💡Okay, so I need to find all prime numbers ( p ) such that ( p^2 + 11 ) has exactly six distinct divisors. Hmm, let me think about how to approach this.First, I remember that the number of divisors of a number is related to its prime factorization. If a number ( n ) has a prime factorization of ( n = p_1^{a_1} p_2^{a_2} dots p_k^{a_k} ), then the total number of divisors is ( (a_1 + 1)(a_2 + 1) dots (a_k + 1) ). So, for ( p^2 + 11 ) to have exactly six divisors, its prime factorization must result in the product of exponents plus one equaling six.Let me recall the possible ways to express six as a product of integers greater than one. Six can be written as ( 6 = 6 times 1 ) or ( 6 = 2 times 3 ). This means that the number ( p^2 + 11 ) could either be of the form ( q^5 ) (since ( 5 + 1 = 6 )) or ( q^2 times r ) (since ( (2 + 1)(1 + 1) = 3 times 2 = 6 )), where ( q ) and ( r ) are distinct primes.Now, let me consider the first case: ( p^2 + 11 = q^5 ). Since ( p ) is a prime number, let's test small primes to see if this equation holds.Starting with ( p = 2 ):( 2^2 + 11 = 4 + 11 = 15 ). Is 15 a fifth power? No, because ( 2^5 = 32 ) and ( 3^5 = 243 ), which are both larger than 15. So, ( p = 2 ) doesn't work.Next, ( p = 3 ):( 3^2 + 11 = 9 + 11 = 20 ). 20 is not a fifth power either. Moving on.( p = 5 ):( 5^2 + 11 = 25 + 11 = 36 ). 36 is ( 6^2 ), not a fifth power.( p = 7 ):( 7^2 + 11 = 49 + 11 = 60 ). 60 isn't a fifth power.( p = 11 ):( 11^2 + 11 = 121 + 11 = 132 ). Not a fifth power.It seems like the first case ( q^5 ) isn't giving us any valid primes. Maybe I should look into the second case where ( p^2 + 11 = q^2 times r ).Let me test the same primes again.Starting with ( p = 2 ):( 2^2 + 11 = 15 ). The prime factors of 15 are 3 and 5, so ( 15 = 3 times 5 ). This is ( q times r ), not ( q^2 times r ). So, it doesn't fit the second case.( p = 3 ):( 3^2 + 11 = 20 ). The prime factors of 20 are 2 and 5, and ( 20 = 2^2 times 5 ). Ah, this fits the second case where ( q = 2 ) and ( r = 5 ). So, ( p = 3 ) works.Let me check ( p = 5 ):( 5^2 + 11 = 36 ). The prime factors of 36 are 2 and 3, and ( 36 = 2^2 times 3^2 ). This gives ( (2 + 1)(2 + 1) = 9 ) divisors, which is more than six. So, ( p = 5 ) doesn't work.( p = 7 ):( 7^2 + 11 = 60 ). The prime factors of 60 are 2, 3, and 5, and ( 60 = 2^2 times 3 times 5 ). This would give ( (2 + 1)(1 + 1)(1 + 1) = 12 ) divisors, which is more than six. So, ( p = 7 ) doesn't work.( p = 11 ):( 11^2 + 11 = 132 ). The prime factors of 132 are 2, 3, and 11, and ( 132 = 2^2 times 3 times 11 ). Again, this would result in ( (2 + 1)(1 + 1)(1 + 1) = 12 ) divisors. Not six.Hmm, so far only ( p = 3 ) works. Let me check a few more primes to be thorough.( p = 13 ):( 13^2 + 11 = 169 + 11 = 180 ). Prime factors of 180 are 2, 3, and 5, with ( 180 = 2^2 times 3^2 times 5 ). Divisors: ( (2 + 1)(2 + 1)(1 + 1) = 18 ). Too many.