🧠:Okay, so I need to figure out if there's a natural number N that satisfies two conditions: first, it must be the sum of 1990 consecutive integers, and second, it can be expressed in exactly 1990 different ways as the sum of consecutive integers. Hmm, let's start by understanding each condition separately and then see how they might intersect.Starting with the first condition: N is the sum of 1990 consecutive integers. Let's recall that the sum of k consecutive integers starting from some integer a can be represented by the formula S = k/2 * [2a + (k - 1)]. So, in this case, k is 1990. Therefore, N would be 1990/2 * [2a + 1989]. Simplifying that, N = 995 * (2a + 1989). Let me check the arithmetic here. If there are 1990 terms starting at a, the last term would be a + 1989, so the average is (a + (a + 1989))/2 = a + 994.5, and multiplying by the number of terms 1990 gives N = 1990*(a + 994.5) = 1990a + 1990*994.5. Wait, but 1990*994.5 is 1990*(994 + 0.5) = 1990*994 + 995. Let me compute that. 1990*994 = (2000 - 10)*(994) = 2000*994 - 10*994 = 1,988,000 - 9,940 = 1,978,060. Then adding 995 gives 1,978,060 + 995 = 1,979,055. So N = 1990a + 1,979,055. Alternatively, from the initial formula, N = 995*(2a + 1989). Since 1990a + 1,979,055 = 995*2a + 995*1989. Wait, 995*2a is indeed 1990a, and 995*1989, let me check that. 1989 is 1990 - 1, so 995*1989 = 995*(1990 - 1) = 995*1990 - 995. 995*1990 is 995*2000 - 995*10 = 1,990,000 - 9,950 = 1,980,050. Then subtract 995 gives 1,980,050 - 995 = 1,979,055. So yes, both ways of computing N give the same result. So N must be equal to 995*(2a + 1989). Since N is a natural number, a must be an integer such that 2a + 1989 is also an integer. But since a is an integer (as we're talking about consecutive integers), 2a is even and 1989 is odd, so 2a + 1989 is odd. Therefore, N must be 995 multiplied by an odd number. So N is of the form 995*(2m + 1) where m is an integer. So N must be an odd multiple of 995. Alternatively, N must be divisible by 995 but not by 2. Wait, 995 is 5*199, so N must be divisible by 5 and 199, but not by 2. So that's the first condition.Now, moving on to the second condition: N can be expressed in exactly 1990 different ways as the sum of consecutive natural numbers. Hmm, I need to recall how the number of ways to express a number as the sum of consecutive integers relates to its divisors. I remember that the number of such representations is equal to the number of odd divisors of N. Wait, is that the case? Let me verify.Suppose we want to write N as the sum of consecutive integers. The formula for the sum of k consecutive integers starting from m is S = k*(2m + k - 1)/2. So N = k*(2m + k - 1)/2. Rearranging, 2N = k*(2m + k - 1). Therefore, k must be a divisor of 2N. But since m must be a positive integer, 2m + k - 1 must also be a positive integer. So for each divisor k of 2N, we can check if (2N)/k - k + 1 is even and positive, which would give us m. So the number of ways to express N as the sum of consecutive positive integers is equal to the number of divisors k of 2N such that k > 0, (2N)/k - k + 1 is positive and even. Alternatively, there's another way to think about this. The number of ways to express N as the sum of consecutive integers (allowing positive and negative integers) is equal to the number of odd divisors of N. But if we restrict to sums of consecutive positive integers, the count might be different. Wait, let me check.Actually, the standard result is that the number of ways to express N as a sum of consecutive positive integers is equal to the number of odd divisors of N greater than 1. Wait, no, that might not be exactly correct. Let me recall.Suppose N is expressed as the sum of k consecutive integers starting from m. Then N = k*m + k*(k-1)/2. So rearranged, m = (N - k*(k-1)/2)/k. For m to be a positive integer, (N - k*(k-1)/2) must be divisible by k and the result must be positive. So (N/k) - (k-1)/2 must be a positive integer. Therefore, (N/k) must be a half-integer if k is even, or an integer if k is odd. Hmm. Alternatively, considering 2N = k*(2m + k - 1), so k must divide 2N. Let’s let k be a positive integer divisor of 2N. Then 2m + k - 1 = 2N/k. Solving for m: m = (2N/k - k + 1)/2. For m to be a positive integer, (2N/k - k + 1) must be even and positive. So two conditions: 2N/k - k + 1 > 0 and (2N/k - k + 1) is even.So the number of ways is the number of positive integer divisors k of 2N where k satisfies those two conditions. But perhaps there's a relation to the number of odd divisors. Let me think. If k is an odd divisor of N, then 2N/k would be even. Let's suppose k is odd. Then 2N/k is even, so 2N/k - k + 1 = even - odd + 1 = even - odd is odd, then +1 is even. So if k is odd, then (2N/k - k + 1) is even, which is good. Then we need 2N/k - k + 1 > 0. Similarly, if k is even, then k divides 2N, so N must be divisible by k/2. Then 2N/k is N/(k/2). So if k is even, then 2N/k is an integer, and then (2N/k - k +1) must be even. Since k is even, 2N/k is integer, let's denote k = 2j. Then 2N/k = N/j. So the expression becomes N/j - 2j + 1. For this to be even: N/j - 2j + 1 must be even. Since 2j is even, N/j +1 must be even. So N/j must be odd. Therefore, if k is even, then j must divide N, and N/j must be odd. Therefore, j must be a divisor of N such that N/j is odd, which implies that j contains all the factors of 2 in N. Therefore, the number of even k's corresponds to the number of divisors of N that contain all the 2's in N. Wait, this is getting a bit complex.But according to some references I recall, the number of ways to express N as the sum of consecutive positive integers is equal to the number of odd divisors of N. Let me check with an example. Take N=9. The divisors of 9 are 1,3,9. All odd. So three divisors, but in the example given, 9 can be expressed as 4+5 and 2+3+4, which is two ways. Hmm, so three divisors but two ways. So perhaps the number of ways is equal to the number of odd divisors greater than 1? For N=9, 3 and 9. That would be two, which matches. Wait, but 1 is also an odd divisor. Let's see if k=1. Then m = (2*9/1 -1 +1)/2 = (18 -1 +1)/2 = 18/2 = 9. So the sum is just 9, which is a single number. But the problem statement says "sum of consecutive natural numbers", which could be a single number. But in the example given, they mention two different ways. So perhaps the count includes the single number as a valid way? Wait, in the example, 9 is written as 4+5 and 2+3+4, but 9 itself is also a sum of one number. So that would be three ways. But the problem statement says "two different ways". Maybe the problem statement is considering sums of two or more numbers? Or perhaps the original example is incorrect. Wait, let me check.Original example: "Additionally, 9 can be written in two different ways as the sum of consecutive natural numbers: 9 = 4 + 5 = 2 + 3 + 4". So the user is excluding the single-term sum. Therefore, perhaps in the problem, the number of ways is considered as the number of ways with two or more terms. Therefore, the formula would be the number of odd divisors minus 1. So for N=9, three odd divisors, subtract 1 gives two, which matches the example. So maybe the number of ways is equal to the number of odd divisors of N minus 1. Then, if the problem says "exactly 1990 different ways", then N must have 1991 odd divisors. Since the number of odd divisors is determined by the prime factorization of N. If N's prime factorization is 2^m * product_{p odd prime} p_i^{e_i}, then the number of odd divisors is product_{i} (e_i +1). Therefore, to have 1991 odd divisors, product (e_i +1) = 1991. Now, 1991 is an odd number, which factors into primes. Let's factor 1991. 1991 ÷ 11 = 181, because 11*180 = 1980, 1980 +11=1991. Then 181 is a prime number. So 1991 =11*181. Therefore, the exponents plus one would be 11 and 181. Therefore, N would have to have two distinct odd primes in its factorization with exponents 10 and 180, or one prime with exponent 1990. Wait, because 1991 is 11*181, so the exponents plus one are 11 and 181, so exponents would be 10 and 180. So N would be of the form 2^m * p^10 * q^180 where p and q are distinct odd primes. Alternatively, if it's a single prime raised to the 1990th power, then e_i +1 =1991, so e_i=1990, but 1991 is not prime. Wait, 1991=11*181, so if we have a single prime factor, then e_i +1=1991 implies e_i=1990, but 1991 is composite, so you can't have a single prime factor unless 1991 is 1*1991, but that would require a prime factor raised to 1990th power. However, since 1991 is composite, that's not possible. Therefore, the only way to have product (e_i +1)=1991 is to have two primes with exponents 10 and 180. Therefore, N must be of the form 2^m * p^10 * q^180 where p and q are distinct odd primes. But also, since N must be expressible as the sum of 1990 consecutive integers, which we established earlier requires N to be 995*(2a +1989). Since 995 is 5*199, so N must be divisible by 5 and 199. Therefore, N must have 5 and 199 as prime factors. Therefore, in the prime factorization of N, p and q must include 5 and 199. So either p=5 and q=199, or p=199 and q=5. The exponents would be 10 and 180. So for example, N=2^m *5^10 *199^180. Alternatively, if there are more prime factors, but since the product of (e_i +1) must be 11*181, which are primes, then we can only have two exponents: 10 and 180. Therefore, N must have exactly two distinct odd prime factors, each raised to exponents 10 and 180. Therefore, the two primes must be 5 and 199, since N must be divisible by 5 and 199. Therefore, N would be 2^m *5^10 *199^180 or 2^m *5^180 *199^10. However, considering the first condition, N must be divisible by 5 and 199, which it is in either case. Additionally, N must be an odd multiple of 995. Since 995=5*199, and N is 2^m *5^10 *199^180 or similar. Wait, but in the first condition, N must be equal to 995*(2a +1989). Since 995 is 5*199, and 2a +1989 is an odd integer (since 1989 is odd and 2a is even), so 2a +1989 is odd, so N is 5*199 multiplied by an odd integer. Therefore, N must be divisible by 5 and 199, but the rest of its factors must be odd. Therefore, in the prime factorization of N, the power of 2 must be zero? Wait, no. Because N is 5*199*(odd integer). The odd integer can have factors of 2? Wait, no, because 2a +1989 is an odd integer (since 1989 is odd, 2a is even, their sum is odd). Therefore, the odd integer (2a +1989) is odd, so it cannot have 2 as a factor. Therefore, N =5*199*(odd integer), so N has exactly one factor of 5 and one factor of 199, and the rest of the factors are odd primes. Wait, but in our second condition, N has to have prime factors 5 and 199 raised to exponents 10 and 180, which is conflicting. Because the first condition requires N to have at least 5 and 199 as primes, but the exponents would need to be at least 1. However, the second condition requires exponents of 5 and 199 to be either 10 and 180 or 180 and 10. Therefore, unless the "odd integer" part in N=5*199*(odd integer) includes additional factors of 5 and 199. Wait, but the "odd integer" is 2a +1989, which is an arbitrary odd integer. So if the odd integer includes factors of 5 and 199, then N would have higher exponents of 5 and 199. Therefore, it is possible. So, N is 5*199*(some odd integer). If we need N to have 5^10 *199^180, then the "some odd integer" must be 5^9 *199^179 * other primes. However, the problem states that N must have exactly 1991 odd divisors, which requires that N's prime factorization has exactly two odd primes with exponents 10 and 180. Therefore, N cannot have any other odd prime factors. Therefore, the "some odd integer" must be 5^9 *199^179. Therefore, N=5*199*5^9*199^179=5^(1+9)*199^(1+179)=5^10*199^180. Similarly, if we have exponents 180 and 10, then N=5^180*199^10. But in either case, N would also need to have 2^m as part of its factorization? Wait, no. Wait, from the first condition, N=995*(2a +1989). 995 is 5*199, and 2a +1989 is an odd integer. Therefore, N=5*199*(odd integer). Therefore, if we set that odd integer to be 5^9*199^179, then N=5^10*199^180. Therefore, in that case, N would not have any factor of 2. Therefore, m=0. Therefore, N is an odd number. So N=5^10*199^180. But in this case, N would have exactly two distinct odd prime factors, 5 and 199, with exponents 10 and 180. Therefore, the number of odd divisors is (10+1)*(180+1)=11*181=1991, which matches the required number of ways (since the number of ways is 1991-1=1990). Therefore, N=5^10*199^180. Additionally, this N is equal to 995*(2a +1989). Let's check if this holds. Since N=5^10*199^180 and 995=5*199, then 5^10*199^180 =5*199*(2a +1989). Therefore, (2a +1989)=5^9*199^179. Since 5^9*199^179 is an integer, and it's odd (as both 5 and 199 are odd), so 2a +1989 is odd, which matches. Then solving for a: a=(5^9*199^179 -1989)/2. Is this an integer? Since 5^9*199^179 is odd and 1989 is odd, their difference is even. Therefore, a is an integer. Therefore, such an N exists, which is 5^10*199^180, and it satisfies both conditions: it's the sum of 1990 consecutive integers starting at a=(5^9*199^179 -1989)/2, and it can be expressed in exactly 1990 ways as the sum of consecutive natural numbers (since it has 1991 odd divisors, leading to 1990 ways when excluding the single-term sum). Therefore, the answer is yes, such an N exists.Wait, but I need to verify whether N=5^10*199^180 indeed has exactly 1991 odd divisors. Let's see. The number of divisors is given by multiplying the exponents incremented by one. For the odd part, since N is 5^10 *199^180, the number of odd divisors is (10+1)*(180+1)=11*181=1991. Therefore, the number of ways to express N as the sum of consecutive positive integers (excluding single-term sums) is 1991 -1=1990. That matches the second condition. For the first condition, since N=5^10*199^180, and 995=5*199, then N=995*(5^9*199^179). As 5^9*199^179 is an odd integer (since both 5 and 199 are odd), then N can be written as the sum of 1990 consecutive integers starting from a=(5^9*199^179 -1989)/2. Since 5^9*199^179 is a very large number, subtracting 1989 will still leave a positive number, so a is a positive integer. Therefore, such an N exists. Hence, the answer is yes, such a natural number N exists.

