answer：Okay, so I have this problem here: Let y be a non-zero natural number. I need to show that y - 1 divides y raised to the power of y squared minus 2 times y raised to the power of y plus 1, and then plus 1. Hmm, that sounds a bit complicated, but let me try to break it down.First, I know that when we say y - 1 divides some expression, it means that when we divide that expression by y - 1, there's no remainder. In other words, the expression is a multiple of y - 1. So, another way to think about it is that if I plug in y = 1 into the expression, it should equal zero because y - 1 would be zero, and if y - 1 divides the expression, then the expression should be zero when y = 1.Let me test that. If y = 1, then the expression becomes 1^(1^2) - 2*1^(1+1) + 1. Simplifying that, 1^1 is 1, 1^2 is also 1, so it becomes 1 - 2*1 + 1, which is 1 - 2 + 1, and that equals 0. Okay, so that checks out. When y = 1, the expression is zero, which means y - 1 divides the expression when y = 1. But I need to show that this is true for any non-zero natural number y, not just y = 1.Maybe I can use modular arithmetic here. If I can show that the expression is congruent to 0 modulo y - 1, then that would mean y - 1 divides the expression. So, let's consider the expression modulo y - 1.I remember that in modular arithmetic, if I have y ≡ 1 mod (y - 1), because y - 1 is the modulus. So, any power of y would also be congruent to 1 raised to that power modulo y - 1. That is, y^k ≡ 1^k ≡ 1 mod (y - 1) for any integer k.Let me apply that to each term in the expression. The first term is y^(y^2). Since y ≡ 1 mod (y - 1), then y^(y^2) ≡ 1^(y^2) ≡ 1 mod (y - 1). Similarly, the second term is 2*y^(y+1). Again, y^(y+1) ≡ 1^(y+1) ≡ 1 mod (y - 1), so 2*y^(y+1) ≡ 2*1 ≡ 2 mod (y - 1). The last term is just 1, so that's 1 mod (y - 1).Putting it all together, the entire expression modulo y - 1 is 1 - 2 + 1. Let's compute that: 1 - 2 is -1, and -1 + 1 is 0. So, the expression is congruent to 0 modulo y - 1, which means that y - 1 divides the expression.Wait, let me double-check that. If I have y ≡ 1 mod (y - 1), then any power of y is also 1 mod (y - 1). So, y^(y^2) is 1, 2*y^(y+1) is 2*1 = 2, and then plus 1. So, 1 - 2 + 1 is indeed 0. Yeah, that seems right.Is there another way to approach this? Maybe using polynomial division or factoring? Let me think. If I consider the expression y^(y^2) - 2y^(y+1) + 1, it's a bit unwieldy because of the exponents. But perhaps I can factor it somehow or find a pattern.Alternatively, I could use the Remainder Theorem, which states that the remainder of a polynomial f(y) divided by y - a is f(a). In this case, if I consider the expression as a polynomial in y, then the remainder when divided by y - 1 should be f(1). As I calculated earlier, f(1) is 0, so y - 1 is a factor of the polynomial. That aligns with what I found using modular arithmetic.Another thought: maybe I can rewrite the expression in terms of (y - 1). Let me see. If I let z = y - 1, then y = z + 1. Substituting that into the expression might make it easier to see if z divides the expression.So, replacing y with z + 1, the expression becomes (z + 1)^((z + 1)^2) - 2*(z + 1)^((z + 1) + 1) + 1. That looks even more complicated, though. Maybe expanding it isn't the best approach.Perhaps instead of substitution, I can look for patterns or use