( p = 17 ):( 17^2 + 11 = 289 + 11 = 300 ). Prime factors: 2, 3, 5. ( 300 = 2^2 times 3 times 5^2 ). Divisors: ( (2 + 1)(1 + 1)(2 + 1) = 18 ). Still too many.Wait, maybe I should think differently. Since ( p ) is a prime, ( p^2 ) is always odd (except when ( p = 2 )), so ( p^2 + 11 ) will be even when ( p ) is odd because an odd plus odd is even. So, ( p^2 + 11 ) is even for all primes ( p ) except ( p = 2 ). For ( p = 2 ), we saw it's 15, which is not suitable.So, for odd primes ( p ), ( p^2 + 11 ) is even, meaning 2 is a prime factor. Let me denote ( p^2 + 11 = 2^k times m ), where ( m ) is an odd integer. Since we need exactly six divisors, the total number of divisors is ( (k + 1) times ) (number of divisors of ( m )).If ( p^2 + 11 = 2^2 times r ), where ( r ) is an odd prime, then the number of divisors is ( (2 + 1)(1 + 1) = 6 ), which is what we want. So, ( p^2 + 11 ) must be equal to ( 4 times r ), where ( r ) is an odd prime.So, let's set ( p^2 + 11 = 4r ). Then, ( r = frac{p^2 + 11}{4} ). Since ( r ) must be an integer, ( p^2 + 11 ) must be divisible by 4. Let's check when ( p^2 equiv -11 mod 4 ). Since ( -11 equiv 1 mod 4 ), we have ( p^2 equiv 1 mod 4 ).But for any odd prime ( p ), ( p ) is congruent to 1 or 3 modulo 4. Squaring these, ( 1^2 equiv 1 mod 4 ) and ( 3^2 equiv 9 equiv 1 mod 4 ). So, ( p^2 equiv 1 mod 4 ) for any odd prime ( p ). Therefore, ( p^2 + 11 equiv 1 + 3 equiv 0 mod 4 ). So, ( p^2 + 11 ) is divisible by 4 for any odd prime ( p ).Therefore, ( r = frac{p^2 + 11}{4} ) must be a prime number. So, our condition reduces to ( frac{p^2 + 11}{4} ) being prime.Let me test this for the primes I checked earlier.For ( p = 3 ):( r = frac{9 + 11}{4} = frac{20}{4} = 5 ), which is prime. So, ( p = 3 ) works.For ( p = 5 ):( r = frac{25 + 11}{4} = frac{36}{4} = 9 ), which is not prime. So, ( p = 5 ) doesn't work.For ( p = 7 ):( r = frac{49 + 11}{4} = frac{60}{4} = 15 ), which is not prime.For ( p = 11 ):( r = frac{121 + 11}{4} = frac{132}{4} = 33 ), which is not prime.For ( p = 13 ):( r = frac{169 + 11}{4} = frac{180}{4} = 45 ), not prime.For ( p = 17 ):( r = frac{289 + 11}{4} = frac{300}{4} = 75 ), not prime.Hmm, seems like only ( p = 3 ) gives a prime ( r ). Let me check ( p = 19 ):( r = frac{361 + 11}{4} = frac{372}{4} = 93 ), not prime.( p = 23 ):( r = frac{529 + 11}{4} = frac{540}{4} = 135 ), not prime.Wait, maybe I should check smaller primes again, but I think I already did that. Let me think if there's another way.Alternatively, maybe ( p^2 + 11 ) could be of the form ( q^5 ). But earlier, when I tested small primes, none of them worked. Let me check ( p = 3 ):( 3^2 + 11 = 20 ), which is not a fifth power. ( p = 2 ): 15, not a fifth power. ( p = 5 ): 36, not a fifth power. ( p = 7 ): 60, not a fifth power. So, seems like this case doesn't give any solutions.Therefore, the only prime ( p ) that satisfies the condition is ( p = 3 ).Wait, just to be thorough, let me check ( p = 7 ) again:( 7^2 + 11 = 60 ). The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. That's 12 divisors, which is more than six. So, no.Similarly, ( p = 11 ): 132 has divisors 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132. Again, 12 divisors.So, yeah, only ( p = 3 ) gives exactly six divisors for ( p^2 + 11 ).I think I've checked all possible cases and the only solution is ( p = 3 ).

