answer：Okay, so I have this system of inequalities to solve, and I need to find the range of the real number ( a ) such that there are exactly three integer solutions for ( x ). The system is:[begin{cases}x + 6 < 2 + 3x frac{a + x}{4} > xend{cases}]Hmm, let me start by solving each inequality separately to find the range of ( x ).Starting with the first inequality: ( x + 6 < 2 + 3x ).I can subtract ( x ) from both sides to get:( 6 < 2 + 2x )Then, subtract 2 from both sides:( 4 < 2x )Divide both sides by 2:( 2 < x )So, the first inequality simplifies to ( x > 2 ).Now, moving on to the second inequality: ( frac{a + x}{4} > x ).I can multiply both sides by 4 to eliminate the denominator:( a + x > 4x )Subtract ( x ) from both sides:( a > 3x )So, this simplifies to ( x < frac{a}{3} ).Putting both inequalities together, I have:( 2 < x < frac{a}{3} )Now, I need this system to have exactly three integer solutions. That means there should be exactly three integers ( x ) that satisfy both inequalities.Since ( x ) has to be greater than 2, the possible integer values for ( x ) start from 3. So, the integers are 3, 4, 5, and so on.But we need exactly three integer solutions. So, the integer values of ( x ) should be 3, 4, and 5. That means the upper bound ( frac{a}{3} ) must be greater than 5 but less than or equal to 6. Because if ( frac{a}{3} ) is greater than 6, then 6 would also be a solution, making it four integer solutions, which is more than we need.So, setting up the inequality:( 5 < frac{a}{3} leq 6 )Multiplying all parts by 3 to solve for ( a ):( 15 < a leq 18 )Therefore, the range of ( a ) is ( 15 < a leq 18 ).Wait, let me double-check. If ( a = 15 ), then ( frac{a}{3} = 5 ). So, the inequality becomes ( 2 < x < 5 ). The integer solutions would be 3, 4. That's only two solutions, which is less than three. So, ( a ) must be greater than 15.If ( a = 18 ), then ( frac{a}{3} = 6 ). The inequality becomes ( 2 < x < 6 ). The integer solutions are 3, 4, 5. That's exactly three solutions. So, ( a ) can be equal to 18.If ( a ) is just above 15, say 15.1, then ( frac{a}{3} ) is just above 5, so ( x ) can be 3, 4, 5, which is three solutions. If ( a ) approaches 18 from below, ( frac{a}{3} ) approaches 