💡Okay, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. The problem says that the sequence {a_n} is arithmetic, and we're given that a₂ = 9 and a₄ = 17. We need to find two things: first, the general formula for the sequence, and second, the sum of the first n terms, denoted as S_n.Alright, starting with part (1). I remember that in an arithmetic sequence, each term is obtained by adding a constant difference to the previous term. This constant is called the common difference, usually denoted as d. So, the general formula for an arithmetic sequence is a_n = a₁ + (n - 1)d, where a₁ is the first term.But in this problem, we aren't given a₁ directly. Instead, we have a₂ and a₄. Let me write down what I know:a₂ = 9a₄ = 17Since it's an arithmetic sequence, the difference between consecutive terms is constant. So, the difference between a₄ and a₂ should be twice the common difference, right? Because from a₂ to a₃ is one difference, and from a₃ to a₄ is another. So, the total difference between a₄ and a₂ is 2d.Let me calculate that:a₄ - a₂ = 17 - 9 = 8So, 2d = 8, which means d = 4. Okay, so the common difference is 4.Now that I have d, I can find a₁. Since a₂ = 9, which is the second term, it should be equal to a₁ + d. So,a₂ = a₁ + d9 = a₁ + 4Therefore, a₁ = 9 - 4 = 5.Got it, so the first term is 5, and the common difference is 4. Now, I can write the general formula for a_n.Using the formula a_n = a₁ + (n - 1)d,a_n = 5 + (n - 1) * 4Let me simplify that:a_n = 5 + 4n - 4a_n = 4n + 1So, the general formula is a_n = 4n + 1. That seems straightforward.Now, moving on to part (2). We need to find the sum of the first n terms, S_n. I remember that the sum of an arithmetic sequence can be calculated using the formula:S_n = n/2 * (a₁ + a_n)Alternatively, it can also be written as:S_n = n/2 * [2a₁ + (n - 1)d]Either formula should work. Let me use the first one because I already have a₁ and a_n.We have a₁ = 5 and a_n = 4n + 1. Plugging these into the formula:S_n = n/2 * (5 + 4n + 1)Simplify inside the parentheses:5 + 1 = 6, so:S_n = n/2 * (4n + 6)Let me factor out a 2 from the numerator:S_n = n/2 * 2*(2n + 3)The 2 in the numerator and the denominator cancel out:S_n = n*(2n + 3)So, expanding that:S_n = 2n² + 3nWait, let me double-check that. Alternatively, using the other formula:S_n = n/2 * [2a₁ + (n - 1)d]Plugging in a₁ = 5 and d = 4:S_n = n/2 * [2*5 + (n - 1)*4]S_n = n/2 * [10 + 4n - 4]Simplify inside the brackets:10 - 4 = 6, so:S_n = n/2 * (4n + 6)Again, factor out a 2:S_n = n/2 * 2*(2n + 3)S_n = n*(2n + 3)S_n = 2n² + 3nSame result. Okay, so that seems consistent.Just to make sure I didn't make a mistake, let me test with n = 2 and n = 4.For n = 2, the sum should be a₁ + a₂ = 5 + 9 = 14.Using the formula S_n = 2n² + 3n,S₂ = 2*(2)² + 3*2 = 2*4 + 6 = 8 + 6 = 14. Correct.For n = 4, the sum should be a₁ + a₂ + a₃ + a₄ = 5 + 9 + 13 + 17.Calculating that: 5 + 9 = 14, 14 + 13 = 27, 27 + 17 = 44.Using the formula:S₄ = 2*(4)² + 3*4 = 2*16 + 12 = 32 + 12 = 44. Correct.Alright, so both parts seem to check out. The general formula for the sequence is a_n = 4n + 1, and the sum of the first n terms is S_n = 2n² + 3n.**Final Answer**(1) The general formula for the sequence is boxed{4n + 1}.(2) The sum of the first n terms of the sequence is boxed{2n^2 + 3n}.