mathematical induction. If I can show that the statement holds for some base case and then assume it holds for y = k and prove it for y = k + 1, that might work. But induction might be tricky here because of the exponents involved.Wait, another idea: maybe I can factor the expression. Let me see if there's a way to factor y^(y^2) - 2y^(y+1) + 1. It kind of resembles a quadratic in terms of y^(y+1). Let me set u = y^(y+1). Then the expression becomes u^y - 2u + 1. Hmm, not sure if that helps. Maybe not.Alternatively, think about the expression as y^(y^2) - 2y^(y+1) + 1. Maybe I can factor out y^(y+1) from the first two terms: y^(y+1)*(y^(y^2 - y - 1) - 2) + 1. That doesn't seem helpful either.Perhaps another approach is to consider the expression modulo y - 1, which I already did, and since it's congruent to 0, that's sufficient. Maybe that's the simplest way.Wait, let me think again about the modular approach. Since y ≡ 1 mod (y - 1), then any power of y is also 1 mod (y - 1). So, y^(y^2) ≡ 1, y^(y+1) ≡ 1, so 2*y^(y+1) ≡ 2. Then the expression becomes 1 - 2 + 1 ≡ 0 mod (y - 1). Yep, that seems solid.I guess I'm overcomplicating it. The modular arithmetic approach is straightforward and sufficient. I don't need to go into polynomial division or factoring because the modular method already shows that the expression is divisible by y - 1.Just to be thorough, let me test it with a specific value of y, say y = 2. Then y - 1 = 1, and the expression becomes 2^(2^2) - 2*2^(2+1) + 1 = 2^4 - 2*2^3 + 1 = 16 - 16 + 1 = 1. Wait, 1 divided by 1 is 1, which is an integer, so it works.Another test with y = 3. Then y - 1 = 2. The expression is 3^(3^2) - 2*3^(3+1) + 1 = 3^9 - 2*3^4 + 1. Let's compute that: 3^9 is 19683, 3^4 is 81, so 2*81 is 162. So, 19683 - 162 + 1 = 19683 - 161 = 19522. Now, divide 19522 by 2: 19522 / 2 = 9761, which is an integer. So, it works for y = 3.Let me try y = 4. y - 1 = 3. The expression is 4^(4^2) - 2*4^(4+1) + 1 = 4^16 - 2*4^5 + 1. 4^16 is a huge number, 4294967296. 4^5 is 1024, so 2*1024 is 2048. So, 4294967296 - 2048 + 1 = 4294965249. Now, divide that by 3: 4294965249 / 3. Let's see, 3 goes into 4 once, remainder 1. Bring down the 2: 12, 3 goes into 12 four times. Bring down the 9: 3 goes into 9 three times. Bring down the 4: 3 goes into 4 once, remainder 1. Bring down the 9: 19, 3 goes into 19 six times, remainder 1. Bring down the 6: 16, 3 goes into 16 five times, remainder 1. Bring down the 5: 15, 3 goes into 15 five times. Bring down the 2: 2, 3 goes into 2 zero times, remainder 2. Bring down the 4: 24, 3 goes into 24 eight times. Bring down the 9: 9, 3 goes into 9 three times. So, the division results in 1431655083 with no remainder. So, it works for y = 4 as well.Okay, so through both the modular arithmetic approach and testing specific cases, it seems that y - 1 indeed divides the given expression. I think I've got it. The key insight was recognizing that y ≡ 1 mod (y - 1), which simplifies the expression significantly when considering it modulo y - 1.

