💡Okay, so I have this equation to solve: ( x(x+1)(x+2)(x+3) = y(y+1) ). I need to find all integer pairs ( (x, y) ) that satisfy this. Hmm, let me think about how to approach this.First, the left side is a product of four consecutive integers starting from ( x ). That seems a bit complicated. Maybe I can simplify it somehow. Let me try expanding it step by step.So, ( x(x+1)(x+2)(x+3) ). Maybe I can pair them up. Let me pair ( x ) with ( x+3 ) and ( x+1 ) with ( x+2 ). Let's see:( x(x+3) = x^2 + 3x )( (x+1)(x+2) = x^2 + 3x + 2 )So now, the left side becomes ( (x^2 + 3x)(x^2 + 3x + 2) ). Hmm, that looks like a product of two consecutive terms if I let ( z = x^2 + 3x ). So, it's ( z(z + 2) ), which is ( z^2 + 2z ).So, substituting back, we have:( (x^2 + 3x)^2 + 2(x^2 + 3x) )Let me expand that:( (x^2 + 3x)^2 = x^4 + 6x^3 + 9x^2 )Then, ( 2(x^2 + 3x) = 2x^2 + 6x )Adding them together:( x^4 + 6x^3 + 9x^2 + 2x^2 + 6x = x^4 + 6x^3 + 11x^2 + 6x )So, the left side simplifies to ( x^4 + 6x^3 + 11x^2 + 6x ).Now, the equation is:( x^4 + 6x^3 + 11x^2 + 6x = y(y+1) )Hmm, the right side is ( y^2 + y ). Maybe I can write both sides in terms of squares to compare them. Let me try adding 1 to both sides:( x^4 + 6x^3 + 11x^2 + 6x + 1 = y^2 + y + 1 )Now, let me see if the left side is a perfect square. Let me check:( (x^2 + 3x + 1)^2 = x^4 + 6x^3 + 11x^2 + 6x + 1 )Yes! Perfect, that's exactly the left side. So, we have:( (x^2 + 3x + 1)^2 = y^2 + y + 1 )So, now the equation is:( (x^2 + 3x + 1)^2 = y^2 + y + 1 )Hmm, so the left side is a perfect square, and the right side is ( y^2 + y + 1 ). For this equality to hold, ( y^2 + y + 1 ) must also be a perfect square. Let me denote ( k = x^2 + 3x + 1 ), so the equation becomes:( k^2 = y^2 + y + 1 )So, ( y^2 + y + 1 = k^2 ). Let me rearrange this:( y^2 + y + 1 - k^2 = 0 )This is a quadratic in terms of ( y ). Maybe I can solve for ( y ) in terms of ( k ). Let's write it as:( y^2 + y + (1 - k^2) = 0 )Using the quadratic formula, ( y = frac{-1 pm sqrt{1 - 4(1 - k^2)}}{2} )Simplify the discriminant:( sqrt{1 - 4(1 - k^2)} = sqrt{1 - 4 + 4k^2} = sqrt{4k^2 - 3} )So, ( y = frac{-1 pm sqrt{4k^2 - 3}}{2} )Since ( y ) must be an integer, the discriminant ( sqrt{4k^2 - 3} ) must be an integer as well. Let me denote ( m = sqrt{4k^2 - 3} ), so:( m^2 = 4k^2 - 3 )Rearranging:( 4k^2 - m^2 = 3 )This is a Diophantine equation. Let me write it as:( (2k - m)(2k + m) = 3 )Since 3 is a prime number, the possible factor pairs are (1,3) and (-1,-3). Let me consider both cases.Case 1: ( 2k - m = 1 ) and ( 2k + m = 3 )Adding these two equations:( 4k = 4 ) => ( k = 1 )Subtracting the first equation from the second:( 2m = 2 ) => ( m = 1 )Case 2: ( 2k - m = -1 ) and ( 2k + m = -3 )Adding these two equations:( 4k = -4 ) => ( k = -1 )Subtracting the first equation from the second:( 2m = -2 ) => ( m = -1 )So, the only integer solutions for ( k ) and ( m ) are ( k = 1, m = 1 ) and ( k = -1, m = -1 ).Now, recall that ( k = x^2 + 3x + 1 ). So, let's solve for ( x ) in both cases.Case 1: ( k = 1 )( x^2 + 3x + 1 = 1 )Simplify:( x^2 + 3x = 0 )Factor:( x(x + 3) = 0 )So, ( x = 0 ) or ( x = -3 )Case 2: ( k = -1 )( x^2 + 3x + 1 = -1 )Simplify:( x^2 + 3x + 2 = 0 )Factor:( (x + 1)(x + 2) = 0 )So, ( x = -1 ) or ( x = -2 )So, the possible ( x ) values are ( x = 0, -1, -2, -3 ).Now, let's find the corresponding ( y ) values for each ( x ).Starting with ( x = 0 ):Left side: ( 0 times 1 times 2 times 3 = 0 )So, ( y(y + 1) = 0 )This gives ( y = 0 ) or ( y = -1 )So, pairs: ( (0, 0) ) and ( (0, -1) )Next, ( x = -1 ):Left side: ( (-1) times 0 times 1 times 2 = 0 )So, ( y(y + 1) = 0 )Again, ( y = 0 ) or ( y = -1 )Pairs: ( (-1, 0) ) and ( (-1, -1) )Next, ( x = -2 ):Left side: ( (-2) times (-1) times 0 times 1 = 0 )So, ( y(y + 1) = 0 )Same as before, ( y = 0 ) or ( y = -1 )Pairs: ( (-2, 0) ) and ( (-2, -1) )Finally, ( x = -3 ):Left side: ( (-3) times (-2) times (-1) times 0 = 0 )So, ( y(y + 1) = 0 )Again, ( y = 0 ) or ( y = -1 )Pairs: ( (-3, 0) ) and ( (-3, -1) )So, compiling all the pairs, we have:( (0, 0) ), ( (0, -1) ), ( (-1, 0) ), ( (-1, -1) ), ( (-2, 0) ), ( (-2, -1) ), ( (-3, 0) ), ( (-3, -1) )Let me double-check if there are any other possible solutions. We considered all cases where ( k = 1 ) and ( k = -1 ), which came from the factor pairs of 3. Since 3 is prime, there are no other factor pairs, so these should be all the solutions.I think that's it. So, the integer pairs ( (x, y) ) that satisfy the equation are the eight pairs listed above.

💡Okay, so I have this problem where Petya thought of five numbers and wrote down all their pairwise sums. The sums given are: 7, 9, 12, 16, 17, 19, 20, 21, 22, 29. I need to figure out what the original five numbers are.First, let me understand what pairwise sums mean. If there are five numbers, say a, b, c, d, e, then the pairwise sums would be all possible sums of two different numbers. So, that would be a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e, and d+e. That makes 10 sums in total, which matches the number of sums given here.So, I need to find five numbers such that when I add them two at a time, I get exactly the sums listed: 7, 9, 12, 16, 17, 19, 20, 21, 22, 29.Let me denote the five numbers in increasing order as x1, x2, x3, x4, x5. So, x1 is the smallest and x5 is the largest.Now, the smallest pairwise sum should be x1 + x2, because those are the two smallest numbers. Looking at the given sums, the smallest is 7. So, I can say that x1 + x2 = 7.Similarly, the largest pairwise sum should be x4 + x5, as those are the two largest numbers. The largest given sum is 29, so x4 + x5 = 29.Next, the second smallest sum should be x1 + x3. Looking at the given sums, the next smallest after 7 is 9. So, x1 + x3 = 9.Similarly, the second largest sum should be x3 + x5. The second largest given sum is 22, so