💡Alright, so I've got this problem here about sets and a new operation defined as "⊗". Let me try to wrap my head around it step by step.First off, the operation is defined as P⊗Q = {x | x ∈ P∪Q and x ∉ P∩Q}. Hmm, okay. So, this operation is taking elements that are in either P or Q but not in both. That sounds familiar—it's like the symmetric difference between two sets. I remember that the symmetric difference includes elements that are in one set or the other but not in both. So, P⊗Q is essentially the symmetric difference of P and Q.Now, the sets given are P = {x | 0 ≤ x ≤ 2} and Q = {x | x > 1}. Let me visualize these sets on the number line to get a better understanding.Set P is all real numbers from 0 to 2, inclusive. So, it's a closed interval [0, 2]. Set Q is all real numbers greater than 1, which is an open interval (1, ∞). To find P⊗Q, which is the symmetric difference, I need to find all elements that are in P or Q but not in both. So, let's break it down:1. **Elements in P but not in Q**: Since Q starts at 1 and goes to infinity, the part of P that's not in Q would be from 0 up to 1. But wait, Q is x > 1, so at x = 1, is it included in Q? No, because Q is strictly greater than 1. So, x = 1 is in P but not in Q. Therefore, this part is [0, 1].2. **Elements in Q but not in P**: Q starts at 1 and goes to infinity, but P only goes up to 2. So, the part of Q that's not in P would be from 2 to infinity. But since P includes 2, does Q include 2? No, because Q is x > 1, which includes numbers just above 1 but not exactly at 2. Wait, actually, Q is all numbers greater than 1, so it includes numbers greater than 2 as well. Therefore, the part of Q not in P is (2, ∞).Putting these two parts together, the symmetric difference P⊗Q is [0, 1] union (2, ∞). Let me double-check to make sure I didn't miss anything. The symmetric difference should exclude the intersection of P and Q. The intersection of P and Q is where both sets overlap. P is [0, 2], and Q is (1, ∞), so their intersection is (1, 2]. Therefore, in the symmetric difference, we exclude (1, 2]. So, from P, we take [0, 1], and from Q, we take (2, ∞). Combining these gives [0, 1] ∪ (2, ∞), which matches what I found earlier.Looking at the options provided:A: [0,1) ∪ (2, +∞)B: [0,1] ∪ (2, +∞)C: [1,2]D: (2, +∞)The correct answer should include 1 because in the symmetric difference, x = 1 is in P but not in Q, so it should be included. Therefore, option B is the right choice.I think I've covered all the steps and checked my reasoning. It seems solid, but just to be thorough, let me consider if there's any overlap or something I might have missed.- P is [0, 2], so it includes 0, 1, and 2.- Q is (1, ∞), so it doesn't include 1 but includes everything above 1.- The intersection is (1, 2], so in the symmetric difference, we remove this part from the union of P and Q.- The union of P and Q is [0, ∞), but removing (1, 2] leaves us with [0, 1] and (2, ∞).Yep, that all checks out. So, I'm confident that option B is correct.**Final Answer**boxed{B}

💡Okay, so I have this problem that says among any 12 consecutive positive integers, there is at least one which is smaller than the sum of its proper divisors. Proper divisors are all the positive integers that divide the number except 1 and the number itself. For example, for 14, the proper divisors are 2 and 7.First, I need to understand what it means for a number to be smaller than the sum of its proper divisors. That sounds like the number is what's called "abundant." An abundant number is one where the sum of its proper divisors is greater than the number itself. So, the problem is essentially saying that in any set of 12 consecutive numbers, there must be at least one abundant number.Alright, so my goal is to show that in any such set, there's always an abundant number. I wonder how I can approach this. Maybe I can look at the properties of numbers and their divisors. Since we're dealing with consecutive numbers, perhaps considering their divisibility by certain numbers would help.Let me think about the numbers from 1 to 12. If I take any 12 consecutive numbers, they will cover all residues modulo 12. That