💡Okay, so I have this problem: For how many positive integers (x) is (x^2 + 6x + 13) between 25 and 50? Hmm, let me try to figure this out step by step. I'm not too confident about my algebra skills, but I'll give it a shot.First, I need to understand what the problem is asking. It wants the number of positive integers (x) such that when you plug (x) into the quadratic expression (x^2 + 6x + 13), the result is greater than 25 and less than 50. So, essentially, I need to find all positive integers (x) where (25 < x^2 + 6x + 13 < 50).Alright, let's start by writing down the inequality:[25 < x^2 + 6x + 13 < 50]I think the first step is to simplify this inequality to make it easier to solve. Maybe I can subtract 13 from all parts to get rid of that constant term. Let's try that:Subtracting 13 from all parts:[25 - 13 < x^2 + 6x + 13 - 13 < 50 - 13]Which simplifies to:[12 < x^2 + 6x < 37]Hmm, okay, so now I have (12 < x^2 + 6x < 37). This still looks a bit complicated. Maybe I can complete the square for the quadratic expression (x^2 + 6x). I remember that completing the square can help rewrite the quadratic in a form that's easier to analyze.To complete the square for (x^2 + 6x), I take half of the coefficient of (x), which is 6, so half of that is 3, and then square it, which gives me 9. So, I can write:[x^2 + 6x = (x + 3)^2 - 9]Let me check that:[(x + 3)^2 = x^2 + 6x + 9]So, subtracting 9 gives (x^2 + 6x), which matches. Great, so now I can substitute this back into the inequality:[12 < (x + 3)^2 - 9 < 37]Now, let's add 9 to all parts of the inequality to isolate the squared term:[12 + 9 < (x + 3)^2 < 37 + 9]Which simplifies to:[21 < (x + 3)^2 < 46]Okay, so now I have (21 < (x + 3)^2 < 46). This means that ((x + 3)^2) is between 21 and 46. Since ((x + 3)^2) is a perfect square, I can think about which perfect squares lie between 21 and 46.Let me list the perfect squares around this range:- (4^2 = 16)- (5^2 = 25)- (6^2 = 36)- (7^2 = 49)So, the perfect squares between 21 and 46 are 25 and 36. That means ((x + 3)^2) can be either 25 or 36.Let's consider each case separately.**Case 1: ((x + 3)^2 = 25)**Taking the square root of both sides:[x + 3 = pm 5]So, solving for (x):1. (x + 3 = 5) → (x = 5 - 3 = 2)2. (x + 3 = -5) → (x = -5 - 3 = -8)But the problem specifies positive integers, so (x = -8) is not valid. Therefore, (x = 2) is a solution.**Case 2: ((x + 3)^2 = 36)**Again, taking the square root of both sides:[x + 3 = pm 6]Solving for (x):1. (x + 3 = 6) → (x = 6 - 3 = 3)2. (x + 3 = -6) → (x = -6 - 3 = -9)Again, since we're only interested in positive integers, (x = -9) is invalid. So, (x = 3) is another solution.Now, I should check if there are any other perfect squares between 21 and 46 that I might have missed. Looking back, the next perfect square after 36 is 49, which is greater than 46, so it's outside our range. Similarly, before 25, the perfect square is 16, which is less than 21, so it's also outside our range. Therefore, the only perfect squares between 21 and 46 are 25 and 36, which correspond to (x = 2) and (x = 3).Just to make sure I didn't make a mistake, let me plug these values back into the original expression to verify.**Checking (x = 2):**[x^2 + 6x + 13 = 2^2 + 6(2) + 13 = 4 + 12 + 13 = 29]29 is between 25 and 50, so that's good.**Checking (x = 3):**[x^2 + 6x + 13 = 3^2 + 6(3) + 13 = 9 + 18 + 13 = 40]40 is also between 25 and 50, so that works too.Let me also check the next integer, (x = 4), to see if it's still within the range.**Checking (x = 4):**[x^2 + 6x + 13 = 4^2 + 6(4) + 13 = 16 + 24 + 13 = 53]53 is greater than 50, so it doesn't satisfy the condition. Similarly, checking (x = 1):**Checking (x = 1):**[x^2 + 6x + 13 = 1^2 + 6(1) + 13 = 1 + 6 + 13 = 20]20 is less than 25, so it doesn't work either.What about (x = 0)? Well, the problem specifies positive integers, so (x = 0) isn't considered. But just for my understanding:**Checking (x = 0):**[x^2 + 6x + 13 = 0 + 0 + 13 = 13]13 is less than 25, so even if it were allowed, it wouldn't satisfy the condition.I think I've covered all possible cases. The only positive integers (x) that satisfy (25 < x^2 + 6x + 13 < 50) are 2 and 3. Therefore, there are 2 such positive integers.**Final Answer**The number of positive integers (x) is boxed{2}.