6 from below, so ( x ) can still be 3, 4, 5.But if ( a ) is exactly 18, ( x ) can be up to 6, but since it's less than 6, 6 is not included. So, 3, 4, 5 are still the solutions.Therefore, the range is ( 15 < a leq 18 ).Looking back at the options, that's option A.I think that's correct. So, the answer should be A: ( 15 < a leq 18 ).**Final Answer**boxed{A}

answer：好，我现在要解决的问题是：找到所有素数对 ((p, q))，使得 (p^2 + 5pq + 4q^2) 是一个完全平方数。首先，我需要理解这个问题。题目给了一个二次表达式，要求它是一个完全平方数，同时 (p) 和 (q) 都是素数。我的目标就是找出所有满足这个条件的素数对。首先，我考虑这个表达式 (p^2 + 5pq + 4q^2)。或许可以尝试将其因式分解，看看是否能找到一些有用的信息。让我试一下因式分解：(p^2 + 5pq + 4q^2)。看起来像是一个二次式，可以尝试分解成两个一次项的乘积。试着分解：(p^2 + 5pq + 4q^2 = (p + a q)(p + b q))，其中 (a) 和 (b) 是常数，满足 (a + b = 5) 和 (a times b = 4)。解这个方程组：(a + b = 5)(ab = 4)解这个方程组，可以得到：(a = 1) 和 (b = 4)，或者 (a = 4) 和 (b = 1)。所以，(p^2 + 5pq + 4q^2 = (p + q)(p + 4q))。现在，这个表达式变成了两个因子的乘积，分别是 (p + q) 和 (p + 4q)。题目要求这个乘积是一个完全平方数，也就是某个整数的平方。接下来，我需要分析这两个因子 (p + q) 和 (p + 4q) 的关系，使得它们的乘积是一个完全平方数。首先，考虑这两个因子的最大公约数（gcd）。设 (d = gcd(p + q, p + 4q))。那么，(d) 必须同时整除 (p + q) 和 (p + 4q)。计算 (d)：(d) 整除 (p + 4q - (p + q) = 3q)，所以 (d) 整除 (3q)。同样，(d) 也整除 (p + q)，而 (p) 和 (q) 都是素数，所以 (d) 的可能值是什么呢？因为 (p) 和 (q) 都是素数，所以 (p + q) 和 (p + 4q) 的最大公约数只能是1或者3，因为 (3q) 的因数可能是1、3、q、3q，但 (p + q) 和 (p + 4q) 都是线性组合，可能不会被更大的数整除，特别是当 (p) 和 (q) 都是素数的时候。所以，考虑两种情况：(d = 1) 或者 (d = 3)。**情况一：(d = 1)**如果 (d = 1)，那么 (p + q) 和 (p + 4q) 必须各自都是完全平方数，因为它们互质，并且它们的乘积是一个完全平方数。设：(p + q = a^2)(p + 4q = b^2)其中，(a) 和 (b) 是正整数，并且 (b > a)。然后，我们可以用第二个方程减第一个方程：( (p + 4q) - (p + q) = b^2 - a^2 )( 3q = (b - a)(b + a) )现在，(3q) 是一个乘积，其中 (q) 是素数，3也是素数，所以可能的因数分解方式有：1. (b - a = 1) 和 (b + a = 3q)2. (b - a = 3) 和 (b + a = q)因为 (b > a)，所以 (b - a) 必须是正整数。接下来，分别分析这两种情况。**子情况1：(b - a = 1) 和 (b + a = 3q)**解这个方程组：(b - a = 1)(b + a = 3q)将两个方程相加：(2b = 3q + 1) ⇒ (b = frac{3q + 1}{2})将两个方程相减：(2a = 3q - 1) ⇒ (a = frac{3q - 1}{2})因为 (a) 和 (b) 必须是整数，所以 (3q + 1) 和 (3q - 1) 都必须是偶数，这意味着 (3q) 必须是奇数，因此 (q) 必须是奇数素数，除了2之外。接下来，代入 (p + q = a^2)：(p + q = left( frac{3q - 1}{2} right)^2)展开：(p + q = frac{(3q - 1)^2}{4})解出 (p)：(p = frac{(3q - 1)^2}{4} - q)化简：(p = frac{9q^2 - 6q + 1 - 4q}{4})(p = frac{9q^2 - 10q + 1}{4})因为 (p) 必须是素数，所以这个表达式必须是一个整数，并且结果必须是素数。