answer：Alright, so I've got this problem here about a school library buying extracurricular books for four different grades. The problem gives me some information about how many books are not for each grade, and I need to figure out how many books were bought for each grade individually. Let me try to unpack this step by step.First, let's list out what we know:1. 110 books are not for the first grade.2. 108 books are not for the second grade.3. 104 books are not for the third grade.4. 119 books are not for the fourth grade.And we need to find the number of books bought for each grade: first, second, third, and fourth. Let's denote the number of books for each grade as follows:- Let ( a ) be the number of books for the first grade.- Let ( b ) be the number of books for the second grade.- Let ( c ) be the number of books for the third grade.- Let ( d ) be the number of books for the fourth grade.Now, the problem states how many books are not for each grade. So, for example, if 110 books are not for the first grade, that means the total number of books for the other three grades (second, third, and fourth) is 110. Similarly, for the other grades.So, we can write the following equations based on the given information:1. ( b + c + d = 110 ) (since these are the books not for the first grade)2. ( a + c + d = 108 ) (books not for the second grade)3. ( a + b + d = 104 ) (books not for the third grade)4. ( a + b + c = 119 ) (books not for the fourth grade)So, now we have four equations with four variables. The goal is to solve for ( a, b, c, ) and ( d ).One approach to solving this system of equations is to add all four equations together. Let's see what that gives us.Adding the left sides:( (b + c + d) + (a + c + d) + (a + b + d) + (a + b + c) )Simplify this:- ( b + c + d ) has one ( b ), one ( c ), and one ( d )- ( a + c + d ) has one ( a ), one ( c ), and one ( d )- ( a + b + d ) has one ( a ), one ( b ), and one ( d )- ( a + b + c ) has one ( a ), one ( b ), and one ( c )Adding them together, we get:- ( a ) appears three times- ( b ) appears three times- ( c ) appears three times- ( d ) appears three timesSo, the sum is ( 3a + 3b + 3c + 3d ), which can be factored as ( 3(a + b + c + d) ).Now, adding the right sides:( 110 + 108 + 104 + 119 )Let's compute that:- ( 110 + 108 = 218 )- ( 218 + 104 = 322 )- ( 322 + 119 = 441 )So, we have:( 3(a + b + c + d) = 441 )Divide both sides by 3:( a + b + c + d = 147 )Okay, so the total number of books bought is 147.Now, we can use this total to find each individual grade's book count by subtracting the given "not for" numbers from the total.Let's start with the first grade. We know that ( b + c + d = 110 ). Since the total is 147, subtracting 110 should give us ( a ):( a = 147 - 110 = 37 )So, there are 37 books for the first grade.Next, for the second grade. We know that ( a + c + d = 108 ). Subtracting this from the total gives:( b = 147 - 108 = 39 )So, there are 39 books for the second grade.Moving on to the third grade. We have ( a + b + d = 104 ). Subtracting this from the total:( c = 147 - 104 = 43 )Therefore, there are 43 books for the third grade.Finally, for the fourth grade. We know that ( a + b + c = 119 ). Subtracting this from the total:( d = 147 - 119 = 28 )So, there are 28 books for the fourth grade.Let me just double-check these numbers to make sure everything adds up correctly.Total books should be ( a + b + c + d = 37 + 39 + 43 + 28 ).Calculating:- ( 37 + 39 = 76 )- ( 76 + 43 = 119 )- ( 119 + 28 = 147 )Yes, that matches the total we found earlier.Also, let's verify each of the original conditions:1. Books not for first grade: ( b + c + d = 39 + 43 + 28 = 110 ) ✔️2. Books not for second grade: ( a + c + d = 37 + 43 + 28 = 108 ) ✔️3. Books not for third grade: ( a + b + d = 37 + 39 + 28 = 104 ) ✔️4. Books not for fourth grade: ( a + b + c = 37 + 39 + 43 = 119 ) ✔️All conditions are satisfied, so the numbers seem correct.I think that's it! So, the number of extracurricular books bought for each grade is 37 for the first grade, 39 for the second grade, 43 for the third grade, and 28 for the fourth grade.**Final Answer**The number of extracurricular books bought for the first grade, second grade, third grade, and fourth grade are respectively: boxed{37}, boxed{39}, boxed{43}, boxed{28} books.