💡To determine the graph of ( y = f(x) + 3 ), we need to understand how adding 3 to the function ( f(x) ) affects its graph. 1. **Vertical Shift**: Adding a constant to a function results in a vertical shift of its graph. Specifically, adding 3 to ( f(x) ) shifts the graph of ( y = f(x) ) upward by 3 units.2. **Analyzing the Original Function**: - For ( x ) in the interval ([-3, 0]), ( f(x) = -2 - x ). This is a linear function with a slope of -1 and a y-intercept at (-2). - For ( x ) in the interval ([0, 2]), ( f(x) = sqrt{4 - (x - 2)^2} - 2 ). This represents the lower half of a circle centered at ((2, -2)) with a radius of 2. - For ( x ) in the interval ([2, 3]), ( f(x) = 2(x - 2) ). This is a linear function with a slope of 2 starting from the point ((2, 0)).3. **Effect of Adding 3**: - The linear segment from ([-3, 0]) will shift upward, changing its y-intercept from (-2) to (1). - The circular segment from ([0, 2]) will shift upward, changing its center from ((2, -2)) to ((2, 1)). - The linear segment from ([2, 3]) will shift upward, changing its starting point from ((2, 0)) to ((2, 3)).4. **Identifying the Correct Graph**: - After the vertical shift, the graph of ( y = f(x) + 3 ) will have all its points moved 3 units higher. - Comparing this transformation with the provided options, the correct graph corresponds to option E.Therefore, the graph of ( y = f(x) + 3 ) is option E.

💡Okay, so I have this problem here about the Cresco family's Hydroflow pump. It says that the pump can move 500 gallons of water every hour. The question is asking how many gallons it will pump in just 30 minutes. Hmm, alright, let's break this down step by step.First, I know that 30 minutes is half of an hour. So, if the pump can do 500 gallons in a full hour, then in half an hour, it should do half of that amount, right? That makes sense because if you're only running it for half the time, it should pump half the water.But wait, let me make sure I'm not missing anything here. Sometimes, these problems can have tricks or additional factors, but in this case, it seems pretty straightforward. It's just about understanding the rate and applying it to a shorter time period.So, the rate is 500 gallons per hour. To find out how much it pumps in 30 minutes, I need to calculate half of 500 gallons. Let me write that down:500 gallons/hour * (30 minutes / 60 minutes) = ?Simplifying the time fraction, 30 minutes is half of 60 minutes, so that's 1/2. Therefore:500 gallons/hour * 1/2 = 250 gallons.Okay, that seems right. But just to double-check, let's think about it differently. If the pump does 500 gallons in 60 minutes, how many gallons does it do in one minute? 500 gallons / 60 minutes = approximately 8.333 gallons per minute.Now, if I multiply that by 30 minutes:8.333 gallons/minute * 30 minutes = 250 gallons.Yep, same answer. So, both methods give me 250 gallons. That makes me more confident that 250 is the correct answer.I guess another way to look at it is by using proportions. If 60 minutes correspond to 500 gallons, then 30 minutes correspond to X gallons. Setting up the proportion:60 minutes / 500 gallons = 30 minutes / X gallons.Cross-multiplying gives:60X = 500 * 30.60X = 15,000.Dividing both sides by 60:X = 15,000 / 60 = 250 gallons.Alright, that's the third method, and it still gives me 250 gallons. I think I'm pretty sure now that the answer is 250 gallons.Just to make sure there's no confusion, let's think about units. The rate is gallons per hour, and we're converting hours to minutes. So, it's important to keep the units consistent. Since 30 minutes is half an hour, it's straightforward to apply that fraction to the total gallons per hour.Another thing to consider is whether the pump's rate changes over time, but the problem doesn't mention anything like that. It just says it pumps at a constant rate of 500 gallons per hour. So, we can safely assume that the rate remains the same throughout the 30 minutes.In summary, whether I break it down by calculating the rate per minute, use proportions, or simply recognize that 30 minutes is half an hour, all methods lead me to the same conclusion: the Hydroflow pump will move 250 gallons of water in 30 minutes.