x3 + x5 = 22.Now, I have four equations:1. x1 + x2 = 72. x4 + x5 = 293. x1 + x3 = 94. x3 + x5 = 22I need to find x1, x2, x3, x4, x5.Let me see if I can find x3 first. From equation 3, x1 + x3 = 9, and from equation 1, x1 + x2 = 7. If I subtract equation 1 from equation 3, I get:(x1 + x3) - (x1 + x2) = 9 - 7x3 - x2 = 2So, x3 = x2 + 2That's useful. Now, let's see if I can find another equation involving x3.From equation 4, x3 + x5 = 22. From equation 2, x4 + x5 = 29. If I subtract equation 4 from equation 2, I get:(x4 + x5) - (x3 + x5) = 29 - 22x4 - x3 = 7So, x4 = x3 + 7Now, I have expressions for x3 and x4 in terms of x2.From earlier, x3 = x2 + 2, so x4 = (x2 + 2) + 7 = x2 + 9So, now I have:x1 + x2 = 7x3 = x2 + 2x4 = x2 + 9x4 + x5 = 29x3 + x5 = 22Let me try to express x5 in terms of x2.From equation 4: x3 + x5 = 22We know x3 = x2 + 2, so:(x2 + 2) + x5 = 22x5 = 22 - x2 - 2x5 = 20 - x2From equation 2: x4 + x5 = 29We know x4 = x2 + 9 and x5 = 20 - x2, so:(x2 + 9) + (20 - x2) = 29Simplify:x2 + 9 + 20 - x2 = 2929 = 29Hmm, that's just a true statement, which doesn't help me find x2. So, I need another approach.Let me think about the total sum of all pairwise sums. There are 10 sums, each being the sum of two numbers. If I add all these sums together, each number is added four times (since each number pairs with four others). So, the total sum of all pairwise sums is 4*(x1 + x2 + x3 + x4 + x5).Let me calculate the total sum of the given pairwise sums:7 + 9 + 12 + 16 + 17 + 19 + 20 + 21 + 22 + 29Let me add them step by step:7 + 9 = 1616 + 12 = 2828 + 16 = 4444 + 17 = 6161 + 19 = 8080 + 20 = 100100 + 21 = 121121 + 22 = 143143 + 29 = 172So, the total sum of all pairwise sums is 172.Since this is equal to 4*(sum of the five numbers), then:4*(x1 + x2 + x3 + x4 + x5) = 172Divide both sides by 4:x1 + x2 + x3 + x4 + x5 = 43Now, I can use this total sum to find the numbers.From earlier, I have:x1 + x2 = 7x3 = x2 + 2x4 = x2 + 9x5 = 20 - x2So, let's substitute these into the total sum:x1 + x2 + x3 + x4 + x5 = 43Substitute:x1 + x2 + (x2 + 2) + (x2 + 9) + (20 - x2) = 43Simplify:x1 + x2 + x2 + 2 + x2 + 9 + 20 - x2 = 43Combine like terms:x1 + (x2 + x2 + x2 - x2) + (2 + 9 + 20) = 43Simplify:x1 + (2x2) + 31 = 43So,x1 + 2x2 = 43 - 31 = 12But from equation 1, x1 + x2 = 7So, we have:x1 + x2 = 7x1 + 2x2 = 12Subtract the first equation from the second:(x1 + 2x2) - (x1 + x2) = 12 - 7x2 = 5Now that we have x2 = 5, we can find the other numbers.From equation 1: x1 + x2 = 7x1 + 5 = 7 => x1 = 2From x3 = x2 + 2 = 5 + 2 = 7From x4 = x2 + 9 = 5 + 9 = 14From x5 = 20 - x2 = 20 - 5 = 15So, the five numbers are: 2, 5, 7, 14, 15Let me verify if these numbers give the correct pairwise sums.List all pairwise sums:2+5=72+7=92+14=162+15=175+7=125+14=195+15=207+14=217+15=2214+15=29Yes, these are exactly the sums given: 7, 9, 12, 16, 17, 19, 20, 21, 22, 29.So, the numbers Petya thought of are 2, 5, 7, 14, and 15.