首先，确保分子能被4整除：(9q^2 - 10q + 1 ≡ 0 mod 4)计算各项模4：- (9q^2 ≡ (1) q^2 mod 4)，因为9 mod4=1- (10q ≡ (2) q mod 4)，因为10 mod4=2- 1 ≡1 mod4所以：(q^2 - 2q + 1 ≡ 0 mod 4)这可以写成：((q - 1)^2 ≡ 0 mod 4)这意味着 (q - 1) 必须是偶数，即 (q) 是奇数，这与之前的结论一致，因为除了2之外，所有素数都是奇数。接下来，考虑 (q) 的可能值，代入计算 (p) 是否为素数。首先，尝试 (q = 3)：(p = frac{9(3)^2 - 10(3) + 1}{4} = frac{81 - 30 + 1}{4} = frac{52}{4} = 13)13是素数，所以 ((p, q) = (13, 3)) 是一个解。接下来，尝试 (q = 5)：(p = frac{9(5)^2 - 10(5) + 1}{4} = frac{225 - 50 + 1}{4} = frac{176}{4} = 44)44不是素数，所以这个情况不成立。再试 (q = 7)：(p = frac{9(7)^2 - 10(7) + 1}{4} = frac{441 - 70 + 1}{4} = frac{372}{4} = 93)93不是素数，所以不行。继续尝试更大的素数，但看起来结果可能不会是素数，因为分母是4，分子增长较快，结果可能不是素数。暂时记录下 (q=3) 的情况，得到一个解。**子情况2：(b - a = 3) 和 (b + a = q)**解这个方程组：(b - a = 3)(b + a = q)相加：(2b = q + 3) ⇒ (b = frac{q + 3}{2})相减：(2a = q - 3) ⇒ (a = frac{q - 3}{2})同样，(a) 和 (b) 必须是整数，所以 (q + 3) 和 (q - 3) 都必须是偶数，这意味着 (q) 必须是奇数素数，除了2之外。代入 (p + q = a^2)：(p + q = left( frac{q - 3}{2} right)^2)展开：(p + q = frac{(q - 3)^2}{4})解出 (p)：(p = frac{(q - 3)^2}{4} - q)化简：(p = frac{q^2 - 6q + 9 - 4q}{4})(p = frac{q^2 - 10q + 9}{4})同样，分子必须能被4整除：(q^2 - 10q + 9 ≡ 0 mod 4)计算各项模4：- (q^2 ≡ (1) q^2 mod 4)，因为q是奇数，所以q^2 ≡1 mod4- (10q ≡ (2) q mod 4)，因为10 mod4=2- 9 ≡1 mod4所以：(1 - 2q + 1 ≡ 0 mod 4)即：(-2q + 2 ≡ 0 mod 4)化简：(-2q ≡ -2 mod 4)两边同时乘以-1：(2q ≡ 2 mod 4)两边同时除以2：(q ≡ 1 mod 2)这与之前的结论一致，即q是奇数素数。接下来，尝试不同的q值：首先，尝试 (q=5)：(p = frac{5^2 - 10*5 + 9}{4} = frac{25 - 50 + 9}{4} = frac{-16}{4} = -4)负数，不是素数，排除。接下来，尝试 (q=7)：(p = frac{7^2 - 10*7 + 9}{4} = frac{49 - 70 + 9}{4} = frac{-12}{4} = -3)同样负数，排除。继续尝试更大的q：(q=11)：(p = frac{121 - 110 + 9}{4} = frac{20}{4} = 5)5是素数，所以 ((p, q) = (5, 11)) 是一个解。再试 (q=13)：(p = frac{169 - 130 + 9}{4} = frac{48}{4} = 12)12不是素数，排除。继续尝试 (q=17)：(p = frac{289 - 170 + 9}{4} = frac{128}{4} = 32)32不是素数，排除。看起来只有在 (q=11) 时，得到一个素数p=5。所以，在情况一中，我们找到了两个解：((13, 3)) 和 ((5, 11))。**情况二：(d = 3)**现在考虑另一种情况，即 (d = 3)。这意味着 (p + q) 和 (p + 4q) 都能被3整除，但它们的商互质。设：(p + q = 3a^2)(p + 4q = 3b^2)其中，(a) 和 (b) 是正整数，并且 (gcd(a, b) = 1)。同样，用第二个方程减第一个方程：(3q = 3b^2 - 3a^2)化简：(q = b^2 - a^2 = (b - a)(b + a))因为 (q) 是素数，所以 (q) 只能分解为1和它本身的乘积。