💡Okay, so there's this store owner who wants to make a 30-pound mixture of two candies. He wants to sell this mixture for 3 per pound. He's planning to use 20 pounds of the first candy and 10 pounds of the second candy. We know that the first candy sells for 2.95 per pound, and we need to find out the price per pound of the second candy.Alright, let's break this down. First, the total weight of the mixture is 30 pounds, and he's using 20 pounds of the first candy and 10 pounds of the second. So, 20 plus 10 equals 30, which checks out.Now, the mixture sells for 3 per pound. So, the total revenue from selling the entire 30-pound mixture would be 30 times 3, which is 90. That makes sense.Next, we need to figure out the total cost of the candies he's using. He's using 20 pounds of the first candy, which costs 2.95 per pound. So, the cost for the first candy is 20 times 2.95. Let me calculate that: 20 times 2.95 is... 20 times 2 is 40, and 20 times 0.95 is 19, so 40 plus 19 is 59. So, the first candy costs him 59 in total.Now, we need to find out how much the second candy costs. Let's call the price per pound of the second candy "x." He's using 10 pounds of it, so the total cost for the second candy is 10 times x, which is 10x.The total cost of both candies together should equal the total revenue from selling the mixture, right? So, the total cost is 59 plus 10x, and that should equal 90.So, setting up the equation: 59 + 10x = 90.Now, let's solve for x. Subtract 59 from both sides: 10x = 90 - 59, which is 31. So, 10x = 31.Then, divide both sides by 10 to find x: x = 31 / 10, which is 3.1.So, the price per pound of the second candy is 3.10.Wait, let me double-check that. If the second candy is 3.10 per pound, then 10 pounds would cost 10 times 3.10, which is 31. Adding that to the 59 for the first candy gives a total of 90, which matches the total revenue from selling the 30-pound mixture at 3 per pound. Yeah, that seems correct.I think I've got it. The second candy should cost 3.10 per pound.