✅:The natural number ( N ) exists and satisfies both conditions. Here's the detailed reasoning:1. Sum of 1990 Consecutive Integers: - The sum of ( k ) consecutive integers starting from ( a ) is ( N = frac{k}{2} times [2a + (k - 1)] ). - For ( k = 1990 ), ( N = 995 times (2a + 1989) ). This implies ( N ) must be an odd multiple of ( 995 ) (since ( 2a + 1989 ) is odd).2. Expressing ( N ) in Exactly 1990 Ways: - The number of ways to express ( N ) as the sum of consecutive natural numbers equals the number of odd divisors of ( N ) minus 1 (excluding single-term sums). - To have exactly 1990 ways, ( N ) must have ( 1991 ) odd divisors. The number of odd divisors is determined by its prime factorization. For ( 1991 ) odd divisors, ( N ) must have a prime factorization of the form ( 2^0 times p^{10} times q^{180} ), where ( p ) and ( q ) are distinct odd primes.3. Combining Both Conditions: - ( N ) must be divisible by ( 995 = 5 times 199 ). Thus, ( p = 5 ) and ( q = 199 ), leading to ( N = 5^{10} times 199^{180} ). - This ( N ) is indeed an odd multiple of ( 995 ), satisfying ( N = 995 times (2a + 1989) ) for some integer ( a ), and has ( (10+1)(180+1) = 1991 ) odd divisors, yielding exactly 1990 valid sums.Conclusion: Yes, such a natural number ( N ) exists. Specifically, ( N = 5^{10} times 199^{180} ) meets both requirements.

🔑:1. We start by analyzing the first condition. We need N to be the sum of 1990 consecutive integers. Let these integers be n, n+1, n+2, ldots, n+1989. The sum N of these integers is: [ N = n + (n+1) + (n+2) + cdots + (n+1989) ] This can be expressed using the formula for the sum of an arithmetic series: [ N = frac{1990}{2} times (2n + 1989) ] Simplifying, we get: [ N = 995 times (2n + 1989) ]2. From the above expression, it is clear that N must be an odd number, since 2n+1989 is always odd.3. Next, consider N to be written as the sum of k+1 consecutive natural numbers, starting with the smallest integer m. Then: [ N = m + (m+1) + cdots + (m+k) ] Using the sum of an arithmetic series again, we get: [ N = frac{k+1}{2} times (2m + k) ] Therefore: [ 2N = (k+1)(2m + k) ] From the initial condition, we also have: [ 1990(2n + 1989) = (k+1)(2m + k) ]4. To satisfy the second condition, we need N to be expressible as the sum of consecutive integers in exactly 1990 different ways. These (m, k) pairs must satisfy the above equation. For any valid pair (m, k), the integer k+1 must be a divisor of 2N. Also, it must hold that: [ k+1 < sqrt{2N} ]5. Since the number of valid pairs (m, k) corresponds to the number of divisors of 2N that are less than sqrt{2N}, it's enough to find an N such that 2N has 1990 divisors. The factors of 995(2n + 1989) play a critical role here.6. Analyze the prime factorization of 2N. Given: [ 2N = 1990 times (2n + 1989) = 2 times 5 times 199 times (2n + 1989) ] To make 2N have 3980 distinct divisors (because counting both divisors less than and greater than sqrt{2N} gives double the number of pairs), we need the following: [ 2N = 2^a times 5^b times 199^c times p_1^{d_1} times p_2^{d_2} times cdots times p_k^{d_k} ] where a, b, c, d_1, d_2, ldots, d_k are such that: [ (a+1)(b+1)(c+1)(d_1+1)(d_2+1)cdots(d_k+1) = 3980 ]7. To satisfy the divisor constraint: [ 3980 = 2 times 1990 = 2 times 2 times 995 = 2^2 times 5 times 199 ] Choose values: [ a = 1, b = 10, c = 180 quad text{or} quad a = 1, b = 180, c = 10 ] Therefore: [ N = 5^{10} times 199^{180} quad text{or} quad N = 5^{180} times 199^{10} ] These solutions meet the criteria for both conditions:[boxed{N = 5^{10} times 199^{180} text{ or } N = 5^{180} times 199^{10}}]Conclusively, this ensures that N is a number that can be expressed as the sum of 1990 consecutive integers and can be represented in exactly 1990 different ways as the sum of consecutive integers.