answer：Alright, so I've got this problem here where I need to find the interval where the zero of the function ( f(x) = a^x + x - b ) lies. The given conditions are that ( 2^a = 3 ) and ( 3^b = 2 ). The options are intervals from -2 to 2, so I need to figure out which one contains the zero of this function.First, let me try to understand what ( a ) and ( b ) are. The equation ( 2^a = 3 ) means that ( a ) is the exponent to which we raise 2 to get 3. In logarithmic terms, that would be ( a = log_2{3} ). Similarly, ( 3^b = 2 ) implies that ( b = log_3{2} ).I remember that logarithms can be converted using the change of base formula, so ( log_2{3} ) is the same as ( frac{ln{3}}{ln{2}} ) and ( log_3{2} ) is ( frac{ln{2}}{ln{3}} ). This might be useful later.Now, let's think about the function ( f(x) = a^x + x - b ). I need to find where this function crosses zero, meaning where ( f(x) = 0 ). To do this, I can evaluate ( f(x) ) at different points and see where it changes sign, which would indicate a zero crossing.Before plugging in values, I should analyze the behavior of ( f(x) ). The function ( a^x ) is an exponential function. Since ( a = log_2{3} ), which is approximately ( 1.58496 ), this is greater than 1. So, ( a^x ) is an increasing function because the base is greater than 1. The term ( x ) is a linear function with a slope of 1, so it's also increasing. Therefore, the sum ( a^x + x ) is increasing. Subtracting ( b ) from this sum will shift the graph down, but it won't affect the monotonicity—meaning the function ( f(x) ) is still increasing.Since ( f(x) ) is monotonically increasing, it can cross the x-axis at most once. So, if I can find two points where ( f(x) ) has opposite signs, the zero must lie between those two points.Let me compute ( f(x) ) at some integer points within the given intervals.First, let's compute ( f(0) ):( f(0) = a^0 + 0 - b = 1 + 0 - b = 1 - b ).Since ( b = log_3{2} ), which is approximately ( 0.6309 ), so ( 1 - b ) is approximately ( 0.3691 ), which is positive. So, ( f(0) > 0 ).Next, let's compute ( f(-1) ):( f(-1) = a^{-1} + (-1) - b = frac{1}{a} - 1 - b ).We know that ( a = log_2{3} ), so ( frac{1}{a} = frac{1}{log_2{3}} = log_3{2} = b ). Therefore, ( f(-1) = b - 1 - b = -1 ), which is negative. So, ( f(-1) < 0 ).Since ( f(-1) ) is negative and ( f(0) ) is positive, by the Intermediate Value Theorem, there must be a zero crossing between ( x = -1 ) and ( x = 0 ). Therefore, the zero of the function lies in the interval ( (-1, 0) ).Just to be thorough, let me check another point, say ( x = 1 ):( f(1) = a^1 + 1 - b = a + 1 - b ).We know ( a = log_2{3} approx 1.58496 ) and ( b = log_3{2} approx 0.6309 ). So, ( f(1) approx 1.58496 + 1 - 0.6309 approx 1.95406 ), which is positive. So, ( f(1) > 0 ).And ( x = -2 ):( f(-2) = a^{-2} + (-2) - b = frac{1}{a^2} - 2 - b ).Calculating ( frac{1}{a^2} approx frac{1}{(1.58496)^2} approx frac{1}{2.51189} approx 0.398 ). So, ( f(-2) approx 0.398 - 2 - 0.6309 approx -2.2329 ), which is negative. So, ( f(-2) < 0 ).But since ( f(-1) ) is already negative and ( f(0) ) is positive, the zero is between ( -1 ) and ( 0 ). Checking ( x = -0.5 ) just to get a better idea:( f(-0.5) = a^{-0.5} + (-0.5) - b = frac{1}{sqrt{a}} - 0.5 - b ).( sqrt{a} approx sqrt{1.58496} approx 1.258 ), so ( frac{1}{sqrt{a}} approx 0.795 ). Thus, ( f(-0.5) approx 0.795 - 0.5 - 0.6309 approx -0.3359 ), which is still negative.So, the function is negative at ( x = -1 ), negative at ( x = -0.5 ), and positive at ( x = 0 ). Therefore, the zero crossing is somewhere between ( x = -0.5 ) and ( x = 0 ), which is still within the interval ( (-1, 0) ).I think this confirms that the zero lies in interval B: ( (-1, 0) ).