means in any such set, there must be a multiple of 12. So, one of the numbers is divisible by 12. Let's denote this number as 12m, where m is some positive integer.Now, let's consider the proper divisors of 12m. The proper divisors would include all the divisors except 1 and 12m. For 12m, the divisors are 1, 2, 3, 4, 6, 12, m, 2m, 3m, 4m, 6m, 12m. So, excluding 1 and 12m, the proper divisors are 2, 3, 4, 6, m, 2m, 3m, 4m, 6m.Wait, that seems a bit complicated. Maybe I should simplify. Since 12m is a multiple of 12, it's also a multiple of 6, 4, 3, and 2. So, 12m has several divisors. The proper divisors would include 2m, 3m, 4m, 6m, and so on. Let me list them out:- 2m- 3m- 4m- 6mAre there more? Well, if m has its own divisors, those would also be divisors of 12m. But since we're only considering proper divisors, we don't include 1 and 12m. So, for simplicity, let's focus on the divisors that come directly from the factors of 12.So, the sum of these proper divisors would be 2m + 3m + 4m + 6m. Let's calculate that:2m + 3m = 5m5m + 4m = 9m9m + 6m = 15mSo, the sum of these proper divisors is 15m. Now, let's compare this to the number itself, which is 12m. Clearly, 15m is greater than 12m because 15m - 12m = 3m, which is positive as long as m is a positive integer.Therefore, the sum of the proper divisors of 12m is greater than 12m itself. This means that 12m is an abundant number. Since in any set of 12 consecutive numbers, there is always a multiple of 12, and that multiple is abundant, it follows that there is at least one abundant number in any such set.Wait, but is this always true? What if m is 1? Then 12m is 12, and the proper divisors are 2, 3, 4, and 6. Their sum is 2 + 3 + 4 + 6 = 15, which is indeed greater than 12. So, 12 is abundant.What if m is 2? Then 12m is 24. The proper divisors of 24 are 2, 3, 4, 6, 8, 12. Their sum is 2 + 3 + 4 + 6 + 8 + 12 = 35, which is greater than 24. So, 24 is also abundant.What about m = 3? 12m = 36. Proper divisors are 2, 3, 4, 6, 9, 12, 18. Sum is 2 + 3 + 4 + 6 + 9 + 12 + 18 = 54, which is greater than 36. So, 36 is abundant.It seems consistent. So, regardless of the value of m, 12m is always abundant because the sum of its proper divisors is 15m, which is greater than 12m.Therefore, in any set of 12 consecutive numbers, there must be a multiple of 12, which is abundant. Hence, there is at least one number in the set that is smaller than the sum of its proper divisors.I think this makes sense. By ensuring that one of the numbers is a multiple of 12, and knowing that such a number is abundant, we can conclude that the condition holds for any 12 consecutive integers.

💡First, I need to convert the number 120,000 into scientific notation. Scientific notation requires the number to be expressed in the form ( a times 10^{n} ), where ( a ) is a number between 1 and 10, and ( n ) is an integer.To achieve this, I will move the decimal point in 120,000 to the left until it is just after the first non-zero digit. Moving the decimal point five places to the left gives me 1.2.Next, I count the number of places the decimal point has been moved, which is 5 in this case. This count becomes the exponent of 10.Therefore, the number 120,000 can be expressed in scientific notation as ( 1.2 times 10^{5} ).Finally, I compare this result with the provided options and find that option C matches the scientific notation I derived.

💡Okay, so I have this problem about complex numbers α, β, and γ. The condition given is that for every natural number k, the sum α^k + β^k + γ^k is an integer. I need to prove that the polynomial (x - α)(x - β)(x - γ) has integer coefficients. Hmm, interesting.First, let me recall that if a polynomial has integer coefficients, then its roots must satisfy certain conditions. Specifically, the sums of the roots, the sums of the products of the roots taken two at a time, and the product of all three roots must all be integers. So, if I can show that α + β + γ, αβ + βγ + γα, and αβγ are all integers, then the polynomial will indeed have integer coefficients.Alright, so let me denote S_k = α^k + β^k + γ^k. The problem states that S_k is an integer for every natural