🧠:Okay, let me try to figure out this problem. So we have two linear functions, f(x) and g(x), whose graphs are parallel lines, not parallel to the coordinate axes. That means they have the same slope but different y-intercepts, right? And we know that the graph of y = (f(x))² touches the graph of y = 11g(x). We need to find all values of A such that the graph of y = (g(x))² touches the graph of y = Af(x). First, let me recall that if two lines are parallel, their slopes are equal. Let me denote f(x) as m x + b and g(x) as m x + c, where m is the common slope, and b and c are the y-intercepts of f and g respectively. Since they're not parallel to the coordinate axes, m is not zero, and neither are the lines vertical or horizontal. Wait, actually, since they are linear functions, they can't be vertical lines, which would not be functions. So, they must have a non-zero slope.Given that, let's write down the equations:f(x) = m x + bg(x) = m x + cNow, the graph of y = (f(x))² is a parabola, right? Because squaring a linear function gives a quadratic, which is a parabola. Similarly, y = 11g(x) is a straight line with slope 11m and y-intercept 11c. Wait, no. Wait, if g(x) is m x + c, then 11g(x) would be 11m x + 11c. But the problem says that the graph of y = (f(x))² touches the graph of y = 11g(x). When a parabola touches a line, that means the line is tangent to the parabola. So, the system of equations y = (f(x))² and y = 11g(x) has exactly one solution. That implies that the quadratic equation (f(x))² = 11g(x) has exactly one real root. Which further means that the discriminant of this quadratic equation is zero.So let's write the equation:(m x + b)² = 11(m x + c)Expanding the left side:m² x² + 2 m b x + b² = 11 m x + 11 cBring all terms to the left side:m² x² + (2 m b - 11 m) x + (b² - 11 c) = 0For this quadratic equation to have exactly one real root, the discriminant must be zero. The discriminant D of a quadratic equation a x² + b x + c = 0 is D = b² - 4 a c. Wait, but here, the quadratic is m² x² + (2 m b - 11 m) x + (b² - 11 c) = 0. So in standard notation:a = m²b (quadratic term coefficient) = (2 m b - 11 m)c (constant term) = (b² - 11 c)Wait, that might get confusing because the linear coefficient is (2 m b - 11 m), which uses the same symbol b as the constant term of f(x). Let me clarify:Let me denote the quadratic equation as:A x² + B x + C = 0Where:A = m²B = 2 m b - 11 mC = b² - 11 cThen discriminant D = B² - 4 A C = 0.So:(2 m b - 11 m)² - 4 * m² * (b² - 11 c) = 0Let me compute that:First, expand (2 m b - 11 m)²:= (2 m b)² - 2 * 2 m b * 11 m + (11 m)²= 4 m² b² - 44 m² b + 121 m²Then compute 4 A C = 4 * m² * (b² - 11 c) = 4 m² b² - 44 m² cTherefore, the discriminant:[4 m² b² - 44 m² b + 121 m²] - [4 m² b² - 44 m² c] = 0Simplify term by term:4 m² b² - 44 m² b + 121 m² - 4 m² b² + 44 m² c = 0The 4 m² b² cancels with -4 m² b², leaving:-44 m² b + 121 m² + 44 m² c = 0Factor out 11 m²:11 m² (-4 b + 11 + 4 c) = 0Since 11 m² is not zero (because m ≠ 0 as the lines are not parallel to the axes, and 11 is a coefficient), we can divide both sides by 11 m²:-4 b + 11 + 4 c = 0Rearranged:-4 b + 4 c + 11 = 0Divide both sides by 4:- b + c + 11/4 = 0So:c - b = -11/4Therefore, c = b - 11/4So the relationship between c and b is c = b - 11/4. That's one equation connecting the constants of the two lines.Now, we need to find all values of A such that the graph of y = (g(x))² touches the graph of y = A f(x). Again, similar to the previous situation. The graph of y = (g(x))² is a parabola, and the graph of y = A f(x) is a straight line. For them to touch each other, the line must be tangent to the parabola. Therefore, the system of equations y = (g(x))² and y = A f(x) must have exactly one solution. Which implies the quadratic equation (g(x))² = A f(x) has exactly one real root, so its discriminant is zero.So let's write that equation:(g(x))² = A f(x)Substituting g(x) = m x + c and f(x) = m x + b:(m x + c)² = A (m x + b)Expanding the left side:m² x² + 2 m c x + c² = A m x + A bBring all terms to left side:m² x² + (2 m c - A m) x + (c² - A b) = 0Again, quadratic equation in x. For it to have exactly one solution, discriminant D = 0.Compute discriminant:D = (2 m c - A m)^2 - 4 * m² * (c² - A b) = 0Let's compute each term:First term: (2 m c - A m)^2= m² (2 c - A)^2Second term: 4 m² (c² - A b)Therefore:m² (2 c - A)^2 - 4 m² (c² - A b) = 0Factor out m²:m² [ (2 c - A)^2 - 4 (c² - A b) ] = 0Since m ≠ 0, we can divide both sides by m²:(2 c - A)^2 - 4 (c² - A b) = 0Expand (2 c - A)^2:= 4 c² - 4 A c + A² - 4 c² + 4 A bSimplify term by term:4 c² - 4 A c + A² - 4 c² + 4 A b = 0The 4 c² cancels with -4 c²:-4 A c + A² + 4 A b = 0Factor out A:A (-4 c + A + 4 b) = 0Therefore, either A = 0 or (-4 c + A + 4 b) = 0But if A = 0, then y = 0, which is a horizontal line. But the parabola y = (g(x))² opens upwards (since it's a square), so the horizontal line y=0 would touch the parabola at its vertex if the vertex is at y=0. The vertex of y = (g(x))² is at the minimum point, which is when g(x)=0. So the vertex is at x = -c/m, y=0. Therefore, y=0 is tangent to the parabola at that point. But in the problem statement, the original lines f(x) and g(x) are not parallel to the coordinate axes. So they have non-zero slope, but A=0 would lead to y=0, which is a coordinate axis. However, the problem doesn't specify that A cannot be zero, just that the original lines are not parallel to coordinate axes. Wait, but the question is about the graph of y=Af(x). If A=0, then y=0, which is the x-axis. But the parabola y=(g(x))² touches the x-axis at its vertex. So that is possible, but we need to check if this is allowed. The problem doesn't restrict A from being zero, so technically, A=0 is a solution. However, we need to verify whether in that case, the lines are not parallel to the coordinate axes. Since f(x) and g(x) are already not parallel to the coordinate axes, but A=0 makes y=0 (the x-axis), which is a coordinate axis, but the problem says the original lines f and g are not parallel to coordinate axes, but doesn't say anything about the line y=Af(x). So perhaps A=0 is acceptable. But wait, the problem says "the graph of the function y = (g(x))² touches the graph of the function y = A f(x)". If A=0, then y=0 is the x-axis, which touches the parabola y=(g(x))² at its vertex. So as long as the parabola is non-degenerate, which it is (since it's a square of a linear function with non-zero slope), then yes, the x-axis is tangent to the parabola at its vertex. Therefore, A=0 is a possible solution. But let me check if there's another solution from the equation -4 c + A + 4 b = 0.So, from the discriminant equation:A (-4 c + A + 4 b) = 0 => A=0 or A = 4 c - 4 b.But from earlier, we found that c = b - 11/4. Let's substitute that into A = 4 c -4 b:A = 4 (b - 11/4) - 4 b = 4 b - 11 -4 b = -11So either A=0 or A=-11.Wait, so possible solutions are A=0 and A=-11.But we need to check if both are valid. Let's see.If A=0, as discussed, the line y=0 is the x-axis, which touches the parabola y=(g(x))² at its vertex. Since g(x) is a linear function with slope m ≠ 0, the parabola y=(g(x))² indeed has a vertex at ( -c/(2m), 0 ), wait, actually, the vertex of y=(g(x))² is at the point where g(x)=0. Since g(x)=m x + c, setting g(x)=0 gives x = -c/m. So the vertex is at ( -c/m, 0 ), and the parabola opens upwards. Therefore, the x-axis is tangent to the parabola at that vertex. Therefore, A=0 is indeed a valid solution.For A=-11, we need to check if this gives a valid tangent. Let's see. If A=-11, then the line is y = -11 f(x). Since f(x) is a linear function with slope m, this line has slope -11 m. The parabola is y=(g(x))². So, we need to check if the line y=-11 f(x) is tangent to the parabola y=(g(x))².But let's verify if A=-11 is indeed a solution. Since we derived A= -11 from the discriminant condition, and we have already established the relation c = b - 11/4 from the first part, substituting back gives A=-11.But let me check if A=-11 actually works. Let's substitute A=-11 into the equation:(g(x))² = -11 f(x)But since (g(x))² is non-negative (as it's a square), the right-hand side must also be non-negative. Therefore, -11 f(x) ≥ 0 => f(x) ≤ 0. So for the equation to have a solution, f(x) must be non-positive where they touch. However, since the line y=-11 f(x) is tangent to the parabola y=(g(x))², which is always non-negative, then y=-11 f(x) must be non-negative at the point of tangency. Therefore, -11 f(x) ≥ 0 implies f(x) ≤ 0. So, unless f(x) is non-positive at that point, there could be a solution. But since we have a tangent point, there must be exactly one solution where this occurs. But since we have already derived A=-11 through the discriminant condition, and we have already used the relationship between c and b from the first condition, which comes from the fact that (f(x))² =11 g(x) is tangent, which requires the discriminant to be zero, so all the conditions are already baked into the equations. Therefore, A=-11 should be valid.But let's double-check with an example. Suppose we choose specific values for m, b, and c that satisfy c = b -11/4, and check if A=-11 works.Let's pick m=1 for simplicity. Let’s choose b=0. Then c=0 -11/4 = -11/4.So f(x)=x, g(x)=x -11/4.First, check if y=(f(x))² touches y=11g(x):y=x² and y=11(x -11/4)Set x² =11x - 121/4x² -11x + 121/4 =0Discriminant D= (11)^2 -4*1*121/4=121 -121=0. So yes, it touches.Now, check y=(g(x))² = (x -11/4)^2 and y=Af(x)=A x.We need to find A such that (x -11/4)^2 = A x has one solution.Expand: x² - (11/2)x + (121/16) = A xx² - (11/2 + A)x + 121/16 =0Discriminant D= [-(11/2 + A)]² -4*1*(121/16)= (11/2 + A)^2 - 121/4Set D=0:(11/2 + A)^2 =121/4Take square roots:11/2 + A = ±11/2Case 1: 11/2 + A =11/2 => A=0Case 2: 11/2 + A = -11/2 => A= -11So A=0 or A=-11. So in this case, both A=0 and A=-11 work. When A=0, the line is y=0, which touches the parabola at x=11/4, y=0. But wait, g(x)=x -11/4, so when x=11/4, g(x)=0, so y=(g(x))²=0, and y=0. So that's the vertex. For A=-11, the line is y=-11x. Let's check where they touch:Solve (x -11/4)^2 = -11xx² - (11/2)x + 121/16 = -11xx² + (11/2)x + 121/16 =0Multiply by 16 to eliminate denominators:16x² +88x +121=0This factors as (4x + 11)^2=0, so x= -11/4. Then y= -11*(-11/4)=121/4. And (g(-11/4))²=( -11/4 -11/4 )²= (-22/4)²=(121/4). So yes, it touches at x=-11/4, y=121/4. So A=-11 is valid.Therefore, both A=0 and A=-11 are solutions. However, the problem says "find all values of A", so both must be included. Wait, but in the earlier general case, when we derived the discriminant, we had A=0 or A=4c -4b. And since c =b -11/4, then 4c -4b=4(b -11/4) -4b=4b -11 -4b=-11. Therefore, the solutions are A=0 and A=-11.But in the problem statement, when it says "the graph of the function y = (g(x))² touches the graph of the function y = Af(x)", in the case of A=0, the graph is the x-axis, which touches the parabola at its vertex. So that's a valid case. However, sometimes problems might have constraints that aren't explicitly stated, but here, there's no restriction on A. Therefore, both A=0 and A=-11 are valid.But let's check if the problem allows A=0. The original problem states that the lines f(x) and g(x) are not parallel to the coordinate axes, but the line y=Af(x) when A=0 is the x-axis, which is a coordinate axis. However, the problem doesn't say that the line y=Af(x) cannot be a coordinate axis, only that f(x) and g(x) are not. So y=Af(x) being a coordinate axis is acceptable. Therefore, A=0 is a valid solution.But wait, in the example above, when A=0, the point of tangency is at (11/4, 0), which is on the x-axis. So yes, it works. Therefore, both A=0 and A=-11 are valid answers.But the problem says "find all values of A", so we should present both. However, let me check the problem statement again:"Find all values of A such that the graph of the function ( y = (g(x))^2 ) touches the graph of the function ( y = Af(x) )."There's no restriction on A, so both solutions are valid. However, I need to verify once more in the general case.In the discriminant derivation, we arrived at A=0 or A=-11. So regardless of the slope m and intercepts b and c (as long as they satisfy c = b -11/4), these are the only solutions. Therefore, the possible values of A are 0 and -11.Wait, but in the example we tried, both worked. So the answer should be A=0 and A=-11. But let me think again. But wait, in the problem statement, it's possible that when A=0, the x-axis touches the parabola, but maybe in the problem's context, they are expecting only non-zero A? The problem didn't specify, so we should include all possible solutions. Therefore, the answer is A=0 and A=-11.But let me check if the original problem in Chinese maybe had different constraints, but as per the translation given, there's no such restriction.Alternatively, maybe there is a miscalculation. Let's go through the discriminant step again.We had:From the first condition, c = b -11/4.For the second discriminant:After expanding, we had:-4 A c + A² + 4 A b = 0Factor out A:A(-4c + A +4b)=0Thus, solutions are A=0 or A=4c -4b.But since c = b -11/4,A=4(b -11/4) -4b =4b -11 -4b= -11Hence, the solutions are A=0 and A=-11.Therefore, the answer should be A=0 and A=-11. So the values of A are 0 and -11.But wait, the problem says "all values of A", so we need to box both. But let me check once more with another example.Let’s take m=2, b=1, then c=1 -11/4= -7/4.So f(x)=2x+1, g(x)=2x -7/4.First condition: (2x+1)^2 =11*(2x -7/4)Expand left:4x² +4x +1 =22x -77/4Bring all terms to left:4x² +4x +1 -22x +77/4=0Simplify:4x² -18x + (1 +77/4)=01 is 4/4, so total constant term 81/4.Thus:4x² -18x +81/4=0Multiply by 4:16x² -72x +81=0Discriminant D=72² -4*16*81=5184 -5184=0, so it's tangent. Good.Now, for the second part, find A such that (g(x))² =A f(x) has one solution.Compute (2x -7/4)^2 =A*(2x +1)Left side:4x² -7x +49/16Set equal to right side:2A x +ABring all terms to left:4x² -7x +49/16 -2A x -A=0Combine like terms:4x² +(-7 -2A)x + (49/16 -A)=0Discriminant D= [ -7 -2A ]² -4*4*(49/16 -A )Compute D:= (7 + 2A)^2 -16*(49/16 -A )=49 +28A +4A² -49 +16A= (49 -49) + (28A +16A) +4A²=44A +4A²Set D=0:4A² +44A=0Factor:4A(A +11)=0Thus, A=0 or A=-11. Same result. Therefore, in this case, A=0 and A=-11.Testing A=0: The equation becomes (2x -7/4)^2=0, which implies 2x -7/4=0 =>x=7/8. Then y=0. So the x-axis is tangent at x=7/8, which is the vertex of the parabola (g(x))². Correct.Testing A=-11: The equation becomes (2x -7/4)^2= -11*(2x +1)Left:4x² -7x +49/16Right:-22x -11Bring all terms to left:4x² -7x +49/16 +22x +11=0Simplify:4x² +15x + (49/16 +176/16)=0=4x² +15x +225/16=0Multiply by 16:64x² +240x +225=0Discriminant D=240² -4*64*225=57600 -57600=0So one solution. Thus, A=-11 works. So in this case, it's valid.Therefore, in both examples, A=0 and A=-11 are solutions, so in general, the answer should be A=0 and A=-11.But the problem might expect only A=-11, considering A=0 trivial? But according to the mathematics, both are valid. Unless there's a restriction in the problem that I missed.Re-reading the problem:"Given two linear functions f(x) and g(x) such that the graphs of y = f(x) and y = g(x) are parallel lines, not parallel to the coordinate axes. It is known that the graph of the function y = (f(x))² touches the graph of the function y = 11g(x). Find all values of A such that the graph of the function y = (g(x))² touches the graph of the function y = Af(x)."No, there's no restriction on A. The problem doesn't say "A is non-zero" or anything. So both A=0 and A=-11 are acceptable. However, looking back at the initial problem statement in Chinese (translated), if the original had any specific constraints, but as given, the translation is as above. Therefore, the correct answers are A=0 and A=-11.But I need to confirm once more. In some contexts, touching might require the point of contact to be above the x-axis, but in the case of A=0, the contact point is on the x-axis, which is still considered touching. So mathematically, it's valid.Therefore, the final answer is A=0 and A=-11.But let me check the problem statement again. It says "the graph of the function y = (f(x))² touches the graph of the function y = 11g(x)". In this case, when they touch, it's at a single point, which could be anywhere, not necessarily on the axis. Similarly, when A=0, the touching point is on the x-axis. So unless there's a reason to exclude A=0, it's valid. Since the problem doesn't specify any such restriction, both solutions are valid. Hence, the values of A are boxed{-11} and boxed{0}. However, sometimes problems expect all answers in one box. Let me check standard conventions. Typically, if multiple solutions, we put them all in one box separated by commas. So boxed{0} and boxed{-11} or boxed{-11} and boxed{0}. But the order might matter. Since the problem says "all values", probably list both inside a single box: boxed{0} and boxed{-11}. But in China's math problems, sometimes they list all solutions in a set. Let me check.Wait, in the original problem, the user hasn't specified the format for multiple answers. In many cases, if multiple answers, they are presented as a list inside a single box. For example, boxed{0} and boxed{-11}. But in some conventions, it's written as boxed{0} and boxed{-11}. Alternatively, if order matters, but here it doesn't. Alternatively, using set notation: boxed{{0, -11}}.But since in the problem statement, it's in Chinese, and in Chinese math problems, often multiple answers are presented in a single box with comma separation or within curly brackets. However, looking at the initial problem's instruction, the user says "put your final answer within boxed{}". So maybe they expect all answers in one box. Let me check the example in the problem statement. The user wrote:"Find all values of A... put your final answer within boxed{}"So likely, multiple answers inside the box, separated by commas. Hence, boxed{0} and boxed{-11} or as boxed{0, -11}. But in the previous example, the quadratic equation had two solutions, and they were presented as boxed{0} and boxed{-11}. However, given that the problem is likely from a competition or textbook, and considering that both solutions are valid, the answer is two values. So in the required format, inside a single box, separated by commas: boxed{0, -11}But in some contexts, especially in China, they might write it as boxed{-11} and boxed{0}, but given that the problem says "all values", putting them in one box as a set is better. Therefore, the final answer is boxed{0} and boxed{-11}, but since the user might expect multiple boxed answers, but according to the instruction, "put your final answer within boxed{}", so maybe all values inside one box. Therefore, boxed{0} and boxed{-11} but how?Wait, looking back at the initial problem statement's instructions: "put your final answer within boxed{}". So the user expects one boxed answer. Therefore, we need to present all answers in one box. So the correct formatting is boxed{0} and boxed{-11} if multiple boxes are allowed, but since the instruction says "within boxed{}", likely one box. Therefore, I should present both values in one box, separated by commas. Therefore, boxed{0, -11}But let me confirm with the examples. In the example above, when solving, we found A=0 and A=-11. So, the answer is both. Hence, boxed together.Yes, I think that's the correct way. Therefore, the final answer is boxed{0} and boxed{-11}, but written as boxed{0} and boxed{-11} if allowed, but since only one box is expected, boxed{0, -11}.But I need to check the original problem's exact wording. The problem says "Find all values of A...". So "values" plural, so the answer is two values. In Chinese math problems, when multiple answers, they are presented in a set or listed with commas inside a single box. Therefore, the correct formatting is boxed{0, -11}.Yes, that makes sense. So even though in the thinking process I considered both, the final answer should be both in one box. Therefore, the final answer is boxed{0, -11}Final AnswerThe values of ( A ) are boxed{0} and boxed{-11}.However, according to the standard convention for multiple answers in a single box, it should be:boxed{0} and boxed{-11}But if the system requires a single box, then:boxed{0, -11}Given the ambiguity, but following the user's initial instruction to "put your final answer within boxed{}", and considering that multiple answers can be included in one box, the correct formatting is:boxed{0} and boxed{-11}But since some systems may require a single box, the safer answer is:boxed{0, -11}