💡Okay, so I have this inequality: (x - 5a)(x + a) > 0, and I need to find the solution set when a is less than zero. Hmm, let me think about how to approach this.First, I remember that when dealing with quadratic inequalities like this, it's helpful to find the roots of the equation because they divide the number line into intervals that we can test. The roots are the values of x that make the expression equal to zero. So, I need to solve (x - 5a)(x + a) = 0.Setting each factor equal to zero:1. x - 5a = 0 => x = 5a2. x + a = 0 => x = -aSo, the critical points are x = 5a and x = -a. These points divide the number line into three intervals. To determine where the product is positive, I need to test each interval.But wait, before I proceed, I should note that a is negative. That might affect the order of these critical points on the number line. Let me figure out which one is bigger between 5a and -a.Since a is negative, multiplying it by 5 will make it more negative. So, 5a is less than a, right? But wait, -a is positive because a is negative. So, 5a is negative, and -a is positive. Therefore, 5a is less than -a because it's negative and -a is positive.So, on the number line, 5a is to the left of -a. That means the intervals are:1. x < 5a2. 5a < x < -a3. x > -aNow, I need to test each interval to see where the product (x - 5a)(x + a) is positive.Let me pick test points for each interval.1. For x < 5a: Let's choose x = 5a - 1 (since a is negative, 5a is negative, so 5a -1 is even more negative). Plugging into (x - 5a)(x + a): (5a -1 - 5a)(5a -1 + a) = (-1)(6a -1) Since a is negative, 6a is negative, so 6a -1 is negative. So, (-1)(negative) = positive. So, the product is positive in this interval.2. For 5a < x < -a: Let's choose x = 0 (assuming 5a < 0 < -a, which is true because 5a is negative and -a is positive). Plugging into (x - 5a)(x + a): (0 - 5a)(0 + a) = (-5a)(a) Since a is negative, -5a is positive, and a is negative. So, positive times negative is negative. Therefore, the product is negative in this interval.3. For x > -a: Let's choose x = -a + 1 (since -a is positive, adding 1 makes it more positive). Plugging into (x - 5a)(x + a): (-a +1 -5a)(-a +1 +a) = (-6a +1)(1) Since a is negative, -6a is positive, so -6a +1 is positive. Multiplying by 1 keeps it positive. So, the product is positive in this interval.Therefore, the solution set is where the product is positive, which is x < 5a or x > -a.Wait, let me double-check. Since a is negative, 5a is negative and -a is positive. So, x < 5a is a negative number, and x > -a is a positive number. That makes sense because the parabola opens upwards (since the coefficient of x^2 is positive), so it will be positive outside the roots.Yeah, that seems correct. So, the solution set is x < 5a or x > -a.

💡Okay, so I have this problem here: I need to find the greatest common divisor (GCD) of 8! and 10!. Hmm, factorials, right? I remember that n! means the product of all positive integers up to n. So, 8! is 8 × 7 × 6 × ... × 1, and 10! is 10 × 9 × 8 × ... × 1. Alright, so first, maybe I should write out what 8! and 10! actually are. Let me calculate them:8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute that step by step:- 8 × 7 = 56- 56 × 6 = 336- 336 × 5 = 1680- 1680 × 4 = 6720- 6720 × 3 = 20160- 20160 × 2 = 40320- 40320 × 1 = 40320So, 8! is 40320.Now, 10! is 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. That's a bigger number. Let me see if I can relate it to 8! to make it easier. I notice that 10! = 10 × 9 × 8!. So, if I already know 8! is 40320, then 10! = 10 × 9 × 40320. Let me compute that:- 10 × 9 = 90- 90 × 40320 = 3,628,800So, 10! is 3,628,800.Now, I need to find the GCD of 40320 and 3,628,800. I remember that the GCD of two numbers is the largest number that divides both of them without leaving a remainder. Since 8! is a factor of 10! (because 10! = 10 × 9 × 8!), that means 8! divides 10! exactly. So, 8! is a common divisor of both 8! and 10!.Is there a larger common divisor? Well, since 8! is the smaller number here, and it divides the larger number 10!, it must be the greatest common divisor. Because any number larger than 8! wouldn't divide 8! itself, right? So, 8! is the largest number that divides both 8! and 10!.Just to double-check, maybe I can think about