因此，有两种可能性：1. (b - a = 1) 和 (b + a = q)2. (b - a = q) 和 (b + a = 1)（但后者不可能，因为 (b > a)，所以 (b + a) 至少为2，而 (q) 是素数，至少为2，所以只有第一种情况成立）所以，解方程组：(b - a = 1)(b + a = q)相加：(2b = q + 1) ⇒ (b = frac{q + 1}{2})相减：(2a = q - 1) ⇒ (a = frac{q - 1}{2})现在，代入 (p + q = 3a^2)：(p + q = 3 left( frac{q - 1}{2} right)^2)展开：(p + q = 3 cdot frac{(q - 1)^2}{4})(p + q = frac{3(q^2 - 2q + 1)}{4})解出 (p)：(p = frac{3(q^2 - 2q + 1)}{4} - q)化简：(p = frac{3q^2 - 6q + 3 - 4q}{4})(p = frac{3q^2 - 10q + 3}{4})同样，分子必须能被4整除：(3q^2 - 10q + 3 ≡ 0 mod 4)计算各项模4：- (3q^2 ≡ 3q^2 mod 4)- (10q ≡ 2q mod 4)- 3 ≡3 mod4所以：(3q^2 - 2q + 3 ≡ 0 mod 4)考虑q为奇数的情况，因为q是素数，除了2之外都是奇数。当q为奇数时，q ≡1或3 mod4。试q=3：(3(3)^2 - 10(3) + 3 = 27 - 30 + 3 = 0)0 mod4=0，符合条件。计算p：(p = frac{0}{4} = 0)，不是素数，排除。试q=5：(3(5)^2 - 10(5) + 3 = 75 - 50 + 3 = 28)28 mod4=0，符合条件。计算p：(p = 28/4 =7)，7是素数，所以 ((p, q) = (7, 5)) 是一个解。试q=7：(3(7)^2 - 10(7) + 3 = 147 - 70 + 3 = 80)80 mod4=0，符合条件。计算p：(p =80/4=20)，20不是素数，排除。试q=11：(3(11)^2 -10(11)+3= 363 -110 +3=256)256 mod4=0，符合条件。计算p：(p=256/4=64)，64不是素数，排除。试q=13：(3(13)^2 -10(13)+3= 507 -130 +3=380)380 mod4=0，符合条件。计算p：(p=380/4=95)，95不是素数，排除。继续尝试更大的q，但结果似乎不会得到素数p，所以目前只找到一个解：((7, 5))。**总结：**在情况一中，找到了两个解：((13, 3)) 和 ((5, 11))。在情况二中，找到了一个解：((7, 5))。现在，我需要检查是否还有其他可能的情况，或者是否遗漏了某些情况。例如，是否还有其他因数分解的方式，或者是否还有其他可能的d值。不过，根据之前的分析，d只能是1或3，所以已经考虑了所有可能的情况。另外，我还需要检查当p或q为2的情况，因为2是唯一的偶素数，可能会导致不同的结果。**检查p=2的情况：**设p=2，代入原式：(2^2 + 5*2*q + 4q^2 = 4 + 10q + 4q^2)这个表达式是否为完全平方数？同样，因式分解：(4 + 10q + 4q^2 = (2 + q)(2 + 4q))检查是否为完全平方数。同样，考虑gcd(2 + q, 2 + 4q)：计算gcd：gcd(2 + q, 2 + 4q) = gcd(2 + q, 2 + 4q - 2*(2 + q)) = gcd(2 + q, 2q - 2)因为q是素数，可能为2或奇数。如果q=2：(2 + 2 =4)，(2 + 4*2=10)，gcd(4,10)=2所以，设：(2 + q = 2a^2)(2 + 4q = 2b^2)其中，a和b互质。解方程组：(2 + q = 2a^2) ⇒ (q = 2a^2 - 2)(2 + 4q = 2b^2) ⇒ (4q = 2b^2 - 2) ⇒ (q = frac{2b^2 - 2}{4} = frac{b^2 -1}{2})将q代入：(2a^2 - 2 = frac{b^2 -1}{2})两边乘以2：(4a^2 -4 = b^2 -1)整理：(b^2 =4a^2 -3)寻找整数解：(b^2 =4a^2 -3)这看起来像是一个Pell方程，但可能没有解，或者只有小解。试a=1：(b^2=4 -3=1) ⇒ b=1所以，q=2*1^2 -2=0，不是素数，排除。试a=2：(b^2=16 -3=13)，13不是平方数，排除。试a=3：(b^2=36 -3=33)，不是平方数，排除。继续下去，似乎没有解，所以p=2时，q无法为素数。**检查q=2的情况：**设q=2，代入原式：(p^2 + 5p*2 +4*2^2 = p^2 +10p +16)这个表达式是否为完全平方数？同样，因式分解：(p^2 +10p +16 = (p + 2)(p + 8))检查是否为完全平方数。考虑gcd(p + 2, p + 8)：gcd(p + 2, p +8) = gcd(p +2, 6)因为p是素数，可能为2或奇数。如果p=2：(2 +2=4)，(2 +8=10)，gcd=2设：(p +2=2a^2)(p +8=2b^2)解方程组：(p=2a^2 -2)(p=2b^2 -8)所以：(2a^2 -2 = 2b^2 -8)化简：(2a^2 -2b^2 = -6)(a^2 -b^2 = -3)((a - b)(a + b) = -3)可能的因数分解：1. (a - b = -1) 和 (a + b = 3)解得：a=1, b=2所以，p=2*1^2 -2=0，不是素数，排除。