of 60.Calculating that: 0.75 * 60 = 45.So, there are 45 marked fish in the lake on September 1st.Therefore, the proportion of marked fish in the lake on September 1st is 45/M.But we also have that this proportion is equal to 3/70, as per the sample caught.So, 45/M = 3/70.We can solve for M:45/M = 3/70Cross-multiplying: 3M = 45 * 703M = 3150M = 3150 / 3M = 1050So, the total number of fish on September 1st is 1050.But earlier, we found that M = 1.25N.So, 1.25N = 1050Solving for N:N = 1050 / 1.25N = 840Therefore, the number of fish in the lake on May 1st is 840.Let me double-check my calculations to make sure I didn't make any mistakes.Starting from the proportion on September 1st: 45 marked fish out of M total fish, which equals 3/70.So, 45/M = 3/70Cross-multiplying: 3M = 45 * 70 = 3150So, M = 1050. That seems correct.Then, since M = 1.25N, N = 1050 / 1.25 = 840. That also checks out.So, I think my answer is correct.

💡First, I need to understand the problem. Contestant A has a 140-meter head start in a 500-meter race and wins by 20 meters. This means that when A finishes the race, B is 20 meters behind.Next, I'll determine the distance each contestant actually runs. Since A starts 140 meters ahead, A only needs to cover 360 meters to finish the race. Contestant B, starting from the beginning, has to cover the full 500 meters.However, A wins by 20 meters, so when A finishes 360 meters, B has only covered 480 meters. This means that in the same amount of time it takes A to run 360 meters, B runs 480 meters.To find the ratio of their speeds, I'll set up the equation for speed, which is distance divided by time. Let ( t ) be the time it takes for A to finish the race. Then:[v_A = frac{360}{t}][v_B = frac{480}{t}]To find the ratio of ( v_A ) to ( v_B ), I'll divide the two equations:[frac{v_A}{v_B} = frac{frac{360}{t}}{frac{480}{t}} = frac{360}{480} = frac{3}{4}]Therefore, the ratio of the speeds of A to B is 3:4.

💡First, I need to understand the relationship between the amounts that A, B, and C receive. The problem states that for every Rs. 1 that A has, B has 65 paisa and C has 40 paisa. This means that the amounts are proportional.I'll start by defining A's share as Rs. ( x ). Based on the given ratios:- B's share is ( 0.65x ) (since 65 paisa is 0.65 of a rupee)- C's share is ( 0.40x ) (since 40 paisa is 0.40 of a rupee)The problem provides that C's share is Rs. 32. Using this information, I can set up the equation:[0.40x = 32]Solving for ( x ):[x = frac{32}{0.40} = 80]So, A's share is Rs. 80.Next, I'll calculate B's share:[0.65x = 0.65 times 80 = 52]B's share is Rs. 52.Finally, I'll find the total sum of money by adding the shares of A, B, and C:[text{Total sum} = 80 + 52 + 32 = 164]Therefore, the total sum of money is Rs. 164.

💡First, I recognize that 80% of the village's population is given as 32,000. To find the total population, I need to determine what number 32,000 represents 80% of.I can set up the equation: 80% of the total population (P) equals 32,000. Converting 80% to a decimal gives me 0.8.So, the equation becomes:0.8 × P = 32,000To solve for P, I divide both sides of the equation by 0.8:P = 32,000 / 0.8Performing the division:32,000 ÷ 0.8 = 40,000Therefore, the total population of the village is 40,000.

💡Okay, so I have this problem here: I need to find the minimum positive period of the function ( y = sinleft(frac{pi}{3} - 2xright) + sin 2x ). Hmm, periodic functions, right? I remember that the period of a sine function is usually ( 2pi ), but when there's a coefficient inside the sine, it affects the period. Let me think.First, let me recall that for a function like ( sin(Bx + C) ), the period is ( frac{2pi}{|B|} ). So, in this case, both terms have ( 2x ) inside the sine function. The first term is ( sinleft(frac{pi}{3} - 2xright) ), which I can rewrite as ( sin(-2x + frac{pi}{3}) ). Since sine is an odd function, ( sin(-theta) = -sintheta ), so this becomes ( -sin(2x - frac{pi}{3}) ).Wait, but does that affect the period? I don't think so because the period is about how often the function repeats, regardless of the phase shift or the amplitude. So, both terms have ( 2x ) inside, which means each of them individually has a period of ( frac{2pi}{2} = pi ).But the function is the sum of two sine functions. So, to find the period of the sum, I need to find the least common multiple (LCM) of the periods of the individual functions. Since both have the same period, ( pi ), the LCM is just ( pi ). So, is the period of the entire function ( pi )?Wait, hold on. Maybe I should simplify the function first before jumping to conclusions. Let me try to combine the two sine terms using a trigonometric identity. I remember that ( sin A + sin B = 2 sinleft(frac{A + B}{2}right) cosleft(frac{A - B}{2}right) ). Maybe that can help.Let me set ( A = frac{pi}{3} - 2x ) and ( B = 2x ). Then,( sin A + sin B = 2 sinleft(frac{A + B}{2}right) cosleft(frac{A - B}{2}right) ).Plugging in the values:( 2 sinleft(frac{frac{pi}{3} - 2x + 2x}{2}right) cosleft(frac{frac{pi}{3} - 2x - 2x}{2}right) ).Simplifying the arguments:For the sine term: ( frac{frac{pi}{3} - 2x + 2x}{2} = frac{frac{pi}{3}}{2} = frac{pi}{6} ).For the cosine term: ( frac{frac{pi}{3} - 4x}{2} = frac{pi}{6} - 2x ).So, the expression becomes:( 2 sinleft(frac{pi}{6}right) cosleft(frac{pi}{6} - 2xright) ).I know that ( sinleft(frac{pi}{6}right) = frac{1}{2} ), so this simplifies to:( 2 times frac{1}{2} times cosleft(frac{pi}{6} - 2xright) = cosleft(frac{pi}{6} - 2xright) ).But cosine is an even function, so ( cos(-theta) = costheta ). Therefore, ( cosleft(frac{pi}{6} - 2xright) = cosleft(2x - frac{pi}{6}right) ).So, the function simplifies to ( cosleft(2x - frac{pi}{6}right) ). Now, let's find the period of this function.Again, for a function ( cos(Bx + C) ), the period is ( frac{2pi}{|B|} ). Here, ( B = 2 ), so the period is ( frac{2pi}{2} = pi ).Wait, so both the original function and the simplified function have the same period, ( pi ). So, the minimum positive period is ( pi ).But just to make sure, let me think again. The original function was a sum of two sine functions with the same frequency, so their sum should also have the same frequency, hence the same period. So, it makes sense that the period is ( pi ).I think that's it. The minimum positive period is ( pi ).**Final Answer**The minimum positive period of the function is boxed{pi}.