✅:Given two linear functions ( f(x) ) and ( g(x) ) with parallel graphs (same slope ( m )) and different y-intercepts ( b ) and ( c ) respectively, we know that the graph of ( y = (f(x))^2 ) touches the graph of ( y = 11g(x) ). This implies the quadratic equation ( (f(x))^2 = 11g(x) ) has a discriminant of zero, leading to the relationship ( c = b - frac{11}{4} ).To find the values of ( A ) such that the graph of ( y = (g(x))^2 ) touches the graph of ( y = Af(x) ), we consider the quadratic equation ( (g(x))^2 = Af(x) ). For tangency, this quadratic equation must have a discriminant of zero. Substituting ( g(x) = mx + c ) and ( f(x) = mx + b ), we derive the discriminant condition and solve for ( A ).Key steps:1. First condition: The discriminant of ( (mx + b)^2 = 11(mx + c) ) leads to ( c = b - frac{11}{4} ).2. Second condition: The discriminant of ( (mx + c)^2 = A(mx + b) ) leads to solving ( A(-4c + A + 4b) = 0 ). Substituting ( c = b - frac{11}{4} ) results in ( A = 0 ) or ( A = -11 ).Verification with examples confirms both solutions are valid. Thus, the values of ( A ) are:[boxed{0} text{ and } boxed{-11}]

🔑:1. Let's consider the given linear functions ( f(x) = ax + b ) and ( g(x) = ax + c ), where ( a neq 0 ). The problem states that the graphs of ( y = f(x) ) and ( y = g(x) ) are parallel lines, which implies they have the same slope, confirming that their slopes (a) are indeed equal and non-zero. Since these lines are not parallel to the coordinate axes, (a neq 0).2. The fact that the graph of ( y = (f(x))^2 ) touches the graph ( y = 11g(x) ) implies that the equation ( (f(x))^2 = 11g(x) ) has exactly one solution. This means the quadratic equation derived from ( (ax + b)^2 = 11(ax + c) ): [ (ax + b)^2 = 11(ax + c) ] can be expanded as: [ a^2x^2 + 2abx + b^2 = 11ax + 11c ] Rearranging this, we get: [ a^2x^2 + 2abx + b^2 - 11ax - 11c = 0 ] Simplifying further, we obtain: [ a^2x^2 + a(2b - 11)x + (b^2 - 11c) = 0 ]3. For this quadratic equation to have exactly one solution, its discriminant ( Delta ) must be zero. The discriminant of the quadratic equation ( ax^2 + bx + c = 0 ) is given by ( Delta = b^2 - 4ac ). Applying this to our equation: [ Delta = [a(2b - 11)]^2 - 4a^2(b^2 - 11c) ] Simplifying: [ Delta = a^2(2b - 11)^2 - 4a^2(b^2 - 11c) ] Factoring out ( a^2 ): [ Delta = a^2[(2b - 11)^2 - 4(b^2 - 11c)] ] Setting ( Delta ) to zero: [ (2b - 11)^2 - 4(b^2 - 11c) = 0 ] Simplifying the equation, we get: [ 4b^2 - 44b + 121 - 4b^2 + 44c = 0 ] Which reduces to: [ 44c - 44b + 121 = 0 ] Solving for ( b ) and ( c ): [ 4c - 4b = 11 ] Thus: [ c = b + frac{11}{4} ]4. Similarly, the graph of ( y = (g(x))^2 ) touching ( y = Af(x) ) implies that the equation ((g(x))^2 = Af(x) ) has exactly one solution. Substituting ( g(x) = ax + c ) and ( f(x) = ax + b ): [ (ax + c)^2 = A(ax + b) ] Expanding: [ a^2x^2 + 2acx + c^2 = Aax + Ab ] Rearranging: [ a^2x^2 + 2acx + c^2 - Aax - Ab = 0 ]5. The discriminant of this quadratic equation must be zero for the equation to have exactly one solution: [ Delta = [a(2c - A)]^2 - 4a^2(c^2 - Ab) ] Simplifying: [ Delta = a^2(2c - A)^2 - 4a^2(c^2 - Ab) ] Factoring out ( a^2 ): [ Delta = a^2[(2c - A)^2 - 4(c^2 - Ab)] ] Setting ( Delta ) to zero: [ (2c - A)^2 - 4(c^2 - Ab) = 0 ] Simplifying the equation, we get: [ 4c^2 - 4Ac + A^2 - 4c^2 + 4Ab = 0 ] Which reduces to: [ A^2 + 4Ab - 4Ac = 0 ] Factoring out: [ A(A + 4b - 4c) = 0 ] Setting each factor to zero: [ A = 0 quad text{or} quad A + 4b - 4c = 0 ] From ( A + 4b - 4c = 0 ), and using ( c = b + frac{11}{4} ): [ A + 4b - 4(b + frac{11}{4}) = 0 ] Simplifying: [ A + 4b - 4b - 11 = 0 ] Therefore: [ A = -11 ]6. Conclusion: Thus, the values of ( A ) for which the graph ( y = (g(x))^2 ) touches the graph ( y = Af(x) ) are ( boxed{0} ) and ( boxed{-11} ).

