answer：First, I need to determine the length of the longer piece of wire. The total length of the wire is 50 cm, and the shorter piece is 14.285714285714285 cm long. By subtracting the length of the shorter piece from the total length, I can find the length of the longer piece.Next, I will calculate the ratio of the shorter piece to the longer piece by dividing the length of the shorter piece by the length of the longer piece. This will give me a numerical ratio.To simplify the ratio, I will look for a common factor in both the shorter and longer piece lengths. In this case, both lengths are divisible by 7. By dividing both lengths by 7, I can express the ratio in its simplest form.Finally, I will present the simplified ratio as the final answer.

answer：Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. It says that Sₙ is the sum of the first n terms of an arithmetic sequence {aₙ}. Then it gives me two pieces of information: a₂ + a₃ = 16 and S₂ = 8. I need to find a₁, which is the first term of the sequence.Alright, let's start by recalling some basic facts about arithmetic sequences. An arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference, which we usually denote as d. So, the nth term of an arithmetic sequence can be written as aₙ = a₁ + (n - 1)d. That makes sense.Now, the sum of the first n terms, Sₙ, of an arithmetic sequence can be calculated using the formula Sₙ = n/2 * (2a₁ + (n - 1)d) or sometimes it's written as Sₙ = n*(a₁ + aₙ)/2. Both formulas are equivalent because aₙ = a₁ + (n - 1)d. So, that's good to remember.Given that, let's look at the information provided. First, a₂ + a₃ = 16. Let me write that down:a₂ + a₃ = 16.Since this is an arithmetic sequence, a₂ = a₁ + d and a₃ = a₁ + 2d. So, substituting these into the equation:(a₁ + d) + (a₁ + 2d) = 16.Let me simplify that:a₁ + d + a₁ + 2d = 16.Combine like terms:2a₁ + 3d = 16.Okay, so that's equation one: 2a₁ + 3d = 16.Next, the problem says S₂ = 8. S₂ is the sum of the first two terms, so that would be a₁ + a₂. Let me write that:S₂ = a₁ + a₂ = 8.But since a₂ = a₁ + d, substituting that in:a₁ + (a₁ + d) = 8.Simplify:2a₁ + d = 8.So, that's equation two: 2a₁ + d = 8.Now, I have a system of two equations:1. 2a₁ + 3d = 16.2. 2a₁ + d = 8.I need to solve for a₁. Let me write them down again:Equation 1: 2a₁ + 3d = 16.Equation 2: 2a₁ + d = 8.Hmm, maybe I can subtract equation 2 from equation 1 to eliminate 2a₁.So, subtracting equation 2 from equation 1:(2a₁ + 3d) - (2a₁ + d) = 16 - 8.Simplify:2a₁ + 3d - 2a₁ - d = 8.The 2a₁ terms cancel out:(3d - d) = 8.Which simplifies to:2d = 8.So, solving for d:d = 8 / 2 = 4.Alright, so the common difference d is 4. Now, I can plug this back into equation 2 to find a₁.Equation 2: 2a₁ + d = 8.Substituting d = 4:2a₁ + 4 = 8.Subtract 4 from both sides:2a₁ = 8 - 4 = 4.Then, divide both sides by 2:a₁ = 4 / 2 = 2.So, the first term a₁ is 2. Let me just double-check to make sure I didn't make any mistakes.First, let's verify a₂ + a₃ = 16.a₁ = 2, d = 4.So, a₂ = a₁ + d = 2 + 4 = 6.a₃ = a₂ + d = 6 + 4 = 10.Adding a₂ + a₃: 6 + 10 = 16. That checks out.Now, S₂ = a₁ + a₂ = 2 + 6 = 8. That also checks out.Okay, so everything seems consistent. I think I did it right. The first term a₁ is 2.**Final Answer**The value of ( a_1 ) is boxed{2}.

answer：First, I need to factorize each of the numbers involved in the expression:- (87 = 3 times 29)- (89) is a prime number.- (91 = 7 times 13)- (93 = 3 times 31)Next, I'll calculate the product (87 cdot 89 cdot 91):[87 cdot 89 cdot 91 = (3 times 29) cdot 89 cdot (7 times 13) = 3 times 29 times 89 times 7 times 13]Now, I'll add 1 to this product:[87 cdot 89 cdot 91 + 1 = 3 times 29 times 89 times 7 times 13 + 1]Since adding 1 to a product of primes results in a number that is not divisible by any of the primes in the product, (87 cdot 89 cdot 91 + 1) is a new prime number.Finally, I'll factorize the entire expression ((87 cdot 89 cdot 91 + 1) cdot 93):[(87 cdot 89 cdot 91 + 1) cdot 93 = (87 cdot 89 cdot 91 + 1) times (3 times 31)]The prime factors in this factorization are:- (3), (29), (89), (7), (13), (31), and the prime number (87 cdot 89 cdot 91 + 1).Therefore, there are a total of 7 different prime factors.