answer：Okay, so I have this problem about a complex number z. The problem says that both z + 2i and z divided by (2 - i) are real numbers. Also, when I take (z + ai) squared, the resulting complex number lies in the first quadrant. I need to find the range of the real number a.Alright, let's start by recalling that a complex number can be written as z = m + ni, where m and n are real numbers. So, let me set z = m + ni.First, the problem states that z + 2i is a real number. Let's compute that:z + 2i = (m + ni) + 2i = m + (n + 2)i.Since this is a real number, the imaginary part must be zero. So, the coefficient of i must be zero:n + 2 = 0.Solving for n, we get n = -2. So, the imaginary part of z is -2. That means z = m - 2i.Next, the problem says that z divided by (2 - i) is also a real number. Let's compute that:z / (2 - i) = (m - 2i) / (2 - i).To simplify this, I'll multiply the numerator and the denominator by the conjugate of the denominator to rationalize it. The conjugate of (2 - i) is (2 + i). So,z / (2 - i) = [(m - 2i)(2 + i)] / [(2 - i)(2 + i)].Let me compute the denominator first:(2 - i)(2 + i) = 2*2 + 2*i - i*2 - i*i = 4 + 2i - 2i - i².Since i² = -1, this becomes 4 - (-1) = 5. So, the denominator is 5.Now, the numerator:(m - 2i)(2 + i) = m*2 + m*i - 2i*2 - 2i*i = 2m + mi - 4i - 2i².Again, since i² = -1, this becomes 2m + mi - 4i - 2*(-1) = 2m + mi - 4i + 2.Let me combine like terms:Real parts: 2m + 2.Imaginary parts: (m - 4)i.So, the numerator is (2m + 2) + (m - 4)i.Putting it all together, z / (2 - i) = [(2m + 2) + (m - 4)i] / 5.Since this is a real number, the imaginary part must be zero. So, the coefficient of i must be zero:(m - 4)/5 = 0.Therefore, m - 4 = 0, which gives m = 4.So, now we know both m and n. z = 4 - 2i.Alright, moving on to the next part. We need to consider (z + ai)² and determine when it lies in the first quadrant.First, let's compute z + ai:z + ai = (4 - 2i) + ai = 4 + (a - 2)i.Now, let's square this:(4 + (a - 2)i)².I'll use the formula (x + y)² = x² + 2xy + y², where x = 4 and y = (a - 2)i.So,(4)² + 2*4*(a - 2)i + [(a - 2)i]².Compute each term:First term: 16.Second term: 8*(a - 2)i.Third term: (a - 2)² * i². Since i² = -1, this becomes -(a - 2)².Putting it all together:16 + 8(a - 2)i - (a - 2)².So, the result is:[16 - (a - 2)²] + 8(a - 2)i.Now, for this complex number to lie in the first quadrant, both the real part and the imaginary part must be positive.So, we have two inequalities:1. 16 - (a - 2)² > 02. 8(a - 2) > 0Let's solve the second inequality first because it seems simpler.Second inequality: 8(a - 2) > 0.Divide both sides by 8:a - 2 > 0So, a > 2.Okay, so a must be greater than 2.Now, the first inequality: 16 - (a - 2)² > 0.Let me rewrite that:16 > (a - 2)²Which is equivalent to:(a - 2)² < 16Taking square roots on both sides, we get:|a - 2| < 4Which means:-4 < a - 2 < 4Adding 2 to all parts:-4 + 2 < a < 4 + 2So,-2 < a < 6But wait, from the second inequality, we already have a > 2. So, combining both results:2 < a < 6But hold on, let me double-check the first inequality.16 - (a - 2)² > 0So, (a - 2)² < 16Which is equivalent to:-4 < a - 2 < 4Which leads to:-2 < a < 6But since a must be greater than 2, the overlapping interval is 2 < a < 6.Wait, but in the initial problem statement, the user wrote:"the point corresponding to the complex number (z + ai)^{2} lies in the first quadrant of the complex plane."So, both the real part and the imaginary part must be positive.Wait, but in my calculation, the real part is 16 - (a - 2)^2 and the imaginary part is 8(a - 2).So, for the real part to be positive:16 - (a - 2)^2 > 0Which gives (a - 2)^2 < 16, so |a - 2| < 4, so a is between -2 and 