2. (a - b = -3) 和 (a + b =1)解得：a=-1, b=2，但a必须为正整数，排除。所以，p=2时，q=2的情况没有解。接下来，考虑p≠2，q=2的情况：设q=2，p为奇素数，检查：(p^2 +10p +16) 是否为完全平方数。设这个表达式等于k^2：(p^2 +10p +16 =k^2)整理：(k^2 - p^2 -10p -16=0)或者：(k^2 = p^2 +10p +16)尝试寻找p使得这个等式成立。试p=3：(3^2 +10*3 +16=9+30+16=55)，不是平方数。p=5：25 +50 +16=91，不是平方数。p=7：49 +70 +16=135，不是平方数。p=11：121 +110 +16=247，不是平方数。p=13：169 +130 +16=315，不是平方数。p=17：289 +170 +16=475，不是平方数。看起来没有解，所以q=2时，p无法为素数。综上所述，除了之前找到的三个解之外，没有其他解。**验证解：**1. ((13, 3))：(13^2 +5*13*3 +4*3^2=169 +195 +36=400=20^2)，正确。2. ((5, 11))：(5^2 +5*5*11 +4*11^2=25 +275 +484=784=28^2)，正确。3. ((7, 5))：(7^2 +5*7*5 +4*5^2=49 +175 +100=324=18^2)，正确。所以，这三个解都是正确的。**最终答案**boxed{(13,3),(5,11),(7,5)}

answer：First, let’s define the initial amount of punch Mark added to the bowl as ( P ) gallons.1. **After the cousin drinks half the punch:** [ text{Remaining punch} = frac{P}{2} ]2. **Mark adds 4 gallons:** [ text{Total punch after adding} = frac{P}{2} + 4 ]3. **Sally drinks 2 gallons:** [ text{Remaining punch after Sally drinks} = frac{P}{2} + 4 - 2 = frac{P}{2} + 2 ]4. **Mark adds 12 gallons to fill the bowl:** [ frac{P}{2} + 2 + 12 = 16 ] 5. **Solve for ( P ):** [ frac{P}{2} + 14 = 16 ] [ frac{P}{2} = 2 ] [ P = 4 ]

answer：To determine how many integer values of ( b ) will result in the same number of valid values of ( n ) for the expression ( 4n + b ), we start by analyzing the original expression ( 4n + 7 ).1. **Find the range of ( n ) for ( 4n + 7 ):** - Solve ( 4n + 7 > 1 ): [ 4n > -6 implies n > -1.5 ] Since ( n ) is an integer, the smallest value ( n ) can take is ( -1 ). - Solve ( 4n + 7 < 200 ): [ 4n < 193 implies n < 48.25 ] Since ( n ) is an integer, the largest value ( n ) can take is ( 48 ). - Therefore, ( n ) ranges from ( -1 ) to ( 48 ), inclusive, giving 50 valid values.2. **Determine the range of ( b ) for ( 4n + b ) to have the same 50 valid values of ( n ):** - For ( n = -1 ): [ 4(-1) + b > 1 implies b - 4 > 1 implies b > 5 ] - For ( n = 48 ): [ 4(48) + b < 200 implies 192 + b < 200 implies b < 8 ] - Combining these, ( b ) must satisfy ( 5 < b < 8 ). The integer values of ( b ) that satisfy this are ( 6 ) and ( 7 ).Thus, there are 2 integer values of ( b ) that result in the same number of valid values of ( n ).