💡Alright, so I have this problem about a function ( y = f(x) ) that's even on ( mathbb{R} ). It also satisfies the equation ( f(x + 6) = f(x) + f(3) ) for any real number ( x ). Additionally, for any ( x_1, x_2 ) in the interval [0, 3], if ( x_1 neq x_2 ), then the difference quotient ( frac{f(x_1) - f(x_2)}{x_1 - x_2} > 0 ). This means the function is increasing on [0, 3]. I need to evaluate four statements about this function and determine which ones are correct. Let me go through each statement one by one, but first, let me try to understand the properties of the function.1. **Even Function**: Since ( f(x) ) is even, ( f(-x) = f(x) ) for all ( x ). This tells me that the graph of the function is symmetric about the y-axis.2. **Functional Equation**: ( f(x + 6) = f(x) + f(3) ). This seems like a recurrence relation or a functional equation that relates the value of the function at ( x + 6 ) to its value at ( x ) plus a constant term ( f(3) ). 3. **Monotonicity on [0, 3]**: The function is increasing on [0, 3] because the difference quotient is positive. So, if ( x_1 > x_2 ), then ( f(x_1) > f(x_2) ).Now, let's look at each statement.**Statement (1): ( f(3) = 0 ).**Hmm, is this true? Let me try plugging in a specific value into the functional equation. If I let ( x = -3 ), then:( f(-3 + 6) = f(-3) + f(3) )Simplifying, that's:( f(3) = f(-3) + f(3) )But since ( f ) is even, ( f(-3) = f(3) ). So substituting that in:( f(3) = f(3) + f(3) )Which simplifies to:( f(3) = 2f(3) )Subtracting ( f(3) ) from both sides:( 0 = f(3) )So, yes, ( f(3) = 0 ). Therefore, statement (1) is correct.**Statement (2): The line ( x = -6 ) is an axis of symmetry for the graph of ( y = f(x) ).**Okay, so if ( x = -6 ) is an axis of symmetry, then for any point ( (x, y) ) on the graph, the point ( (-12 - x, y) ) should also be on the graph. In other words, ( f(-12 - x) = f(x) ).But wait, let's see if we can derive this from the given properties. We know that ( f(x + 6) = f(x) + f(3) ). Since we already found that ( f(3) = 0 ), this simplifies to:( f(x + 6) = f(x) )So, the function is periodic with period 6. That means ( f(x + 6) = f(x) ) for all ( x ).Since the function is even, ( f(-x) = f(x) ). Combining this with periodicity, let's see what happens when we shift by 6:( f(-x) = f(x) )But ( f(x + 6) = f(x) ), so replacing ( x ) with ( -x ):( f(-x + 6) = f(-x) = f(x) )So, ( f(-x + 6) = f(x) ). Let me rewrite this as:( f(6 - x) = f(x) )This suggests that the function is symmetric about the line ( x = 3 ), because reflecting around ( x = 3 ) would take ( x ) to ( 6 - x ).But the statement is about symmetry around ( x = -6 ). Hmm, is that necessarily true?Wait, let's think about the periodicity. If the function has period 6, then shifting by 6 doesn't change the function. So, if it's symmetric about ( x = 3 ), then shifting that symmetry line by 6 units to the left would give another axis of symmetry at ( x = 3 - 6 = -3 ). Similarly, shifting it further left by another 6 units would give ( x = -9 ), and so on.But the statement is about ( x = -6 ). Is that an axis of symmetry?Alternatively, maybe I can use the functional equation and the evenness to find another symmetry.Given ( f(x + 6) = f(x) ) and ( f(-x) = f(x) ), let's see:( f(-x) = f(x) )But ( f(x) = f(x - 6 + 6) = f(x - 6) + f(3) ). Wait, since ( f(3) = 0 ), this simplifies to ( f(x) = f(x - 6) ). So, ( f(x - 6) = f(x) ).Therefore, ( f(-x) = f(x) = f(x - 6) ). So, ( f(-x) = f(x - 6) ).Let me replace ( x ) with ( x + 6 ):( f(-(x + 6)) = f((x + 6) - 6) )Simplifying:( f(-x - 6) = f(x) )But since ( f ) is even, ( f(-x - 6) = f(x + 6) ). So:( f(x + 6) = f(x) )Which we already knew.Hmm, not sure if that helps. Maybe another approach.If the function is periodic with period 6 and even, then it's symmetric about every line ( x = 6k ) for integer ( k ). Wait, is that the case?Wait, no. For a periodic function, if it's symmetric about a line, then it's symmetric about all lines shifted by the period. So, if it's symmetric about ( x = 0 ) (since it's even), then it's symmetric about ( x = 6k ) for any integer ( k ).But ( x = -6 ) is one such line, since ( -6 = 6 times (-1) ). So, yes, the function should be symmetric about ( x = -6 ).Wait, let me think again. If a function is even, it's symmetric about ( x = 0 ). If it's periodic with period 6, then it's symmetric about all lines ( x = 6k ). So, ( x = -6 ) is just another line of symmetry, 6 units to the left of ( x = 0 ). Therefore, statement (2) should be correct.