🧠:Okay, let's see. The problem is about a soccer team that has won 60% of their 45 matches so far. We need to find out the minimum number of additional matches they need to win to reach a 75% winning percentage. Hmm, percentage problems can sometimes be tricky, so I need to take it step by step.First, let's figure out how many matches the team has already won. They've played 45 matches and won 60% of them. So 60% of 45 is 0.6 * 45. Let me calculate that: 0.6 * 45 = 27. So they've won 27 matches out of 45. That leaves them with 45 - 27 = 18 matches that they didn't win (either lost or drew).Now, the question is about reaching a 75% winning percentage. But wait, we need to clarify: is the 75% winning percentage based on the total matches played so far plus the additional matches they will play? The problem says "the minimum number of matches that this team still needs to win to reach a winning percentage of 75%." It doesn't specify the total number of matches remaining, so I think we have to assume that they can play as many additional matches as needed, and we need to find the minimum number of wins required in those additional matches to make their overall winning percentage 75%.Wait, but maybe there's another interpretation. If the total number of matches is fixed, but that doesn't seem to be the case here. The problem states "the minimum number of matches that this team still needs to win," implying that they will play some number of matches in the future, and we have to find how many they must win out of those to get their overall percentage to 75%. But the problem doesn't specify how many more matches they will play, so the minimum number of wins would depend on how many matches they have left. Wait, but maybe the problem is assuming that they can play any number of matches, and we need to find the minimal number of matches they need to play and win such that their total wins over total matches is 75%. Because if they play more matches, they have more opportunities to win, but they want the minimum number of wins required, which might mean playing as few matches as possible, but needing to win all of them. Wait, but that might not be correct. Let me think again.Wait, maybe the problem is phrased as "the minimum number of matches that this team still needs to win" regardless of how many matches they play. So they can play any number of additional matches, but they need to win a certain number of them, and we need the smallest number of wins required to achieve 75% overall. But how?Let me set up an equation. Let's let x be the number of additional matches they need to win. But wait, if they play y additional matches, and win x of them, then their total wins become 27 + x, and total matches become 45 + y. We need (27 + x)/(45 + y) >= 0.75. But we need to find the minimal x such that there exists a y >= x (since you can't win more matches than you play), and x is minimized.So the problem reduces to finding the smallest x where there exists some y >= x such that (27 + x)/(45 + y) >= 0.75. We need to minimize x. So to minimize x, we can assume that y is as small as possible, which would be y = x (i.e., they win all their remaining matches). Because if they win all the additional matches they play, then y = x, and that would give the minimal y for a given x. Therefore, we can set y = x, leading to the equation:(27 + x)/(45 + x) >= 0.75Solving for x:Multiply both sides by (45 + x):27 + x >= 0.75*(45 + x)Compute 0.75*(45 + x):0.75*45 = 33.75, and 0.75x, so 33.75 + 0.75xThus:27 + x >= 33.75 + 0.75xSubtract 0.75x from both sides:27 + 0.25x >= 33.75Subtract 27:0.25x >= 6.75Multiply both sides by 4:x >= 27So x must be at least 27. So they need to win 27 more matches, and if they play 27 more matches and win all of them, their total wins would be 27 + 27 = 54, and total matches would be 45 + 27 = 72. Then 54/72 = 0.75, which is 75%.Wait, but 27 seems like a lot. Let me check the calculations again.Original wins: 27 out of 45.If they win x more matches out of x played, then total wins: 27 + x, total matches: 45 + x.Set (27 + x)/(45 + x) = 0.75.Cross-multiplying: 27 + x = 0.75*(45 + x)27 + x = 33.75 + 0.75xx - 0.75x = 33.75 - 270.25x = 6.75x = 6.75 / 0.25x = 27.Yes, that's correct. So they need to play 27 more matches and win all of them. That gives 54 wins out of 72 matches, which is exactly 75%.But is there a way to have fewer wins if they play more matches? For example, maybe if they play more matches but don't win all of them, but still reach the 75%? Wait, but the problem asks for the minimum number of matches they still need to win, regardless of how many they play. So even if they play more matches, they might need to win fewer, but perhaps not. Wait, let's think again.Suppose they play y matches, winning x of them, where y >= x. We need (27 + x)/(45 + y) >= 0.75. We need to minimize x. So perhaps playing more matches (y > x) but winning x could lead to a lower x? Let's test with x=26.If x=26, then (27 + 26)/(45 + y) = 53/(45 + y) >= 0.75. So 53 >= 0.75*(45 + y). Let's solve 53 = 0.75*(45 + y).53 / 0.75 = 45 + y53 / 0.75 ≈ 70.6667So 70.6667 = 45 + y → y ≈ 25.6667. But y must be at least x=26, but here y≈25.6667, which is less than 26, which is impossible. So even if we take y=26, then 53/(45 +26) = 53/71 ≈ 0.7465, which is less than 75%. So x=26 is not enough.Similarly, trying x=27, as before, with y=27 gives 54/72=0.75.If we take x=27 and y=28, then (27+27)/(45+28)=54/73≈0.7397, which is less than 75%, so they need to win more. Wait, no, if they play y=28 matches and win x=27, then total wins are 27+27=54, total matches 45+28=73. 54/73≈0.7397, which is still under 75%. So even if they play more matches but don't win all, they might need more wins. But the problem is to find the minimal number of wins required, regardless of how many matches they play. Therefore, the minimal x is when y is as small as possible, which is y=x, so that they win all additional matches. Therefore, the answer is 27.Wait, but let me check for x=27 and y=27, which gives exactly 75%. If they play 27 more matches, win all, then it's 54/72=0.75. If they play more than 27 matches and lose some, they might need more wins. For example, if they play 28 matches, and win 27, then total wins 54, total matches 73, which is 54/73≈73.97%, which is less than 75%. So they would need to win more. So in that case, to get 75%, they would need (27 + x)/(45 + y) >=0.75. Let's solve for x if y=28.(27 + x)/73 >=0.75 → 27 + x >= 54.75 → x >=54.75 -27=27.75. So x must be at least 28 (since x must be integer). So if they play 28 matches, they need to win 28 to get (27+28)/(45+28)=55/73≈75.34%, which is over 75%. But that's 28 wins, which is more than 27. So the minimal x is 27 when y=27.Similarly, if they play y=26 matches and have to win x=27, that's impossible because y can't be less than x. Therefore, the minimal number of wins required is 27, achieved by playing 27 matches and winning all of them.Another way to think about it: the current number of losses/draws is 45 -27=18. If we want the losses/draws to be 25% of the total matches (since 75% are wins), then 18 + (y -x) = 0.25*(45 + y). Because the total losses/draws would be original 18 plus the new losses/draws, which is y -x (since they played y matches and won x, so lost y -x). So:18 + (y -x) = 0.25*(45 + y)Simplify:18 + y -x = 0.25*45 + 0.25y18 + y -x = 11.25 + 0.25yBring all terms to left:18 + y -x -11.25 -0.25y =06.75 +0.75y -x =0So:x=6.75 +0.75yWe want to minimize x. Since x and y must be integers (number of matches), and y >=x (can't win more than played), so substitute x=0.75y +6.75. Since x must be an integer, 0.75y +6.75 must be integer. Let's solve for y.Let me express 0.75y as (3/4)y. So:(3/4)y + 6.75 = xMultiply both sides by 4:3y +27 =4xThus, 4x =3y +27. So 3y =4x -27. Therefore, y=(4x -27)/3. Since y must be an integer >=x, let's find x such that (4x -27) is divisible by 3, and (4x -27)/3 >=x.First, 4x -27 ≡0 mod3 →4x ≡27 mod3 →4x ≡0 mod3 (since 27≡0 mod3). Since 4≡1 mod3, so x≡0 mod3. So x must be a multiple of 3.Let x=3k. Then y=(4*3k -27)/3=(12k -27)/3=4k -9.And we need y >=x →4k -9 >=3k →k >=9.So k must be at least 9, which makes x=3*9=27, y=4*9 -9=27. So x=27, y=27.If k=10, x=30, y=40 -9=31, which is y=31 >=30. But x=30 is larger than 27, so 27 is the minimal x.Therefore, the minimum number of matches they need to win is 27.

✅:The soccer team has won 60% of 45 matches, totaling 27 wins. To achieve a 75% overall winning percentage, let ( x ) be the minimum additional wins needed, assuming they play ( y geq x ) more matches. The equation becomes:[frac{27 + x}{45 + y} geq 0.75]To minimize ( x ), assume ( y = x ) (winning all remaining matches):[frac{27 + x}{45 + x} = 0.75]Solving for ( x ):1. Cross-multiply: ( 27 + x = 0.75(45 + x) ) 2. Expand: ( 27 + x = 33.75 + 0.75x ) 3. Rearrange: ( 0.25x = 6.75 ) 4. Solve: ( x = 27 )Final Answer: The team must win at least (boxed{27}) more matches.

🔑:Given: A football team has won (60%) of 45 matches played. To find: The minimum number of matches the team still needs to win to achieve a (75%) win rate.1. Initially, calculate the number of matches won so far: [ text{Number of matches won} = 0.60 times 45 = 27 ]2. Let (n) be the additional number of matches the team needs to win. The total number of matches played would then be (45 + n) and the total number of matches won would be (27 + n).3. Set up the equation to express the desired win rate of (75%): [ frac{27 + n}{45 + n} = 0.75 ]4. Convert the percentage to a fraction: [ 0.75 = frac{3}{4} ]5. Substitute this into the equation: [ frac{27 + n}{45 + n} = frac{3}{4} ]6. To eliminate the fraction, cross-multiply: [ 4(27 + n) = 3(45 + n) ]7. Simplify and solve for (n): [ 108 + 4n = 135 + 3n ] Subtract (3n) from both sides: [ 108 + n = 135 ] Subtract 108 from both sides: [ n = 27 ]Conclusion: The team needs to win at least ( boxed{27} ) more matches to achieve a (75%) success rate.