answer：First, determine how many math problems Marvin practiced today. Since he practiced three times as many problems today as he did yesterday, multiply the number of problems he solved yesterday by 3.Next, calculate the total number of problems Marvin practiced over the two days by adding the problems from yesterday and today.Then, determine how many problems Arvin practiced each day. Arvin practiced twice as many problems as Marvin on each day, so multiply Marvin's daily problems by 2 for both yesterday and today.Finally, add up all the problems practiced by both Marvin and Arvin over the two days to find the total number of problems they have practiced altogether.

answer：Okay, so I have this problem here: Given that log base 5 of a is greater than log base 5 of b, which of the following inequalities must hold true? The options are A, B, C, and D. Let me try to figure this out step by step.First, I remember that logarithms can be tricky, but I also recall that if the base of the logarithm is greater than 1, the logarithmic function is increasing. That means if log base 5 of a is greater than log base 5 of b, then a must be greater than b. So, from the given inequality, I can deduce that a > b. Also, since logarithms are only defined for positive numbers, both a and b must be positive. So, a > b > 0.Now, let's look at each option one by one.**Option A: sqrt(a) < sqrt(b)**Hmm, if a is greater than b, and both are positive, then taking the square root of both should preserve the inequality, right? Because the square root function is also increasing. So, if a > b, then sqrt(a) should be greater than sqrt(b). That would mean sqrt(a) < sqrt(b) is actually false. So, option A is incorrect.**Option B: log base 5 of (a - b) > 0**Okay, let's see. Since a > b, a - b is positive. So, log base 5 of (a - b) is defined. Now, log base 5 of something is greater than 0 means that something is greater than 5^0, which is 1. So, log base 5 of (a - b) > 0 implies that a - b > 1. But wait, do we know that a - b is greater than 1? Not necessarily. We only know that a > b, but the difference could be less than 1 or greater than 1. For example, if a = 2 and b = 1, then a - b = 1, and log base 5 of 1 is 0, which is not greater than 0. So, option B doesn't have to be true.**Option C: 5^(a - b) > 1**Alright, let's think about this. Since a > b, a - b is positive. Now, 5 raised to any positive power is greater than 5^0, which is 1. So, 5^(a - b) > 1 must be true because a - b is positive. That makes sense because the exponential function with base greater than 1 is increasing. So, option C is correct.**Option D: a*c > b*c**Hmm, this one depends on the value of c. If c is positive, then multiplying both sides of a > b by c would preserve the inequality, so a*c > b*c. But if c is negative, then multiplying both sides by c would reverse the inequality, making a*c < b*c. Also, if c is zero, then a*c = b*c, which is not greater. Since we don't know the value of c, we can't be sure that a*c > b*c. So, option D is not necessarily true.So, after going through each option, only option C must hold true based on the given information.