number k. I need to use this to show that the elementary symmetric sums are integers.Let me start with k = 1. Then S_1 = α + β + γ. So, S_1 is an integer. That takes care of the first symmetric sum.Next, let's consider k = 2. Then S_2 = α^2 + β^2 + γ^2. I know that S_2 can be expressed in terms of S_1 and the second symmetric sum. Specifically, S_2 = (α + β + γ)^2 - 2(αβ + βγ + γα). Since S_2 is an integer and S_1 is an integer, it follows that 2(αβ + βγ + γα) must also be an integer. Therefore, αβ + βγ + γα is either an integer or a half-integer. Hmm, so it's either integer or half-integer.Wait, but let's see. If αβ + βγ + γα is a half-integer, that might complicate things. Maybe I can get more information by considering higher k.Let's try k = 3. Then S_3 = α^3 + β^3 + γ^3. I remember that there's an identity for the sum of cubes: α^3 + β^3 + γ^3 = (α + β + γ)^3 - 3(α + β + γ)(αβ + βγ + γα) + 3αβγ. So, plugging in S_1, S_2, and S_3, we can write:S_3 = S_1^3 - 3S_1(αβ + βγ + γα) + 3αβγ.We know S_3 is an integer, S_1 is an integer, and we already have that 2(αβ + βγ + γα) is an integer. Let me denote P = αβ + βγ + γα and Q = αβγ. Then the equation becomes:S_3 = S_1^3 - 3S_1 P + 3Q.Since S_3 and S_1 are integers, and 3S_1 P is a multiple of 3 times an integer or half-integer, depending on P. But since 2P is an integer, P is either integer or half-integer. Let me write P = m/2 where m is an integer. Then 3S_1 P = 3S_1 (m/2) = (3S_1 m)/2. For this term to be such that when subtracted from S_1^3 and added to 3Q, the result is integer, we need (3S_1 m)/2 to be either integer or half-integer. But S_1 is integer, so 3S_1 m is integer, so (3S_1 m)/2 is either integer or half-integer.But S_3 is integer, so 3Q must compensate for any half-integer parts. Let's see:S_3 = integer - (3S_1 m)/2 + 3Q.So, if (3S_1 m)/2 is integer, then 3Q must be integer, so Q is rational with denominator dividing 3. If (3S_1 m)/2 is half-integer, then 3Q must be half-integer as well, so Q is rational with denominator dividing 6.But I need to show that Q is integer. Hmm, maybe I need more information.Let me consider k = 4. Then S_4 = α^4 + β^4 + γ^4. I can express S_4 in terms of S_1, S_2, S_3, and S_4. Wait, actually, using Newton's identities, which relate power sums to elementary symmetric sums.Newton's identities state that:S_1 = e_1,S_2 = e_1 S_1 - 2e_2,S_3 = e_1 S_2 - e_2 S_1 + 3e_3,S_4 = e_1 S_3 - e_2 S_2 + e_3 S_1,and so on.In our case, e_1 = S_1, e_2 = P, e_3 = Q.So, let's write down the expressions:S_1 = e_1,S_2 = e_1 S_1 - 2e_2,S_3 = e_1 S_2 - e_2 S_1 + 3e_3,S_4 = e_1 S_3 - e_2 S_2 + e_3 S_1.We know S_1, S_2, S_3, S_4 are all integers.From S_2: S_2 = e_1^2 - 2e_2. Since S_2 is integer and e_1 is integer, 2e_2 must be integer, so e_2 is either integer or half-integer.From S_3: S_3 = e_1 S_2 - e_2 e_1 + 3e_3.Plugging in e_1 = S_1, e_2 = P, e_3 = Q:S_3 = S_1 S_2 - P S_1 + 3Q.We already have that 2P is integer, so P = m/2, m integer.Then, S_3 = S_1 S_2 - (m/2) S_1 + 3Q.Since S_3 is integer, and S_1 S_2 is integer, and (m/2) S_1 is either integer or half-integer, 3Q must compensate for any fractional part.If (m/2) S_1 is integer, then 3Q must be integer, so Q is rational with denominator dividing 3.If (m/2) S_1 is half-integer, then 3Q must be half-integer, so Q is rational with denominator dividing 6.But we need to show Q is integer. Maybe considering S_4 will help.From S_4: S_4 = e_1 S_3 - e_2 S_2 + e_3 S_1.Plugging in e_1 = S_1, e_2 = P, e_3 = Q:S_4 = S_1 S_3 - P S_2 + Q S_1.Again, S_4 is integer, S_1 S_3 is integer, P S_2 is (m/2) S_2, which is either integer or half-integer, and Q S_1 is Q times integer.So, similar to before, the term P S_2 is either integer or half-integer, so Q S_1 must compensate to make the entire expression integer.If P S_2 is integer, then Q S_1 must be integer. Since S_1 is integer, Q must be rational with denominator dividing S_1. But S_1 is integer, so Q must be rational with denominator dividing 1, i.e., integer.If P S_2 is half-integer, then Q S_1 must be half-integer. So, Q must be rational with denominator dividing 2. But from earlier, Q is either denominator 3 or 6. Hmm, this is getting a