answer：First, I need to determine the dimensions of the original square pattern. Since there are 9 black tiles and 16 white tiles, the total number of tiles is 25. This suggests that the original square is 5x5.Next, I'll consider the extension process. Adding a border with two alternating layers around the original square will increase the size of the square. The first layer will add a black border, and the second layer will add a white border.For the first layer (black border), the new square becomes 7x7. The number of tiles in this layer is 49, and subtracting the original 25 tiles gives 24 new black tiles.For the second layer (white border), the new square becomes 9x9. The number of tiles in this layer is 81, and subtracting the previous total of 49 tiles gives 32 new white tiles.Now, I'll calculate the total number of black and white tiles after the extension. The total black tiles are 9 (original) + 24 (new) = 33. The total white tiles are 16 (original) + 32 (new) = 48.Finally, the ratio of black tiles to white tiles in the extended pattern is 33:48.

answer：Alright, so I've got this system of inequalities to solve, and I need to figure out the range of values for ( m ) such that there are exactly 4 integer solutions. The system is:[begin{cases}x < m 7 - 2x leq 1end{cases}]Okay, let's start by solving each inequality separately to understand the constraints on ( x ).First, the second inequality: ( 7 - 2x leq 1 ). I need to solve for ( x ). Let me subtract 7 from both sides:[-2x leq 1 - 7 -2x leq -6]Now, I need to divide both sides by -2. Hmm, wait a second, when I divide or multiply both sides of an inequality by a negative number, the direction of the inequality sign flips. So, dividing both sides by -2 gives:[x geq 3]Alright, so from the second inequality, ( x ) must be greater than or equal to 3.Now, looking at the first inequality: ( x < m ). So, ( x ) must be less than ( m ).Combining both inequalities, the solution set for ( x ) is:[3 leq x < m]So, ( x ) is between 3 and ( m ), including 3 but not including ( m ).The problem states that there are exactly 4 integer solutions. Let's think about what integers satisfy ( 3 leq x < m ). The integers would be 3, 4, 5, and 6. That's four integers. So, the next integer after 6 is 7, but since ( x ) must be less than ( m ), ( m ) must be greater than 6 but not greater than or equal to 7, because if ( m ) were 7 or more, then 7 would also be a solution, making it the fifth integer solution.Wait, let me double-check that. If ( m ) is exactly 7, then ( x ) can be up to just below 7, so 6 is still the largest integer solution. So, ( m ) can be 7, right? Because if ( m = 7 ), then ( x ) is less than 7, so 6 is still included, and 7 is not. So, 3, 4, 5, 6 are the four integer solutions.But if ( m ) is greater than 7, say 8, then ( x ) can be up to just below 8, which would include 7 as an integer solution, making it five solutions: 3, 4, 5, 6, 7. That's more than four, so ( m ) can't be greater than 7.On the lower end, if ( m ) is less than or equal to 6, say 6, then ( x ) has to be less than 6, so the integer solutions would be 3, 4, 5. That's only three solutions, which is less than four. So, ( m ) has to be greater than 6 to include 6 as a solution.Wait, but if ( m ) is exactly 6, then ( x ) is less than 6, so the integer solutions are 3, 4, 5. That's three solutions. So, ( m ) needs to be greater than 6 to include 6 as a solution, making the integer solutions 3, 4, 5, 6.But if ( m ) is just a little bit more than 6, say 6.1, then ( x ) can be up to 6.1, so 6 is still included as an integer solution. Similarly, if ( m ) is 7, as I thought earlier, 6 is still the largest integer solution.So, putting it all together, ( m ) has to be greater than 6 but less than or equal to 7. Because if ( m ) is greater than 7, we get an extra integer solution, and if ( m ) is 6 or less, we don't have enough integer solutions.So, the range of ( m ) is ( 6 < m leq 7 ).Looking back at the options:A: ( 6 < m < 7 )B: ( 6 leq m < 7 )C: ( 6 leq m leq 7 )D: ( 6 < m leq 7 )So, the correct answer should be D: ( 6 < m leq 7 ).Wait, just to make sure, let's test the boundaries.If ( m = 6 ), then ( x < 6 ), so integer solutions are 3, 4, 5. That's three solutions, which is not enough.If ( m = 7 ), then ( x < 7 ), so integer solutions are 3, 4, 5, 6. That's four solutions, which is exactly what we need.If ( m = 6.5 ), then ( x < 6.5 ), so integer solutions are 3, 4, 5, 6. Still four solutions.If ( m = 7.5 ), then ( x < 7.5 ), so integer solutions are 3, 4, 5, 6, 7. That's five solutions, which is too many.If ( m = 6.1 ), ( x < 6.1 ), integer solutions are 3, 4, 5, 6. Four solutions.So, yes, ( m ) must be greater than 6 and up to 7, including 7.Therefore, the correct answer is D: ( 6 < m leq 7 ).