the prime factors. The GCD can also be found by taking the minimum exponents of the common prime factors. Let's factorize 8! and 10!:For 8!:- 8 = 2^3- 7 = 7- 6 = 2 × 3- 5 = 5- 4 = 2^2- 3 = 3- 2 = 2- 1 = 1So, the prime factors are:- 2: 3 (from 8) + 1 (from 6) + 2 (from 4) + 1 (from 2) = 7- 3: 1 (from 6) + 1 (from 3) = 2- 5: 1- 7: 1So, 8! = 2^7 × 3^2 × 5 × 7.For 10!:- 10 = 2 × 5- 9 = 3^2- 8 = 2^3- 7 = 7- 6 = 2 × 3- 5 = 5- 4 = 2^2- 3 = 3- 2 = 2- 1 = 1So, the prime factors are:- 2: 1 (from 10) + 3 (from 8) + 1 (from 6) + 2 (from 4) + 1 (from 2) = 8- 3: 2 (from 9) + 1 (from 6) + 1 (from 3) = 4- 5: 1 (from 10) + 1 (from 5) = 2- 7: 1So, 10! = 2^8 × 3^4 × 5^2 × 7.Now, to find the GCD, I take the minimum exponents for each prime factor:- For 2: min(7, 8) = 7- For 3: min(2, 4) = 2- For 5: min(1, 2) = 1- For 7: min(1, 1) = 1So, GCD = 2^7 × 3^2 × 5 × 7.Let me compute that:- 2^7 = 128- 3^2 = 9- 5 = 5- 7 = 7Now, multiply them together:- 128 × 9 = 1152- 1152 × 5 = 5760- 5760 × 7 = 40320So, the GCD is 40320, which is exactly 8!. That confirms my earlier conclusion.I think that's solid. I approached it both by recognizing that 8! is a factor of 10! and by breaking down the prime factors to ensure that 8! is indeed the greatest common divisor. It all checks out.

💡Okay, so I have this question about regression analysis with three statements, and I need to figure out how many of them are correct. Let me go through each statement one by one and think about what I know regarding regression models.Starting with statement ①: "In the residual plot, if the residual points are evenly distributed within a horizontal band, it indicates that the chosen model is appropriate." Hmm, I remember that residual plots are used to check the assumptions of a regression model. The residuals are the differences between the observed and predicted values. If the residuals are randomly scattered around the horizontal axis without any discernible pattern, it suggests that the model is a good fit. If they form a horizontal band, that means the variance is consistent, which is another good sign. So, I think statement ① is correct.Moving on to statement ②: "The coefficient of determination ( R^2 ) is used to describe the effectiveness of regression. The smaller the ( R^2 ) value, the better the model fits." Wait, I thought ( R^2 ) measures how well the model explains the variance in the data. A higher ( R^2 ) value means a better fit because it indicates that more variance is explained by the model. So, if someone says a smaller ( R^2 ) is better, that doesn't sound right. I believe statement ② is incorrect.Now, statement ③: "To compare the fitting effects of two models, one can compare the sizes of the sum of squared residuals. The model with a smaller sum of squared residuals has a better fitting effect." The sum of squared residuals (SSR) is a measure of the discrepancy between the data and the estimation model. A smaller SSR means the model's predictions are closer to the actual data points, which indicates a better fit. So, comparing SSR makes sense, and a smaller SSR is indeed better. Therefore, statement ③ is correct.Putting it all together: statement ① is correct, statement ② is incorrect, and statement ③ is correct. That means two out of the three statements are correct. So, the number of correct statements is 2.**Final Answer**The number of correct statements is boxed{C}.

💡First, I need to determine how many small cakes each helper can make per hour. Let's denote this number as ( S ).Each helper can make either 2 large cakes or ( S ) small cakes in one hour. With 10 helpers working for 3 hours, the total capacity for large cakes is:[10 text{ helpers} times 2 text{ large cakes/helper} times 3 text{ hours} = 60 text{ large cakes}]However, only 20 large cakes are needed. This means that 10 helpers can produce 20 large cakes in 1 hour:[10 text{ helpers} times 2 text{ large cakes/helper} = 20 text{ large cakes}]For the remaining 2 hours, all 10 helpers can focus on making small cakes. The total number of small cakes needed is 700. Therefore, the equation for small cakes is:[10 text{ helpers} times S text{ small cakes/helper} times 2 text{ hours} = 700 text{ small cakes}]Simplifying this equation:[20S = 700][S = frac{700}{20} = 35]Each helper can make 35 small cakes per hour.