6.But from the imaginary part, 8(a - 2) > 0, so a > 2.So, combining these, a must be between 2 and 6.Wait, but in the initial solution, the user had a different result. Wait, let me check.Wait, in the initial solution, the user had:16 - (a - 2)^2 > 0Which simplifies to:(a - 2)^2 < 16Which gives a between -2 and 6.But the imaginary part is 8(a - 2) > 0, so a > 2.Hence, the range is 2 < a < 6.But wait, in the initial problem, the user had a different result. Wait, let me check again.Wait, in the initial solution, the user had:16 - (a - 2)^2 > 0Which is correct.Then, they wrote:a^2 - 4a < 0Wait, that seems incorrect.Wait, 16 - (a - 2)^2 > 0Let me expand (a - 2)^2:(a - 2)^2 = a² - 4a + 4So, 16 - (a² - 4a + 4) > 0Which is 16 - a² + 4a - 4 > 0Simplify:(16 - 4) + 4a - a² > 012 + 4a - a² > 0Which can be rewritten as:-a² + 4a + 12 > 0Multiply both sides by -1 (remember to reverse the inequality):a² - 4a - 12 < 0Now, factor this quadratic:Looking for two numbers that multiply to -12 and add to -4.Hmm, -6 and +2: (-6)*(+2) = -12, (-6) + (+2) = -4.So, (a - 6)(a + 2) < 0So, the critical points are a = -2 and a = 6.We can make a sign chart:- For a < -2: (a - 6) is negative, (a + 2) is negative. Negative * negative = positive.- For -2 < a < 6: (a - 6) is negative, (a + 2) is positive. Negative * positive = negative.- For a > 6: (a - 6) is positive, (a + 2) is positive. Positive * positive = positive.Since we have (a - 6)(a + 2) < 0, the solution is where the product is negative, which is -2 < a < 6.So, that's consistent with my earlier result.But wait, in the initial solution, the user had:16 - (a - 2)^2 > 0Which they rewrote as a² - 4a < 0, which is incorrect.Because 16 - (a² - 4a + 4) = 16 - a² + 4a - 4 = 12 + 4a - a², which is not equal to a² - 4a.So, that was a mistake in the initial solution.Therefore, the correct range is 2 < a < 6.But wait, in the initial solution, they had 2 < a < 4.Hmm, so that must have been an error.Wait, let me check the initial solution again.In the initial solution, they had:16 - (a - 2)^2 > 0Which they rewrote as a² - 4a < 0, which is incorrect.Because 16 - (a - 2)^2 = 16 - (a² - 4a + 4) = 16 - a² + 4a - 4 = 12 + 4a - a².So, 12 + 4a - a² > 0Which is equivalent to -a² + 4a + 12 > 0Which is equivalent to a² - 4a - 12 < 0Which factors to (a - 6)(a + 2) < 0, so -2 < a < 6.So, the correct range is -2 < a < 6, but since from the imaginary part, a > 2, the overlapping interval is 2 < a < 6.Therefore, the correct range is 2 < a < 6.Wait, but in the initial solution, they had 2 < a < 4.So, that must have been a mistake.Therefore, the correct answer is 2 < a < 6.But let me double-check.Wait, let's compute (z + ai)^2 again.z = 4 - 2iz + ai = 4 - 2i + ai = 4 + (a - 2)iSquaring this:(4 + (a - 2)i)^2 = 4^2 + 2*4*(a - 2)i + (a - 2)^2 i^2= 16 + 8(a - 2)i - (a - 2)^2So, real part: 16 - (a - 2)^2Imaginary part: 8(a - 2)So, for the real part to be positive:16 - (a - 2)^2 > 0Which is equivalent to (a - 2)^2 < 16So, |a - 2| < 4Which gives -4 < a - 2 < 4So, -2 < a < 6Imaginary part positive:8(a - 2) > 0 => a > 2So, combining both, 2 < a < 6Therefore, the correct range is 2 < a < 6.Hence, the initial solution had an error in the algebra when expanding 16 - (a - 2)^2, leading to an incorrect range.So, the correct range is 2 < a < 6.But wait, let me check if a = 6 is allowed.At a = 6, the real part is 16 - (6 - 2)^2 = 16 - 16 = 0, which is not positive. So, a must be less than 6.Similarly, at a = 2, the imaginary part is 8(2 - 2) = 0, which is not positive. So, a must be greater than 2.Therefore, the range is 2 < a < 6.So, the correct answer is 2 < a < 6.**Final Answer**The range of the real number ( a ) is boxed{2 < a < 6}.