**Statement (3): The function ( y = f(x) ) is increasing on ( [-9, -6] ).**Hmm. Let's analyze this. We know that ( f ) is increasing on [0, 3]. Since ( f ) is even, it should be decreasing on [-3, 0]. Because if you reflect an increasing function over the y-axis, it becomes decreasing.Now, since the function has period 6, the behavior on [0, 3] repeats every 6 units. So, on [6, 9], the function should be increasing as well, just like on [0, 3]. Similarly, on [-6, -3], the function should be decreasing, just like on [-3, 0].But the statement is about [-9, -6]. Let's see. Since the function is periodic with period 6, the interval [-9, -6] is the same as shifting [-3, 0] by -6. So, if on [-3, 0], the function is decreasing, then on [-9, -6], which is just [-3, 0] shifted left by 6, the function should also be decreasing.But the statement says it's increasing on [-9, -6]. That contradicts what I just thought. So, is statement (3) correct or not?Wait, let me double-check.We know that on [0, 3], the function is increasing. Since it's even, on [-3, 0], it's decreasing. Now, because of periodicity, the function repeats every 6 units. So, on [6, 9], it's increasing, and on [-6, -3], it's decreasing.But what about [-9, -6]? That's another interval of 3 units. Let me see.If I take the interval [-9, -6], which is 3 units long, and since the function has period 6, this interval is equivalent to shifting [ -3, 0 ] by -6. So, the behavior on [-9, -6] should be the same as on [-3, 0], which is decreasing. Therefore, the function is decreasing on [-9, -6], not increasing. So, statement (3) is incorrect.Wait, but hold on. Let me think again. Maybe I'm confusing the direction.If the function is periodic with period 6, then f(x + 6) = f(x). So, the graph of the function repeats every 6 units. So, if on [0, 3], it's increasing, then on [6, 9], it's also increasing. Similarly, on [-6, -3], it's decreasing, just like on [-3, 0].But [-9, -6] is another interval. Let's see, if I take x in [-9, -6], then x + 6 is in [-3, 0]. So, f(x) = f(x + 6). Since f is decreasing on [-3, 0], f(x + 6) is decreasing as x increases. Therefore, f(x) is decreasing on [-9, -6]. So, statement (3) is incorrect.**Statement (4): The function ( y = f(x) ) has four zeros on ( [-9, 9] ).**We know that ( f(3) = 0 ). Since the function is even, ( f(-3) = 0 ) as well. Also, because the function is periodic with period 6, then ( f(3 + 6k) = 0 ) for any integer k. So, let's list the zeros in [-9, 9].Starting from -9:- ( f(-9) = f(-9 + 6) = f(-3) = 0 )- ( f(-3) = 0 )- ( f(3) = 0 )- ( f(9) = f(9 - 6) = f(3) = 0 )So, zeros at x = -9, -3, 3, 9. That's four zeros. But wait, let me check if there are any more zeros.Is there a possibility of more zeros within the interval? For example, between -9 and -3, or between 3 and 9, are there any other zeros?Given that the function is increasing on [0, 3], and since f(3) = 0, then on [0, 3), the function is increasing from f(0) to f(3) = 0. So, if f(0) is some value, and it's increasing to 0 at x = 3, then f(0) must be less than 0? Wait, no, because if it's increasing on [0, 3], and f(3) = 0, then f(0) must be less than 0.Similarly, on [-3, 0], since it's decreasing, f(-3) = 0, and it's decreasing to f(0). So, f(0) is the minimum point on [-3, 3].But wait, if f(0) is less than 0, then the function crosses zero at x = -3, 3, and maybe somewhere else?Wait, no. Because