answer：To find the greatest common factor (GCF) of 60, 90, and 150, I will start by performing the prime factorization of each number.First, for 60:- 60 divided by 2 is 30.- 30 divided by 2 is 15.- 15 divided by 3 is 5.- 5 divided by 5 is 1.So, the prime factors of 60 are (2^2 times 3 times 5).Next, for 90:- 90 divided by 2 is 45.- 45 divided by 3 is 15.- 15 divided by 3 is 5.- 5 divided by 5 is 1.So, the prime factors of 90 are (2 times 3^2 times 5).Then, for 150:- 150 divided by 2 is 75.- 75 divided by 3 is 25.- 25 divided by 5 is 5.- 5 divided by 5 is 1.So, the prime factors of 150 are (2 times 3 times 5^2).Now, I will identify the common prime factors across all three numbers:- The common prime factors are 2, 3, and 5.To determine the GCF, I will take the lowest power of each common prime factor:- The lowest power of 2 is (2^1).- The lowest power of 3 is (3^1).- The lowest power of 5 is (5^1).Finally, I will multiply these together to find the GCF:[ 2^1 times 3^1 times 5^1 = 2 times 3 times 5 = 30 ]Therefore, the greatest common factor of 60, 90, and 150 is 30.

answer：First, calculate the total distance driven in a month. The one-way commute is 21 miles, so the round trip is 42 miles. Over 5 days a week, the weekly distance is 42 miles/day multiplied by 5 days, which equals 210 miles per week. For 4 weeks in a month, the total monthly distance is 210 miles/week multiplied by 4 weeks, resulting in 840 miles/month.Next, determine the total monthly contribution for gas. Each of the five friends pays 14/month, so the total contribution is 5 friends multiplied by 14, totaling 70/month.Then, calculate the number of gallons of gas purchased with the 70. Given that gas costs 2.50 per gallon, divide 70 by 2.50/gallon to get 28 gallons.Finally, find the miles per gallon (mpg) by dividing the total miles driven in a month by the number of gallons of gas used. So, 840 miles/month divided by 28 gallons equals 30 mpg.

answer：Okay, so I have this problem where I need to find all four-digit positive integers such that the sum of the squares of the digits equals twice the sum of the digits. Hmm, let me try to understand what this means.First, let's break down the problem. A four-digit number can be represented as ( overline{a_3a_2a_1a_0} ), where ( a_3 ) is the thousands digit, ( a_2 ) is the hundreds digit, ( a_1 ) is the tens digit, and ( a_0 ) is the units digit. Each of these digits can range from 0 to 9, except ( a_3 ), which can't be 0 because it's a four-digit number.The equation given is:[sum_{i=0}^3 a_i^2 = 2 sum_{i=0}^3 a_i]So, the sum of the squares of each digit equals twice the sum of the digits themselves. Let me rewrite this equation to see if I can simplify it or find a pattern.Subtracting ( 2 sum a_i ) from both sides, I get:[sum_{i=0}^3 (a_i^2 - 2a_i) = 0]Hmm, maybe I can complete the square for each term ( a_i^2 - 2a_i ). Completing the square for a quadratic expression ( x^2 - 2x ) gives ( (x - 1)^2 - 1 ). So applying this to each term:[sum_{i=0}^3 ((a_i - 1)^2 - 1) = 0]Which simplifies to:[sum_{i=0}^3 (a_i - 1)^2 - 4 = 0]So,[sum_{i=0}^3 (a_i - 1)^2 = 4]Alright, so the sum of the squares of each digit minus one equals 4. That seems more manageable. Now, each term ( (a_i - 1)^2 ) is a non-negative integer because squares are always non-negative. So, I need to find all combinations of ( a_3, a_2, a_1, a_0 ) such that when each is decreased by 1 and squared, their