🧠:Okay, let's tackle this problem step by step. So, we have two boys, Ruslan and Kolya, who saved coins of 1, 2, and 5 rubles. But here's the catch: Ruslan's piggy bank doesn't contain any coins of the same denomination as Kolya's. That means if Kolya has, say, 1-ruble coins, Ruslan doesn't have any 1-ruble coins, and similarly for 2 and 5 rubles. So their coin denominations are completely disjoint. The question is whether both of them can pay exactly 2006 rubles using the same number of coins. We need to figure out if such a scenario is possible given the constraints on their coin denominations.First, let me rephrase the problem to make sure I understand it. Ruslan and Kolya each have coins, but they don't share any denominations. For example, if Ruslan has 1 and 2 ruble coins, then Kolya can only have 5 ruble coins. Or if Kolya has 1 and 5 ruble coins, Ruslan must have only 2 ruble coins. The key is that the denominations they have are entirely different. Now, both need to pay 2006 rubles. The challenge is to see if they can do this using the same number of coins, regardless of the denominations they have. So, even though they have different coins, can they each make up 2006 rubles with an identical number of coins?Let me start by considering the possible combinations of denominations each could have. Since there are three denominations (1, 2, 5), and they must have non-overlapping sets, each boy must have a subset of these denominations, with no overlap. Possible cases for their denominations:1. Ruslan has 1 and 2 ruble coins; Kolya has 5 ruble coins.2. Ruslan has 1 and 5 ruble coins; Kolya has 2 ruble coins.3. Ruslan has 2 and 5 ruble coins; Kolya has 1 ruble coins.4. Ruslan has only 1 ruble coins; Kolya has 2 and 5.5. Ruslan has only 2 ruble coins; Kolya has 1 and 5.6. Ruslan has only 5 ruble coins; Kolya has 1 and 2.But wait, since they can't have any overlapping denominations, each must have a subset of the three, and their subsets must be complements of each other. So, for example, if one has two denominations, the other must have the remaining one. If one has one denomination, the other must have the remaining two. So these are the possible scenarios.Now, the problem is for each of these scenarios, can both boys make 2006 rubles with the same number of coins? If even one scenario allows this, then the answer is yes. Otherwise, no.Let me consider each case one by one.Case 1: Ruslan has 1 and 2; Kolya has 5.So Ruslan can use 1s and 2s. To make 2006, he can use various combinations. Let's see the minimal number of coins he would need. The maximum coins would be 2006 1-ruble coins, but we need to see if he can use the same number of coins as Kolya.Kolya can only use 5s. So for Kolya to make 2006, he needs to use 5-ruble coins. But 2006 divided by 5 is 401.2, which is not an integer. So Kolya can't make 2006 with only 5s. Therefore, this case is impossible. Because Kolya can't form 2006 with only 5s. So this case is invalid.Case 2: Ruslan has 1 and 5; Kolya has 2.Ruslan can use 1s and 5s. Let's see if he can make 2006. Similarly, Kolya can only use 2s. For Kolya to make 2006, he needs 2006 / 2 = 1003 coins. So Kolya needs exactly 1003 coins. So the question is, can Ruslan make 2006 rubles using 1003 coins with 1s and 5s?Let's check. Let x be the number of 5s, y the number of 1s. Then:5x + y = 2006x + y = 1003Subtract the second equation from the first:4x = 2006 - 1003 = 1003So x = 1003 / 4 = 250.75But x must be an integer. So no solution here. Therefore, Ruslan cannot make 2006 with 1003 coins in this case. So this case also doesn't work.Case 3: Ruslan has 2 and 5; Kolya has 1.Kolya has only 1s, so to make 2006, he needs 2006 coins. So the number of coins here is 2006. Then Ruslan needs to make 2006 rubles with 2s and 5s using 2006 coins. Let's check if that's possible.Let x be number of 5s, y number of 2s.5x + 2y = 2006x + y = 2006Subtracting:5x + 2y - (x + y) = 2006 - 20064x + y = 0But x and y are non-negative integers. The only solution is x=0, y=0, but that gives 0 rubles. Not possible. So no solution here. So this case also fails.Case 4: Ruslan has only 1s; Kolya has 2 and 5.Ruslan has only 1s, so to pay 2006 he needs 2006 coins. Kolya has 2s and 5s and needs to make 2006 with 2006 coins. Let's check if Kolya can do that.Let x be 5s, y be 2s.5x + 2y = 2006x + y = 2006Subtracting:5x + 2y - x - y = 2006 - 20064x + y = 0Again, same as before, only solution x=0, y=0. Not possible. So no.Case 5: Ruslan has only 2s; Kolya has 1 and 5.Ruslan needs to make 2006 with only 2s. 2006 /2 = 1003 coins. So he uses 1003 coins. Kolya needs to make 2006 with 1s and 5s using 1003 coins. Let's check.Let x be 5s, y be 1s.5x + y = 2006x + y = 1003Subtract:4x = 2006 - 1003 = 1003x = 1003 /4 = 250.75. Not integer. So no solution.Case 6: Ruslan has only 5s; Kolya has 1 and 2.Ruslan needs to make 2006 with 5s. 2006 /5 = 401.2. Not integer. So impossible. So this case fails.Hmm. So all the cases where one has two denominations and the other has one seem to fail. Let's check if there's another possibility where each has two denominations. Wait, but the problem states that Ruslan's piggy bank does not contain coins of the same denomination as Kolya's. That means their sets of denominations are disjoint. Since there are only three denominations, the maximum they can have is two each, but that would require four different denominations, which doesn't exist. Therefore, one must have two denominations, and the other must have the remaining one. So the six cases above are exhaustive.But in all those cases, we saw that it's impossible. Wait, but let me double-check. Maybe there's a case where they have different numbers, but the same number of coins. Wait, for example, in case 2, Kolya has 2s, needs to make 2006 with 1003 coins. Ruslan has 1 and 5s, tried to make 2006 with 1003 coins but couldn't. But maybe there's a different approach?Wait, maybe there are other possibilities where the number of coins is the same but not necessarily minimal? For example, maybe they both use more coins than the minimal required. But in the cases above, we considered the required number based on the other's minimal coins. Wait, no, actually the problem says "using the same number of coins", not necessarily the minimal. So perhaps there's a way for both to use the same number, even if it's more than the minimal required.Wait, let's reconsider case 2. Ruslan has 1 and 5s, Kolya has 2s. Suppose Kolya uses 1003 coins of 2s to make 2006. So Kolya's coins: 1003. Then Ruslan needs to make 2006 with 1003 coins using 1s and 5s. But when we tried that, we found that it's impossible because x would need to be 250.75. But maybe there's another way? Wait, but the equations are exact. If you have x + y = 1003 and 5x + y = 2006, then subtracting gives 4x = 1003, which is not divisible by 4. Therefore, no solution. So that's impossible.Similarly in case 5, where Ruslan has 2s, needs 1003 coins, Kolya has 1 and 5s and also needs to make 2006 in 1003 coins. But that leads to the same problem. So no.What about if we consider that maybe they use more coins than the minimal? For example, suppose in case 1: Ruslan has 1 and 2s, Kolya has 5s. But Kolya can't make 2006 with 5s. So that case is out.Alternatively, maybe there's a case where both have two denominations each? Wait, but the problem states that their denominations are disjoint. Since there are only three denominations, if each has two, that would require four different denominations, which don't exist. So no, that's impossible. So only cases where one has two denominations and the other has one.But all those cases lead to impossible equations. Therefore, the answer is no, it's not possible.Wait, but let's think again. Maybe there's a trick here. Let's think modulo some number.For example, in case where Kolya has 2s and 5s (if that's possible). Wait, but according to the problem, the denominations are disjoint. Wait, no, in the possible cases, if one has two denominations, the other has the remaining one. Wait, but in case 4, Ruslan has 1s, Kolya has 2 and 5. Then Kolya can use 2s and 5s. Let's see if he can make 2006 with some number of coins, and Ruslan makes 2006 with 1s, which would require 2006 coins. So if Kolya can make 2006 with 2006 coins using 2s and 5s, then yes. But let's check:Kolya needs 5x + 2y = 2006 and x + y = 2006.Then:From x + y = 2006, y = 2006 - x.Substitute into the first equation:5x + 2(2006 - x) = 20065x + 4012 - 2x = 20063x = 2006 - 4012 = -2006x = -2006/3, which is negative. Impossible. So no solution here.Similarly, other cases also lead to negative or fractional solutions.Wait, maybe there's a different way. Suppose they don't have to use all their coins, but just use some coins to sum up to 2006. The problem doesn't state that they have to use all their coins, just that they can pay 2006. So maybe the denominations they have are such that they can form 2006 with some combination.But the problem is, given their denominations are complementary, whether the same number of coins can be used.Wait, let's consider another angle. Let's suppose that both boys can use exactly N coins to make 2006. We need to find if such an N exists where both can make 2006 with N coins given their disjoint denominations.So instead of looking at individual cases, perhaps we can model this as a system of equations.Let me denote the denominations Ruslan has as R1, R2 (if he has two) and Kolya's as K1, K2, etc., but since they are disjoint, and from the set {1,2,5}.Suppose Ruslan has denominations A and B, and Kolya has denomination C, where {A,B,C} = {1,2,5}.Then, we need:For Ruslan: a*A + b*B = 2006, with a + b = NFor Kolya: c*C = 2006, with c = NSo Kolya's equation is straightforward: C*N = 2006. Therefore, N must be 2006 / C. Since C is either 1, 2, or 5. So N would have to be 2006, 1003, or 401.2. But N must be integer, so if C=5, it's impossible. If C=2, then N=1003. If C=1, N=2006.Similarly, for Ruslan, if he has denominations A and B (two coins), then he needs to make 2006 with N coins. So we can write equations for each possible pair.Case when Kolya has C=2 (so N=1003):Ruslan would have denominations 1 and 5. Then:a + 5b = 2006a + b = 1003Subtract: 4b = 1003 => b = 1003/4 = 250.75. Not integer. Impossible.Case when Kolya has C=1 (N=2006):Ruslan has denominations 2 and 5.Then:2a + 5b = 2006a + b = 2006Multiply the second equation by 2: 2a + 2b = 4012Subtract from first equation: 3b = -2006 => b negative. Impossible.Case when Kolya has C=5 (N=401.2): Not integer. Impossible.Alternatively, if Kolya has two denominations, say C and D, then Ruslan has the remaining one.Wait, but earlier analysis showed that if Kolya has two denominations, Ruslan has the remaining one, which would require Ruslan to make 2006 with one denomination. Let's check.Suppose Ruslan has only 1s (denomination 1), Kolya has 2 and 5.Ruslan needs 2006 coins. Kolya needs to make 2006 with 2006 coins using 2 and 5. But:5x + 2y = 2006x + y = 2006Then as before, 3x = -2006, impossible.Similarly, if Ruslan has only 2s: 2006 /2 = 1003 coins. Kolya has 1 and 5, needs to make 2006 with 1003 coins:x + 5y = 2006x + y = 1003Subtract: 4y = 1003, y=250.75. No.If Ruslan has only 5s: 2006/5 = 401.2. Not possible.Alternatively, maybe there's a case where both have one denomination each, but different. For example, Ruslan has 1s, Kolya has 2s. Then Ruslan needs 2006 coins, Kolya needs 1003 coins. Not same number. If Ruslan has 2s, Kolya has 5s. Ruslan needs 1003 coins, Kolya can't make 2006 with 5s. Similarly other combinations.Wait, but perhaps there's a way where they use different combinations of their available denominations but still end up with the same number of coins. Let's think outside the cases.Suppose Ruslan has 1 and 5, Kolya has 2. Can they both use N coins to make 2006?For Kolya: 2*N = 2006 => N=1003For Ruslan: 1*a +5*b =2006, a +b=1003Which gives a=1003 -bSo substituting: (1003 -b) +5b =2006 =>4b=1003 =>b=250.75. Not integer. No.If Ruslan has 2 and 5, Kolya has 1. Kolya needs 2006 coins. Ruslan needs 2006 coins of 2 and 5. But:2a +5b=2006a +b=2006Multiply second by 2: 2a +2b=4012Subtract: 3b= -2006 => Impossible.If Ruslan has 1 and 2, Kolya has5. Kolya can't make 2006. So no.Alternatively, suppose they have different combinations. Wait, but given the constraints, their denominations must be complementary. So the only possible pairs are as previously analyzed.Another angle: parity. 2006 is even. So if Kolya has 2s, he can make 2006. If he has 1s and 5s, then since 5 is odd, to make even total, he needs even number of 5s. Similarly, for Ruslan, if he has 1s and 5s, to make 2006 (even), he needs even number of 5s. But maybe this helps in some way.But we need to match the number of coins. Let's consider if both use even number of coins or odd. But I'm not sure.Wait, let's think about modulo 4.Suppose N is the number of coins for both.If Kolya has 2s: N=1003. Then Ruslan has 1 and 5s. The total for Ruslan is 1*a +5*b=2006, a +b=1003.So a=1003 -b. Substitute into total:(1003 -b) +5b=2006 =>4b=1003. But 1003 modulo4 is 3, since 4*250=1000, 1003=4*250+3. So 4b=1003 =>b=1003/4=250.75. Not integer.If Kolya has 1s: N=2006. Ruslan has 2 and5s. Then:2a +5b=2006a +b=2006 =>a=2006 -bSubstitute:2*(2006 -b) +5b=2006 =>4012 -2b +5b=2006 =>3b= -2006 =>Impossible.If Kolya has 5s: impossible, since 2006/5 is not integer.Another approach: since the problem is asking if it's possible, maybe there's a way by considering that one of them can have two denominations and the other one, but with some specific combinations.Wait, let's suppose that Ruslan has 1 and 2, Kolya has 5. But Kolya can't make 2006. So that's out.If Ruslan has 1 and 5, Kolya has 2. We saw that N=1003 doesn't work.If Ruslan has 2 and 5, Kolya has 1. N=2006, but equations lead to negative.Alternatively, maybe the answer is no, but I need to check again.Wait, perhaps there's a non-integer solution, but coins have to be integer. So all cases lead to impossibility. Therefore, the answer is no, it's not possible.But wait, the question is "Can the boys each pay 2006 rubles from their piggy banks using the same number of coins?" So the answer is no.But wait, let me check one more time. Suppose that the denominations are different. For example, Ruslan has 1 and 5, Kolya has 2. Is there a way for both to use the same number of coins? For Kolya, using 2s, he needs 1003 coins. Ruslan needs to use 1003 coins with 1 and5. But equations show it's impossible. Similarly other cases.Therefore, the conclusion is that it's impossible. So the answer is no.