bit tangled.Wait, maybe instead of going step by step, I can consider that since all S_k are integers, the generating function for S_k must be a rational function with integer coefficients. The generating function is G(t) = Σ_{k=1}^∞ S_k t^{k-1}.But I'm not sure if that's the right approach. Maybe another way is to consider that the minimal polynomial of α, β, γ must have integer coefficients, but I'm not sure.Alternatively, perhaps I can use induction. Suppose that for all k ≤ n, S_k is integer, then show that S_{n+1} is integer. But I'm not sure how that helps in proving the coefficients are integers.Wait, another idea: if α, β, γ are roots of a monic polynomial with integer coefficients, then they must be algebraic integers. But the problem doesn't state that α, β, γ are algebraic, just that their power sums are integers. Hmm.But maybe I can use the fact that if the power sums are integers, then the elementary symmetric sums must be rational numbers, and with some conditions, they must be integers.Wait, from S_1, S_2, S_3, we can express e_1, e_2, e_3 in terms of S_1, S_2, S_3.From S_1 = e_1,S_2 = e_1^2 - 2e_2 ⇒ e_2 = (e_1^2 - S_2)/2,S_3 = e_1 S_2 - e_2 e_1 + 3e_3 ⇒ e_3 = (S_3 - e_1 S_2 + e_2 e_1)/3.Since e_1 is integer, e_2 is (integer^2 - integer)/2, so e_2 is either integer or half-integer.Similarly, e_3 is (integer - integer + (integer or half-integer)*integer)/3.So, e_3 is either integer or something like (integer + half-integer)/3, which could be a fraction with denominator 6.But we need to show e_3 is integer.Wait, maybe considering higher k will help us get more constraints on e_2 and e_3.Let me compute S_4 using Newton's identities:S_4 = e_1 S_3 - e_2 S_2 + e_3 S_1.We know S_4 is integer, e_1 is integer, S_3 is integer, e_2 is (integer or half-integer), S_2 is integer, e_3 is (integer or fraction), S_1 is integer.So, S_4 = integer * integer - (integer or half-integer) * integer + (integer or fraction) * integer.So, the first term is integer, the second term is integer or half-integer, the third term is integer or fraction.Thus, S_4 must be integer, so the sum of these terms must be integer. Therefore, any fractional parts must cancel out.If e_2 is integer, then the second term is integer, and e_3 must be such that the third term is integer, so e_3 must be integer.If e_2 is half-integer, then the second term is half-integer, so the third term must be half-integer as well to make the total sum integer. Therefore, e_3 must be half-integer divided by integer, which would be a fraction with denominator 2.But wait, e_3 is (S_3 - e_1 S_2 + e_2 e_1)/3. If e_2 is half-integer, then e_2 e_1 is half-integer, so S_3 - e_1 S_2 + e_2 e_1 is integer - integer + half-integer = half-integer. Therefore, e_3 = half-integer / 3, which is a fraction with denominator 6.But then, in S_4, e_3 S_1 would be (fraction with denominator 6) * integer, which could be a fraction with denominator 6. But S_4 is integer, so the sum must cancel out any fractions. This seems complicated.Maybe I need to consider more terms or find another approach.Wait, another idea: if the power sums S_k are integers for all k, then the generating function G(t) = Σ_{k=1}^∞ S_k t^{k-1} must be a rational function with integer coefficients. Because the generating function for the power sums of roots of a polynomial is related to the polynomial itself.Specifically, if the polynomial is x^3 - e_1 x^2 + e_2 x - e_3, then the generating function G(t) can be expressed as (e_1 - e_2 t + e_3 t^2)/(1 - e_1 t + e_2 t^2 - e_3 t^3). Wait, is that right?Let me recall that for a polynomial P(x) = (x - α)(x - β)(x - γ), the generating function for the power sums is G(t) = Σ_{k=1}^∞ S_k t^{k-1} = P'(t)/P(t). Wait, no, actually, it's related to the derivative.Wait, more accurately, the generating function for the power sums is G(t) = Σ_{k=1}^∞ S_k t^{k} = (α t)/(1 - α t) + (β t)/(1 - β t) + (γ t)/(1 - γ t). Hmm, that might be more complicated.Alternatively, using the fact that the power sums satisfy a linear recurrence relation with coefficients given by the elementary symmetric sums. Specifically, for a cubic polynomial, the power sums satisfy S_k = e_1 S_{k-1} - e_2 