sum is 4.Let me think about the possible values each ( (a_i - 1)^2 ) can take. Since each ( a_i ) is a digit from 0 to 9, ( a_i - 1 ) can range from -1 to 8, so ( (a_i - 1)^2 ) can be 0, 1, 4, 9, 16, 25, 36, 49, 64, or 81. But since the total sum is 4, each individual term can't be larger than 4. So, the possible values for each ( (a_i - 1)^2 ) are 0, 1, or 4.Therefore, each ( (a_i - 1)^2 ) can be 0, 1, or 4. Let's consider the different ways to get a sum of 4 from four such terms.Case 1: One of the terms is 4, and the rest are 0.Case 2: Four of the terms are 1, and the rest are 0.Case 3: Two of the terms are 1, and one term is 2, but wait, 2 isn't a possible value because ( (a_i - 1)^2 ) can only be 0, 1, or 4. So, scratch that.Wait, actually, since each term can only be 0, 1, or 4, the only possible combinations are:- One 4 and three 0s.- Four 1s and zero 4s.Because 4 can be achieved by either one 4 and the rest 0s or four 1s. Let me check if there are other possibilities.If I have two 2s, but since 2 isn't a possible value, that's not allowed. Similarly, any other combination would involve numbers larger than 4, which we can't have. So, only the two cases above.Let me analyze each case.**Case 1: One term is 4, the rest are 0.**This means that for one digit, ( (a_i - 1)^2 = 4 ), so ( a_i - 1 = pm 2 ), which gives ( a_i = 3 ) or ( a_i = -1 ). But since digits can't be negative, ( a_i = 3 ).The other digits must satisfy ( (a_j - 1)^2 = 0 ), so ( a_j = 1 ) for ( j neq i ).So, in this case, one digit is 3, and the other three digits are 1. However, we have to remember that ( a_3 ) cannot be 0, but in this case, it's either 1 or 3, so it's fine.Now, how many such numbers are there? It depends on which digit is 3. Since it's a four-digit number, the thousands digit ( a_3 ) can be 3 or 1. If ( a_3 = 3 ), then the other three digits are 1. If ( a_3 = 1 ), then one of the other digits is 3, and the rest are 1.Wait, actually, in this case, only one digit is 3, and the rest are 1. So, the number can have the 3 in any of the four positions. However, since it's a four-digit number, the thousands digit can be 3 or 1. So, if 3 is in the thousands place, we have 3111. If 3 is in the hundreds, tens, or units place, we have 1311, 1131, or 1113.So, that's four numbers: 3111, 1311, 1131, 1113.Wait, but earlier I thought only three numbers because I considered only when ( a_3 = 1 ). But actually, when ( a_3 = 3 ), it's also a valid number. So, total four numbers in this case.Wait, but hold on, let me verify. If ( a_3 = 3 ), then the other digits are 1, so the number is 3111. If ( a_3 = 1 ), then one of the other digits is 3, so the numbers are 1311, 1131, 1113. So, that's four numbers in total.**Case 2: Four terms are 1, and zero terms are 4.**This means that each ( (a_i - 1)^2 = 1 ), so ( a_i - 1 = pm 1 ), which gives ( a_i = 2 ) or ( a_i = 0 ).So, each digit is either 0 or 2. But again, ( a_3 ) can't be 0, so ( a_3 ) must be 2. The other digits ( a_2, a_1, a_0 ) can be either 0 or 2.So, how many such numbers are there? Since ( a_3 = 2 ), and each of the other three digits can be 0 or 2, that's ( 2^3 = 8 ) possibilities.Let me list them:- 2000- 2002- 2020- 2022- 2200- 2202- 2220- 2222So, eight numbers in this case.Wait, but hold on, let me check if these numbers satisfy the original equation.Take 2000:Sum of squares: ( 2^2 + 0^2 + 0^2 + 0^2 = 4 )Twice the sum: ( 2*(2 + 0 + 0 + 