answer：Okay, so I have this problem about a sequence of positive integers defined by g₁ = 1, and for every positive integer n, g_{n+1} = g_n² + g_n + 1. I need to show that g_n² + 1 divides g_{n+1}² + 1 for every positive integer n. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me write down what I know. The sequence starts with g₁ = 1. Then, using the recurrence relation, I can compute the next few terms to see if I can spot a pattern or get some intuition.So, g₂ = g₁² + g₁ + 1 = 1² + 1 + 1 = 3.Then, g₃ = g₂² + g₂ + 1 = 3² + 3 + 1 = 9 + 3 + 1 = 13.Next, g₄ = g₃² + g₃ + 1 = 13² + 13 + 1 = 169 + 13 + 1 = 183.Hmm, okay, so the sequence is growing pretty quickly. Now, I need to show that g_n² + 1 divides g_{n+1}² + 1. Let me write that out in terms of the given recurrence.Given that g_{n+1} = g_n² + g_n + 1, I can express g_{n+1}² + 1 as (g_n² + g_n + 1)² + 1. Maybe if I expand this, I can see if it's a multiple of g_n² + 1.Let me compute (g_n² + g_n + 1)² first. Expanding that, I get:(g_n² + g_n + 1)² = (g_n²)² + 2g_n²(g_n) + 2g_n²(1) + (g_n)² + 2g_n(1) + 1²Wait, that seems a bit messy. Maybe I should use the formula (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc.Applying that here, where a = g_n², b = g_n, and c = 1:(g_n² + g_n + 1)² = (g_n²)² + (g_n)² + (1)² + 2(g_n²)(g_n) + 2(g_n²)(1) + 2(g_n)(1)Simplifying each term:= g_n⁴ + g_n² + 1 + 2g_n³ + 2g_n² + 2g_nNow, combine like terms:= g_n⁴ + 2g_n³ + (g_n² + 2g_n²) + 2g_n + 1= g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 1Okay, so (g_n² + g_n + 1)² = g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 1.Now, adding 1 to this, as per the problem statement, gives:g_{n+1}² + 1 = (g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 1) + 1 = g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 2.So, I need to show that g_n² + 1 divides this expression: g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 2.Hmm, maybe I can factor this expression in terms of g_n² + 1. Let me try to see if I can factor it.Let me denote D = g_n² + 1. I want to see if D divides the expression E = g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 2.So, if I can write E as D multiplied by some polynomial, then D divides E.Let me try polynomial division or see if I can factor E as (g_n² + 1)(something).Let me assume that E = (g_n² + 1)(A g_n² + B g_n + C). Let's find A, B, C such that this holds.Multiplying out the right-hand side:(g_n² + 1)(A g_n² + B g_n + C) = A g_n⁴ + B g_n³ + C g_n² + A g_n² + B g_n + CCombine like terms:= A g_n⁴ + B g_n³ + (C + A) g_n² + B g_n + CNow, set this equal to E = g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 2.So, equate coefficients:For g_n⁴: A = 1For g_n³: B = 2For g_n²: C + A = 3. Since A = 1, then C = 2.For g_n: B = 2, which matches.For the constant term: C = 2, which also matches.So, E = (g_n² + 1)(g_n² + 2g_n + 2).Wow, that worked out nicely! So, g_{n+1}² + 1 = (g_n² + 1)(g_n² + 2g_n + 2).Therefore, since g_n² + 1 is a factor of g_{n+1}² + 1, it follows that g_n² + 1 divides g_{n+1}² + 1.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I computed g_{n+1}² + 1 correctly by expanding (g_n² + g_n + 1)² and adding 1. The expansion gave me g_n⁴ + 2g_n³ + 3g_n² + 2g_n + 2.2. Then, I set up the factorization of this expression as (g_n² + 1)(A g_n² + B g_n + C) and solved for A, B, C by equating coefficients.3. I found A=1, B=2, C=2, which worked out perfectly.4. So, the factorization is correct, and thus g_n² + 1 divides g_{n+1}² + 1.I think that makes sense. Maybe I can test it with the values I computed earlier to see if it holds.Let's take n=1:g₁ = 1, so g₁² + 1 = 1 + 1 = 2.g₂ = 3, so g₂² + 1 = 9 + 1 = 10.Does 2 divide 10? Yes, because 10 ÷ 2 = 5.Now, n=2:g₂ = 3, so g₂² + 1 = 9 + 1 = 10.g₃ = 13, so g₃² + 1 = 169 + 1 = 170.Does 10 divide 170? Yes, because 170 ÷ 10 = 17.n=3:g₃ = 13, so g₃² + 1 = 169 + 1 = 170.g₄ = 183, so g₄² + 1 = 183² + 1 = 33489 + 1 = 33490.Does 170 divide 33490? Let's check: 33490 ÷ 170 = 197. So yes, it divides exactly.Okay, so the pattern holds for the first few terms. That gives me more confidence that the proof is correct.I think I've got it. By expanding g_{n+1}² + 1 and factoring it as (g_n² + 1)(g_n² + 2g_n + 2), I've shown that g_n² + 1 is indeed a divisor of g_{n+1}² + 1 for every positive integer n.