S_{k-2} + e_3 S_{k-3} for k ≥ 4.Since all S_k are integers, and the recurrence has coefficients e_1, e_2, e_3, which are rational numbers (since e_1 is integer, e_2 is integer or half-integer, e_3 is more complicated), but the recurrence must produce integers for all k.This suggests that the coefficients e_1, e_2, e_3 must be such that the recurrence preserves integrality. For that, e_1, e_2, e_3 must be integers. Because if any of them were non-integers, the recurrence could introduce fractions.Wait, let me think about that. Suppose e_1 is integer, e_2 is half-integer, e_3 is something. Then, starting from S_1, S_2, S_3 integers, S_4 = e_1 S_3 - e_2 S_2 + e_3 S_1.If e_2 is half-integer, then e_2 S_2 is half-integer, and e_3 S_1 could be something. For S_4 to be integer, the sum must cancel out the half-integer parts.But then, S_5 = e_1 S_4 - e_2 S_3 + e_3 S_2.Again, e_2 S_3 is half-integer, and e_3 S_2 could be something. For S_5 to be integer, again, the fractions must cancel.This seems to impose that e_2 must actually be integer, because otherwise, the half-integer terms would keep appearing, and unless e_3 is chosen to cancel them, which would require e_3 to have specific fractional parts, but then e_3 would have to be rational with denominator dividing some power of 2, but we also have S_3 involving e_3, which might conflict.Alternatively, maybe e_2 must be integer to avoid introducing fractions in the recurrence.Wait, let's suppose e_2 is half-integer. Then, S_4 = e_1 S_3 - (m/2) S_2 + e_3 S_1.For S_4 to be integer, (m/2) S_2 must be canceled by e_3 S_1. So, e_3 S_1 must be equal to (m/2) S_2 + integer.But e_3 is (S_3 - e_1 S_2 + e_2 e_1)/3.If e_2 is m/2, then e_3 = (S_3 - e_1 S_2 + (m/2) e_1)/3.So, e_3 is (integer - integer + (m/2) integer)/3.Which is (integer + (m/2) integer)/3.So, e_3 is (integer + half-integer)/3, which is a fraction with denominator 6.Then, e_3 S_1 is (fraction with denominator 6) * integer, which is a fraction with denominator 6.But in S_4, we have S_4 = integer - (m/2) S_2 + e_3 S_1.So, S_4 = integer - half-integer + fraction with denominator 6.To get S_4 integer, the sum of -half-integer + fraction with denominator 6 must be integer.But -1/2 + 1/6 = -1/3, which is not integer. So, unless m is even, making (m/2) integer, then e_2 is integer.Wait, if m is even, then e_2 is integer. So, perhaps m must be even, which would make e_2 integer.Wait, let me think. If e_2 is m/2, and m is even, say m = 2n, then e_2 = n, which is integer.So, perhaps e_2 must be integer to avoid fractions in the recurrence.Therefore, e_2 must be integer.Similarly, from e_3 = (S_3 - e_1 S_2 + e_2 e_1)/3.Since e_1, e_2, S_3, S_2 are all integers, e_3 must be integer as well.Therefore, all elementary symmetric sums e_1, e_2, e_3 are integers.Hence, the polynomial (x - α)(x - β)(x - γ) = x^3 - e_1 x^2 + e_2 x - e_3 has integer coefficients.So, I think that's the way to go. By using the recurrence relation from Newton's identities and the fact that all power sums are integers, we can conclude that the elementary symmetric sums must be integers, hence the polynomial has integer coefficients.**Final Answer**boxed{(x - alpha)(x - beta)(x - gamma) text{ has integer coefficients.}}

💡To solve the problem, we start by noting that (C - D), (C), (D), and (C + D) are all prime numbers. 1. **Identifying Even Primes**: - Since (C - D) and (C + D) are primes and not even (except for 2), they must be odd. This implies that one of (C) or (D) must be even. The only even prime number is 2. Therefore, (D = 2).2. **Determining (C)**: - With (D = 2), we have (C - 2), (C), and (C + 2) as primes. These are three consecutive odd primes. - The smallest set of three consecutive odd primes is 3, 5, and 7. Thus, (C = 5).3. **Listing the Primes**: - The primes are (C - D = 3), (C = 5), (C + D = 7), and (D = 2).4. **Calculating the Product**: - The product of these primes is (3 times 5 times 7 times 2). - Calculating step-by-step: - (3 times 5 = 15) - (15 times 7 = 105) - (105 times 2 = 210)5. **Analyzing the Product**: - The product 210 is divisible by 7.Thus, the product of the four primes is divisible by 7.(boxed{text{D}})