the function is increasing on [0, 3], starting from f(0) < 0 and reaching f(3) = 0. So, it only crosses zero at x = 3 on [0, 3]. Similarly, on [-3, 0], it's decreasing from f(-3) = 0 to f(0) < 0, so it only crosses zero at x = -3.Now, considering the periodicity, on [3, 9], the function repeats its behavior from [-3, 3]. So, on [3, 9], it's increasing from f(3) = 0 to f(9) = 0, but wait, that would mean it's increasing from 0 to 0, which would imply it's constant? But we know it's increasing on [0, 3], so on [3, 9], it's increasing as well, but since f(3) = 0 and f(9) = 0, it must stay at 0? That can't be, unless the function is zero on [3, 9], which contradicts it being increasing.Wait, maybe I made a mistake here. Let me think again.Since f(x + 6) = f(x), the function is periodic with period 6. So, the behavior on [0, 3] is the same as on [6, 9]. So, on [6, 9], the function is increasing, just like on [0, 3]. But f(6) = f(0), and f(9) = f(3) = 0. So, if f(0) < 0, then on [6, 9], the function goes from f(6) = f(0) < 0 to f(9) = 0, increasing. So, it only crosses zero at x = 9.Similarly, on [-9, -6], the function is decreasing, as we saw in statement (3). So, f(-9) = f(-3) = 0, and f(-6) = f(0) < 0. So, it goes from 0 at x = -9 to f(-6) < 0, decreasing. So, it only crosses zero at x = -9.Therefore, the zeros are only at x = -9, -3, 3, 9. So, four zeros in total on [-9, 9]. Therefore, statement (4) is correct.Wait, but let me check if there are any other zeros. For example, could the function cross zero somewhere else?Suppose, for instance, that f(0) is negative, as we concluded earlier. Then, on [0, 3], it goes from f(0) < 0 to f(3) = 0, increasing. So, it doesn't cross zero in between, just at x = 3. Similarly, on [-3, 0], it goes from f(-3) = 0 to f(0) < 0, decreasing. So, no zero crossings in between.Similarly, on [3, 6], which is the same as [-3, 0] shifted by 6, the function is decreasing from f(3) = 0 to f(6) = f(0) < 0. So, no zero crossings there except at x = 3.On [6, 9], it's increasing from f(6) = f(0) < 0 to f(9) = 0, so again, only crossing zero at x = 9.Similarly, on [-6, -3], it's decreasing from f(-6) = f(0) < 0 to f(-3) = 0, so only crossing zero at x = -3.And on [-9, -6], it's decreasing from f(-9) = 0 to f(-6) = f(0) < 0, so only crossing zero at x = -9.Therefore, the only zeros are at x = -9, -3, 3, 9. So, four zeros in total on [-9, 9]. So, statement (4) is correct.Wait, but hold on. If the function is periodic with period 6, then f(-9) = f(-9 + 6) = f(-3) = 0, and f(9) = f(9 - 6) = f(3) = 0. So, that's consistent.But just to be thorough, let me consider if the function could have more zeros.Suppose, for example, that f(0) = 0. Then, since f is increasing on [0, 3], starting from 0, it would go up, but we know f(3) = 0. That would mean the function goes up and then back down to 0, which would imply a maximum somewhere in between, but that would contradict the function being increasing on [0, 3]. Therefore, f(0) cannot be 0; it must be less than 0.Similarly, if f(0) were positive, then on [0, 3], the function would have to decrease to reach 0 at x = 3, but it's given that the function is increasing on [0, 3], so f(0) must be less than f(3) = 0. Therefore, f(0) < 0.Thus, the function only crosses zero at x = -9, -3, 3, 9 on the interval [-9, 9]. So, four zeros. Therefore, statement (4) is correct.**Summary of Thoughts:**- Statement (1): Correct, because plugging x = -3 into the functional equation and using evenness leads to f(3) = 0.- Statement (2): Correct, because the function is periodic with period 6 and even, so it's symmetric about all lines x = 6k, including x = -6.- Statement (3): Incorrect, because on [-9, -6], the function is decreasing, not increasing.- Statement (4): Correct, because the function has zeros at x = -9, -3, 3, 9, totaling four zeros on [-9, 9].**Final Answer**The correct statements are (1), (2), and (4). So, the answer is boxed{1,2,4}.