0) = 4 )Yes, 4 = 4, so it works.Take 2022:Sum of squares: ( 2^2 + 0^2 + 2^2 + 2^2 = 4 + 0 + 4 + 4 = 12 )Twice the sum: ( 2*(2 + 0 + 2 + 2) = 2*6 = 12 )Yes, 12 = 12, so it works.Take 2222:Sum of squares: ( 2^2 + 2^2 + 2^2 + 2^2 = 16 )Twice the sum: ( 2*(2 + 2 + 2 + 2) = 16 )Yes, 16 = 16, so it works.Okay, so all eight numbers in this case satisfy the condition.Wait, but earlier in Case 1, I had four numbers: 3111, 1311, 1131, 1113. Let me check one of them.Take 3111:Sum of squares: ( 3^2 + 1^2 + 1^2 + 1^2 = 9 + 1 + 1 + 1 = 12 )Twice the sum: ( 2*(3 + 1 + 1 + 1) = 2*6 = 12 )Yes, 12 = 12, so it works.Take 1311:Sum of squares: ( 1^2 + 3^2 + 1^2 + 1^2 = 1 + 9 + 1 + 1 = 12 )Twice the sum: ( 2*(1 + 3 + 1 + 1) = 2*6 = 12 )Yes, 12 = 12, so it works.Okay, so all four numbers in Case 1 also satisfy the condition.Wait, but earlier I thought only three numbers in Case 1, but actually, it's four because 3111 is also valid.So, in total, we have 4 (from Case 1) + 8 (from Case 2) = 12 numbers.But hold on, let me make sure I didn't miss any other cases.Earlier, I considered two cases: one where one of the ( (a_i - 1)^2 ) is 4 and the rest are 0, and another where all four ( (a_i - 1)^2 ) are 1. But wait, could there be a case where two of the ( (a_i - 1)^2 ) are 2 and the rest are something else? But no, because ( (a_i - 1)^2 ) can only be 0, 1, or 4 as established earlier. So, 2 isn't a possible value, so that case isn't possible.Alternatively, could we have two 2s and two 0s? But again, since ( (a_i - 1)^2 ) can't be 2, that's not possible.Wait, another thought: what if one of the ( (a_i - 1)^2 ) is 4 and another is 1, and the rest are 0? Then the total sum would be 5, which is more than 4, so that's not allowed.Similarly, if two of them are 1 and one is 2, but again, 2 isn't allowed. So, no, that's not possible.Therefore, the only possible cases are the two I considered: one 4 and three 0s, or four 1s.So, in total, 4 + 8 = 12 four-digit numbers satisfy the given condition.Wait, but let me double-check if there are any other possibilities. For example, could we have one digit where ( (a_i - 1)^2 = 4 ) and another where ( (a_j - 1)^2 = 0 ), but that would still fall under Case 1, right? Because the sum would still be 4.Wait, no, if we have one 4 and one 0, the sum would be 4 + 0 = 4, but we have four digits, so the other two digits would also have to contribute 0 each. So, that's still covered in Case 1.Similarly, if we have four 1s, that's covered in Case 2.So, I think I've covered all possible cases.Therefore, the total number of four-digit numbers satisfying the condition is 12.But just to be thorough, let me list all the numbers I found:From Case 1:- 3111- 1311- 1131- 1113From Case 2:- 2000- 2002- 2020- 2022- 2200- 2202- 2220- 2222That's 4 + 8 = 12 numbers. Let me count them:1. 31112. 13113. 11314. 11135. 20006. 20027. 20208. 20229. 220010. 220211. 222012. 2222Yes, that's 12 numbers. I think that's all of them.Wait, but hold on, let me check if 3000 is a valid number.Sum of squares: ( 3^2 + 0^2 + 0^2 + 0^2 = 9 )Twice the sum: ( 2*(3 + 0 + 0 + 0) = 6 )9 ≠ 6, so 3000 doesn't satisfy the condition. So, it's not included in our list, which is correct.Similarly, let me check 1111:Sum of squares: ( 1 + 1 + 1 + 1 = 4 )Twice the sum: ( 2*(1 + 1 + 1 + 1) = 8 )4 ≠ 8, so 1111 doesn't satisfy the condition.Okay, so our list seems correct.Therefore, the final answer is 12 four-digit numbers.