✅:The boys cannot each pay 2006 rubles using the same number of coins. Here's why:1. Disjoint Denominations: Ruslan and Kolya have non-overlapping coin denominations. Since there are only three denominations (1, 2, 5), one must use two denominations, and the other must use the remaining one.2. Case Analysis: - Case 1: If Kolya uses 5-ruble coins, he cannot form 2006 rubles (2006 is not divisible by 5). - Case 2: If Kolya uses 2-ruble coins, he needs 1003 coins. Ruslan (with 1 and 5) would need to solve (5x + y = 2006) and (x + y = 1003). This leads to (4x = 1003), which is not an integer. - Case 3: If Kolya uses 1-ruble coins, he needs 2006 coins. Ruslan (with 2 and 5) would need to solve (2a + 5b = 2006) and (a + b = 2006). This results in (3b = -2006), impossible for non-negative integers.3. All Other Cases similarly fail due to non-integer solutions or negative coin counts when matching the required total and coin numbers.Conclusion: No combination of disjoint denominations allows both boys to use the same number of coins to sum to 2006 rubles. boxed{text{No}}

🔑:To check if Ruslan and Kolya can each pay 2006 rubles using the same number of coins but of different denominations (1, 2, 5 rubles), we need to analyze the conditions provided.1. Ruslan and Kolya have no coins of the same denomination: This means that if one has coins of 1 ruble, the other doesn't have any coins of 1 ruble, and similarly for other denominations.2. Let's assume they both need to make 2006 rubles using the same number of coins: Let ( k ) be the number of coins each of them uses.3. Analysis assuming possible denominations: - If Kolya has 5-ruble coins, then: [ 5 times k = 2006 ] Clearly, ( 2006 / 5 = 401.2 ), which is not an integer. Thus, Kolya cannot pay using only 5-ruble coins. - If Kolya has 1-ruble coins, then: [ 1 times k = 2006 ] This makes ( k = 2006 ), meaning Kolya has 2006 coins of 1 ruble each. - Then Ruslan must use a combination of 2-ruble and 5-ruble coins without any 1-ruble coins. The total number of coins must still be 2006. But the total value cannot be formed precisely because: - The smallest denomination (2 rubles) combined would quickly reach a number exceeding 2006 rubles if other coins are included. - Alternatively, Ruslan could not just use 5-ruble coins, as previously shown. - Assume Kolya has 2-ruble coins: [ 2 times k = 2006 ] This makes ( k = 1003 ). So Kolya needs 1003 coins of 2 rubles each.4. Result of each scenario: - Kolya successfully creates 2006 rubles using 1003 coins of 2 rubles each. - If Ruslan needs to make a combination of 1 and 5-ruble coins: [ x times 1 + y times 5 = 2006 ] [ x + y = 1003 ] Expressing it in terms of even and odd counts: - With a mixed count of both 1-ruble (odd) and 5-ruble (odd) coins, the resulting ( x ) and ( y ) must alternatively sum to 2006 rubles, which is an even value. However: [ x + y = text{odd } k = 1003 ] This contradiction signifies the impossibility of forming 2006 rubles with an exact even mix meeting odd counts of either denominations. Conclusion:Based on the above reasoning, it's clear that Ruslan and Kolya cannot both provide 2006 rubles using the same number of coins which are of different denominations.# Final answer:[boxed{text{No, they cannot.}}]

🧠:Okay, so I need to figure out the number of ways to go from (0,0) to (m,n) by only moving right or up one unit at a time. Let me start by visualizing the grid. If I'm at (0,0), and I want to get to (m,n), I have to move a total of m steps to the right and n steps up, right? Because each right step increases the x-coordinate by 1, and each up step increases the y-coordinate by 1. So, to get from (0,0) to (m,n), I need exactly m right steps and n up steps in some order. Wait, so the total number of steps is m + n. And the problem is essentially asking how many different sequences of these steps there are. Since each step is either right (R) or up (U), the number of unique paths should be the number of ways to arrange m R's and n U's in a sequence. Hmm, that sounds like a permutation problem with repeated elements. The formula for permutations of multiset is (m + n)! divided by (m!n!), right? Let me confirm that. For example, if I have a total of k items, where there are k1 of one type, k2 of another, etc., the number of distinct permutations is k! divided by the product of the factorials of each type. So in this case, we have (m + n) steps, with m of one type (R) and n of another (U). So the number of ways should be (m + n)! / (m!n!). But let me make sure I'm not missing something. Maybe I can test with a small example. Let's say m = 2 and n = 2. So moving from (0,0) to (2,2). How many paths are there? Let's list them. Each path would be a sequence of 2 R's and 2 U's. The possible permutations are:1. R, R, U, U2. R, U, R, U3. R, U, U, R4. U, R, R, U5. U, R, U, R6. U, U, R, RSo that's 6 paths. Applying the formula: (2 + 2)! / (2!2!) = 4! / (2!2!) = 24 / 4 = 6. That matches. Okay, another example: m = 1, n = 1. Then the number of paths should be (1 + 1)! / (1!1!) = 2 / 1 = 2. Indeed, from (0,0) you can go right then up, or up then right. That works too.Wait, maybe another example where m and n are different. Let's take m = 3, n = 2. Then the formula gives (3 + 2)! / (3!2!) = 120 / (6 * 2) = 10. Let me see if that makes sense. The number of sequences with 3 R's and 2 U's. For example, the first position could be R, then the remaining steps are 2 R's and 2 U's. The number of permutations would be 5 choose 3 (since we need to choose positions for the R's), which is 10. Yep, that's the same as 5! / (3!2!). So that checks out.Alternatively, thinking in terms of combinations: choosing which steps are right (or equivalently which are up). To get from (0,0) to (m,n), you have to make a total of m + n moves, and you need to choose m of them to be right moves (the rest will automatically be up moves). The number of ways to choose m steps from (m + n) is the binomial coefficient C(m + n, m) which is equal to (m + n)! / (m!n!). So both approaches confirm the same formula.But maybe there's another way to think about it recursively. Suppose the number of ways to get to (m,n) is equal to the number of ways to get to (m-1,n) plus the number of ways to get to (m,n-1). Because the last step before reaching (m,n) must be either from the left (m-1,n) or from below (m,n-1). So the total number is the sum of those two. This recursive approach makes sense. For example, in the case of (2,2), the number of ways to get there would be the number of ways to (1,2) plus the number of ways to (2,1). Let's see: number of ways to (1,2) is C(3,1) = 3, and number of ways to (2,1) is C(3,2) = 3. So 3 + 3 = 6, which matches. Similarly, for (1,1), it's the sum of (0,1) and (1,0). Each of those has only 1 way, so 1 + 1 = 2. That works too. This recursive formula is actually the same as the combinatorial formula because the binomial coefficients satisfy the same recurrence: C(m + n, m) = C(m + n - 1, m - 1) + C(m + n - 1, m). So that's consistent. But maybe the user is expecting a more detailed explanation of why the combination formula works here. Let me elaborate. Each path can be represented as a sequence of moves. Since we have to make exactly m right moves and n up moves, the problem reduces to arranging these moves in any order. The number of distinct arrangements is given by the binomial coefficient, which counts the number of ways to choose positions for the right moves (or equivalently the up moves) in the sequence of total moves. Another perspective: imagine you have a string composed of m R's and n U's. Each unique string corresponds to a unique path. The number of such strings is the same as the number of combinations of (m + n) moves taken m at a time (or n at a time), which is again (m + n)! / (m!n!). Alternatively, thinking about the grid. To reach (m,n), every path must cross a certain number of vertical and horizontal lines. But each path is a lattice path consisting of right and up moves only. The number of such paths is a well-known result in combinatorics, often referred to as the number of "lattice paths" or "Dyck paths" when considering certain restrictions, but in this case without restrictions, it's straightforward. Wait, Dyck paths usually refer to paths that stay above the diagonal, but in this case, we're not restricted. So it's just a general lattice path problem. The formula is indeed the binomial coefficient. Is there a way to see this using Pascal's triangle? For example, each point (m,n) can be associated with the entry in Pascal's triangle, but since we're moving in two dimensions, the number of ways to reach (m,n) is the sum of the ways to reach (m-1,n) and (m,n-1), which is the same recursive formula as Pascal's triangle. Therefore, building up the grid, each entry is the sum of the entry to the left and the entry below, leading to the binomial coefficient at position (m + n choose m). Yes, that's another way to visualize it. If we create a grid where each cell (i,j) contains the number of ways to reach (i,j) from (0,0), then the grid would be constructed such that each cell is the sum of the cell to the left and the cell below. This grid is essentially Pascal's triangle extended to two dimensions, and the value at (m,n) is indeed C(m + n, m). Let me confirm with another example. Suppose m = 3, n = 1. Then the number of paths should be C(4,3) = 4. The possible sequences are: R, R, R, U; R, R, U, R; R, U, R, R; U, R, R, R. That's 4 paths. Correct. Similarly, if m = 0 or n = 0, there's only 1 path. For example, from (0,0) to (0,5), you can only move up 5 times, no choices involved. Similarly, to (5,0), only right moves. So the formula works there too: C(n,0) = 1, which is 1 way. Another edge case: m = 0 and n = 0. The number of paths should be 1, since you're already at the destination. Applying the formula: C(0,0) = 1. Which is correct. So, through various examples and reasoning approaches—combinatorial, recursive, Pascal's triangle—it seems that the formula holds up. Therefore, the number of ways to go from (0,0) to (m,n) moving only right or up is the binomial coefficient (m + n choose m), which equals (m + n)! / (m!n!). I think that covers it. But just to be thorough, is there a scenario where this might not apply? For example, if there are obstacles on the grid, but the problem statement doesn't mention any. So in the standard case without obstacles, the formula should be correct. Therefore, the final answer is the binomial coefficient C(m + n, m), which simplifies to (m + n)! divided by (m!n!).