answer：Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. It says that the sum of the first n terms is Sₙ, and specifically, S₃ is 6. Then, I need to find the value of 5a₁ + a₇. The options are A: 12, B: 10, C: 24, D: 6.Hmm, arithmetic sequences. I remember that in an arithmetic sequence, each term increases by a common difference, d. So, the nth term can be written as aₙ = a₁ + (n-1)d, where a₁ is the first term.The sum of the first n terms, Sₙ, is given by the formula Sₙ = n/2 * (2a₁ + (n-1)d). Alternatively, it can also be written as Sₙ = n(a₁ + aₙ)/2. Both formulas are equivalent.Given that S₃ = 6, let's plug n = 3 into the sum formula. So, S₃ = 3/2 * (2a₁ + 2d) = 6. Let me write that out:S₃ = (3/2)(2a₁ + 2d) = 6.Simplify that equation. First, multiply both sides by 2 to eliminate the fraction:3*(2a₁ + 2d) = 12.Then, divide both sides by 3:2a₁ + 2d = 4.I can factor out a 2:2(a₁ + d) = 4.Divide both sides by 2:a₁ + d = 2.Okay, so from this, I know that a₁ + d = 2. That's one equation, but I have two variables here: a₁ and d. So, I might need another equation to solve for both variables, but the problem doesn't give me more information directly. Hmm.Wait, the question is asking for 5a₁ + a₇. Maybe I can express a₇ in terms of a₁ and d and then see if I can find 5a₁ + a₇ using the information I have.Let's recall that a₇ = a₁ + 6d, because each term increases by d. So, a₇ = a₁ + 6d.Therefore, 5a₁ + a₇ = 5a₁ + (a₁ + 6d) = 6a₁ + 6d.Hmm, that simplifies to 6(a₁ + d). Wait a minute, from earlier, I found that a₁ + d = 2. So, substituting that in:6(a₁ + d) = 6*2 = 12.Oh, so 5a₁ + a₇ equals 12. That's option A.Wait, let me double-check to make sure I didn't make any mistakes. So, starting from S₃ = 6:S₃ = (3/2)(2a₁ + 2d) = 6.Multiplying both sides by 2: 3*(2a₁ + 2d) = 12.Divide by 3: 2a₁ + 2d = 4.Factor: 2(a₁ + d) = 4.Divide by 2: a₁ + d = 2.Then, 5a₁ + a₇ = 5a₁ + (a₁ + 6d) = 6a₁ + 6d = 6(a₁ + d) = 6*2 = 12.Yes, that seems correct. So, the answer should be 12, which is option A.I don't think I made any calculation errors here. It all seems to follow logically. I used the formula for the sum of an arithmetic sequence, simplified it to find a relationship between a₁ and d, and then used that relationship to find the value of 5a₁ + a₇. Everything checks out.**Final Answer**boxed{A}

answer：Okay, so I'm trying to figure out the right sampling methods for these two investigations. Let me start by understanding what each investigation entails.First, there's this company with sales points spread across four regions: A, B, C, and D. The numbers are 150, 120, 180, and 150 respectively. So, in total, that's 600 sales points. They want to investigate product sales by taking a sample of 100 from these 600. That's investigation (1).Then, in region C, which has 180 sales points, there are 20 very large ones. They want to sample 7 of these to look into sales revenue and after-sales service. That's investigation (2).Alright, so for investigation (1), I need to choose a sampling method that allows the company to get a representative sample from all four regions. Since the regions have different numbers of sales points, a simple random sample might not give a proportional representation from each region. That makes me think of stratified sampling. In stratified sampling, the population is divided into subgroups (strata) and samples are taken from each stratum. This way, each region is adequately represented in the sample.For investigation (2), it's specifically about the 20 very large sales points in region C. They want to sample 7 out of these 20. Since these are distinct and specifically identified, a simple random sampling method would work here. In simple random sampling, every individual in the population has an equal chance of being selected, ensuring fairness and reducing bias.Wait, but could there be another method? Maybe systematic sampling? In systematic sampling, you select every nth individual from a list. But for investigation (2), since it's a small, specific group, simple random sampling seems more straightforward and appropriate. Systematic sampling might not be as reliable if there's an underlying pattern in the list.So, putting it all together, investigation (1) should use stratified sampling to ensure proportional representation from each region, and investigation (2) should use simple random sampling to fairly select 7 out of the 20 large sales points in region C.