✅:The number of ways to move from point ((0,0)) to ((m,n)) using only one-unit steps to the right (R) or upward (U) is determined by the number of distinct sequences of (m) R's and (n) U's. This is equivalent to choosing (m) positions for R's (or (n) positions for U's) out of (m + n) total steps, leading to the binomial coefficient:[binom{m+n}{m} = frac{(m + n)!}{m! , n!}]Step-by-Step Explanation:1. Total Steps: To reach ((m,n)), you need (m) right steps and (n) up steps, totaling (m + n) steps.2. Arrangements: Each unique path corresponds to a distinct permutation of these steps. The number of such permutations is given by the multinomial coefficient for two categories (R and U).3. Combinatorial Formula: The number of ways to choose (m) positions for R's (or equivalently (n) positions for U's) in a sequence of (m + n) steps is: [ binom{m+n}{m} = frac{(m + n)!}{m! , n!} ]4. Recursive Validation: The count also satisfies the recurrence relation: [ text{Paths}(m,n) = text{Paths}(m-1,n) + text{Paths}(m,n-1) ] with base cases (text{Paths}(k,0) = text{Paths}(0,k) = 1) for any (k).Final Answer:[boxed{dfrac{(m + n)!}{m! , n!}}]

🔑:1. Understand the Problem: - We begin on a grid at point (0, 0). - We must reach point (m, n) by making steps of one unit either to the right (denoted by "D") or upward (denoted by "H").2. Determine the Total Number of Steps: - To reach (m, n) from (0, 0), the total number of steps required is comprised of m steps to the right and n steps upwards. - Thus, we need a total of m + n steps.3. Formulating the Problem in Terms of Combinations: - Each path can be represented as a sequence of "D" and "H" where "D" represents a step to the right and "H" represents a step upwards. - We need exactly m "D" steps and n "H" steps in our path. - Therefore, the problem reduces to finding the number of sequences of length m + n that have exactly m "D" steps (and hence, by default, n "H" steps).4. Using Combinatorics: - The number of such sequences (or paths) is the number of ways to choose m positions out of m + n total positions for the "D" steps. The rest of the positions will automatically be filled with "H" steps. - Mathematically, this is given by the binomial coefficient: [ binom{m+n}{m} ] - Alternatively, it can also be expressed as binom{m+n}{n}, since choosing m positions out of m+n for "D" is equivalent to choosing n positions out of m+n for "H".5. Conclusion: - The number of distinct paths from (0, 0) to (m, n) on the grid, moving only right or up, is given by the binomial coefficient binom{m+n}{m}.[boxed{binom{m+n}{m}}]

🧠:Okay, so I need to show that for any integer n where n is at least 2, the expression 4n² - 1 is not a prime number. Hmm, let me start by understanding what the problem is asking. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So if 4n² - 1 isn't prime, that means it must be composite, right? Which implies it has factors other than 1 and itself. First, maybe I should try plugging in some values of n starting from 2 to see if the pattern holds. Let me test n=2: 4*(2)^2 -1 = 4*4 -1 = 16 -1 =15. 15 is 3*5, so composite. Okay, that's not prime. Next, n=3: 4*9 -1=36-1=35. 35 is 5*7, composite. n=4: 4*16 -1=64-1=63, which is 7*9=63. Also composite. n=5: 4*25 -1=100-1=99=9*11. Hmm, so far all these are composite. So the expression seems to generate composite numbers for n≥2. But how do I prove this in general? Maybe there's a way to factor the expression 4n² -1. Let me think. 4n² is a square term, so perhaps this is a difference of squares? Wait, 4n² is (2n)^2. So 4n² -1 is (2n)^2 - (1)^2. Oh! That's a difference of squares. And the difference of squares can be factored as (a - b)(a + b). So in this case, that would be (2n -1)(2n +1). Therefore, 4n² -1 factors into (2n -1)(2n +1). So unless one of those factors is 1, the number would be composite. But when n ≥2, let's check the factors. For n=2, (2*2 -1)=3 and (2*2 +1)=5, both greater than 1. For n=3, 5 and 7, etc. So both factors are integers greater than 1 when n≥2. Therefore, 4n² -1 is the product of two numbers each greater than 1, which means it's composite. Hence, it can't be prime. Wait, let me verify if there's any case where one of the factors could be 1. Suppose 2n -1 =1. Solving for n gives 2n=2 => n=1. But n has to be at least 2, so n=1 is excluded. Similarly, 2n +1=1 would imply 2n=0 => n=0, which is also not in the domain. Therefore, for n≥2, both factors are at least 3 and 5, respectively. So the product is definitely composite. Is there another way to approach this? Maybe by induction? Let's see. Base case: n=2, as above, 15 which is composite. Assume that for some k≥2, 4k² -1 is composite. Then for k+1, 4(k+1)^2 -1. But expanding this gives 4(k² + 2k +1) -1 =4k² +8k +4 -1=4k² +8k +3. Hmm, not sure if that helps. The inductive step might not be straightforward here. Alternatively, maybe using modular arithmetic? Let's see. 4n² -1 mod 3. Let's compute for some n. Wait, but perhaps that's complicating things. The factoring approach seems much more straightforward. Wait, another thought: since 4n² -1 = (2n -1)(2n +1), and both factors are consecutive odd numbers. For n≥2, the factors are 3 and 5, 5 and 7, 7 and 9, etc. Since they are two apart, they are both odd and their product is 4n² -1. Since both factors are greater than 1, their product cannot be prime. So the key insight here is recognizing the expression as a difference of squares, factoring it, and then showing that for the given domain of n, both factors are necessarily greater than 1. Therefore, the product must be composite. I should also check if there's a possibility that one of the factors could be -1, but since n is a positive integer greater than or equal to 2, both (2n -1) and (2n +1) are positive. So no issues there. Another edge case: when n=1, 4(1)^2 -1=3, which is prime. But the problem specifies n≥2, so that's okay. Therefore, the proof is straightforward once the factoring is recognized. So maybe the answer is simply factoring the expression into two integers greater than 1 for n≥2, hence it's composite.

✅:To demonstrate that ( 4n^2 - 1 ) is not prime for any integer ( n geq 2 ), observe the following:1. Factor the expression: ( 4n^2 - 1 ) is a difference of squares, which factors as: [ 4n^2 - 1 = (2n)^2 - 1^2 = (2n - 1)(2n + 1). ]2. Analyze the factors: For ( n geq 2 ): - ( 2n - 1 geq 3 ) (since ( 2(2) - 1 = 3 )), - ( 2n + 1 geq 5 ). Both factors are integers greater than 1.3. Conclusion: Since ( 4n^2 - 1 ) is the product of two integers ( (2n - 1) ) and ( (2n + 1) ), each exceeding 1 for ( n geq 2 ), the expression cannot be prime. It is necessarily composite.Final Answer: For all integers ( n geq 2 ), ( 4n^2 - 1 ) factors as ( (2n - 1)(2n + 1) ), both of which are greater than 1, proving it is not prime. (boxed{4n^2 - 1 text{ is composite for } n geq 2}).

🔑:1. Consider the expression (4n^2 - 1). Observe that it can be factored using the difference of squares formula: [ 4n^2 - 1 = (2n)^2 - 1^2 = (2n-1)(2n+1). ]2. Analyze the factors (2n-1) and (2n+1): - Since (n) is an integer such that (n geqslant 2), then (2n-1) is calculated as follows: [ 2n - 1 geqslant 2 times 2 - 1 = 4 - 1 = 3. ] - Also, for (2n+1), we have: [ 2n + 1 geqslant 2 times 2 + 1 = 4 + 1 = 5. ] 3. Notice that both (2n-1) and (2n+1) are greater than 1: - Specifically, we find that: [ 2n-1 geqslant 3 > 1 quad text{and} quad 2n+1 geqslant 5 > 1. ]4. Since (4n^2 - 1) is a product of two integers, (2n-1) and (2n+1), both greater than 1, it implies that: [ 4n^2 - 1 = (2n-1)(2n+1) ] cannot be a prime number because a prime number must have exactly two distinct positive divisors, 1 and itself. Conclusion:[boxed{4n^2 - 1 text{ is not prime}}]