answer：Okay, so I have this problem: Determine the positive integers that are one and a half times the product of their digits. Hmm, that sounds interesting. Let me try to wrap my head around it.First, let me make sure I understand the problem correctly. We need to find all positive integers ( n ) such that ( n = frac{3}{2} times ) (product of its digits). So, if ( n ) is a number with digits ( a_k, a_{k-1}, ldots, a_1 ), then:[n = frac{3}{2} times (a_k times a_{k-1} times ldots times a_1)]Alright, so the number itself is one and a half times the product of its digits. That means the product of the digits has to be a significant portion of the number itself. Since multiplying by ( frac{3}{2} ) increases the product, the number can't be too large, otherwise, the product of its digits might not keep up.Let me start by considering the number of digits in ( n ). If ( n ) is a single-digit number, say ( n = d ), then the equation becomes:[d = frac{3}{2} times d]Subtracting ( frac{3}{2}d ) from both sides:[d - frac{3}{2}d = 0 implies -frac{1}{2}d = 0 implies d = 0]But ( d ) is a positive integer, so there are no single-digit solutions. Got it.Moving on to two-digit numbers. Let ( n = 10a + b ), where ( a ) and ( b ) are digits from 1 to 9 (since digits can't be zero, otherwise the product would be zero, and ( n ) wouldn't be zero). The equation becomes:[10a + b = frac{3}{2} times (a times b)]Let me rearrange this equation to make it easier to handle. Multiply both sides by 2 to eliminate the fraction:[20a + 2b = 3ab]Bring all terms to one side:[3ab - 20a - 2b = 0]Hmm, this looks a bit tricky. Maybe I can factor this somehow. Let me try to factor by grouping. Let's see:[3ab - 20a - 2b = 0]Factor ( a ) from the first two terms:[a(3b - 20) - 2b = 0]Hmm, not sure if that helps. Maybe I can solve for one variable in terms of the other. Let's solve for ( a ):[3ab - 20a = 2b]Factor ( a ):[a(3b - 20) = 2b]So,[a = frac{2b}{3b - 20}]Since ( a ) and ( b ) are digits from 1 to 9, let's see for which values of ( b ) this fraction gives an integer between 1 and 9.First, the denominator ( 3b - 20 ) must be positive because ( a ) has to be positive. So,[3b - 20 > 0 implies 3b > 20 implies b > frac{20}{3} approx 6.666]So ( b ) must be 7, 8, or 9.Let's check each case:1. **Case ( b = 7 ):**[a = frac{2 times 7}{3 times 7 - 20} = frac{14}{21 - 20} = frac{14}{1} = 14]But ( a ) must be a single digit, so 14 is invalid.2. **Case ( b = 8 ):**[a = frac{2 times 8}{3 times 8 - 20} = frac{16}{24 - 20} = frac{16}{4} = 4]Okay, ( a = 4 ) is valid. So the number is 48. Let's check:Product of digits: ( 4 times 8 = 32 )One and a half times the product: ( frac{3}{2} times 32 = 48 )Yes, that works.3. **Case ( b = 9 ):**[a = frac{2 times 9}{3 times 9 - 20} = frac{18}{27 - 20} = frac{18}{7} approx 2.571]Not an integer, so invalid.So, the only two-digit number that works is 48.Now, let's consider three-digit numbers. Let ( n = 100a + 10b + c ), where ( a, b, c ) are digits from 1 to 9. The equation becomes:[100a + 10b + c = frac{3}{2} times (a times b times c)]Again, multiply both sides by 2:[200a + 20b + 2c = 3abc]This seems more complicated. Let me see if I can find any solutions here.First, note that ( a ) is at least 1, so the left side is at least 200 + 20 + 2 = 222, while the right side is ( 3abc ). Let's see what the maximum possible product is. The maximum product for three digits is ( 9 times 9 times 9 = 729 ), so ( 3abc ) can be up to 2187. But the left side is 200a + 20b + 2c, which for ( a=9 ) is 1800 + 20b + 2c, which is up to 1800 + 180 + 18 = 1998. So, the right side can be larger, but it's not guaranteed.Let me try to find bounds. Let's see:Given ( 200a + 20b + 2c = 3abc ), we can write:[3abc = 200a + 20b + 2c]Divide both sides by ( a ) (since ( a geq 1 )):[3bc = 200 + frac{20b}{a} + frac{2c}{a}]Since ( a geq 1 ), the terms ( frac{20b}{a} ) and ( frac{2c}{a} ) are at least 20b and 2c, respectively. Wait, that doesn't make sense because dividing by a larger ( a ) would make those terms smaller. Hmm, maybe another approach.Alternatively, let's consider that ( 3abc geq 200a ), so:[3bc geq 200]Which implies:[bc geq frac{200}{3} approx 66.666]So ( bc geq 67 ). Since ( b ) and ( c ) are digits from 1 to 9, their product can be at most 81 (9x9). So ( bc ) is between 67 and 81.Let me list all possible pairs ( (b, c) ) such that ( bc geq 67 ):Possible pairs:- ( b=9 ): ( c geq lceil 67/9 rceil = 8 ). So ( c=8,9 )- ( b=8 ): ( c geq lceil 67/8 rceil = 9 ). So ( c=9 )- ( b=7 ): ( c geq lceil 67/7 rceil = 10 ). But ( c ) can't be 10, so no solutions here.- Similarly, ( b leq 6 ): ( c ) would have to be greater than 9, which isn't possible.So the possible pairs are:- ( (9,8) ), ( (9,9) ), ( (8,9) )Let me check each case.1. **Case ( b=9, c=8 ):**Plug into the equation:[3a times 9 times 8 = 200a + 20 times 9 + 2 times 8][216a = 200a + 180 + 16][216a = 200a + 196][16a = 196][a = frac{196}{16} = 12.25]Not an integer, invalid.2. **Case ( b=9, c=9 ):**[3a times 9 times 9 = 200a + 20 times 9 + 2 times 9][243a = 200a + 180 + 18][243a = 200a + 198][43a = 198][a = frac{198}{43} approx 4.604]Not an integer, invalid.3. **Case ( b=8, c=9 ):**[3a times 8 times 9 = 200a + 20 times 8 + 2 times 9][216a = 200a + 160 + 18][216a = 200a + 178][16a = 178][a = frac{178}{16} = 11.125]Again, not an integer, invalid.So, no three-digit numbers satisfy the condition.Moving on to four-digit numbers. Let ( n = 1000a + 100b + 10c + d ), with ( a, b, c, d ) from 1 to 9. The equation becomes:[1000a + 100b + 10c + d = frac{3}{2} times (a times b times c times d)]Multiply both sides by 2:[2000a + 200b + 20c + 2d = 3abcd]This seems even more complex. Let me see if I can find any solutions here.First, note that the left side is at least 2000 + 200 + 20 + 2 = 2222, while the right side is ( 3abcd ). The maximum product for four digits is ( 9^4 = 6561 ), so ( 3abcd ) can be up to 19683. But the left side is 2000a + 200b + 20c + 2d, which for ( a=9 ) is 18000 + 200b + 20c + 2d, up to 18000 + 1800 + 180 + 18 = 19998. So, the right side can be larger, but again, it's not guaranteed.Let me try to find bounds. From the equation:[3abcd = 2000a + 200b + 20c + 2d]Divide both sides by ( a ):[3bcd = 2000 + frac{200b}{a} + frac{20c}{a} + frac{2d}{a}]Since ( a geq 1 ), the terms ( frac{200b}{a} ), ( frac{20c}{a} ), and ( frac{2d}{a} ) are at least 200b, 20c, and 2d, respectively. Wait, that doesn't make sense because dividing by a larger ( a ) would make those terms smaller. Hmm, maybe another approach.Alternatively, let's consider that ( 3abcd geq 2000a ), so:[3bcd geq 2000]Which implies:[bcd geq frac{2000}{3} approx 666.666]So ( bcd geq 667 ). Since ( b, c, d ) are digits from 1 to 9, their product can be at most 729 (9x9x9). So ( bcd ) is between 667 and 729.Let me see what combinations of ( b, c, d ) can give a product in that range.Given ( b, c, d geq 1 ), and ( bcd geq 667 ), let's see:The maximum product is 729, so ( bcd ) can be 667 to 729.Let me try to find such triplets.First, note that 9x9x9=729, which is the maximum. So, to get 667, we need to find triplets where the product is just below 729.Let me try ( b=9 ):Then ( c times d geq frac{667}{9} approx 74.111 ). So ( c times d geq 75 ).Possible pairs ( (c,d) ) with ( c times d geq 75 ):- ( c=9 ): ( d geq lceil 75/9 rceil = 9 ). So ( d=9 )- ( c=8 ): ( d geq lceil 75/8 rceil = 10 ). Not possible- Similarly, lower ( c ) would require higher ( d ), which isn't possible.So, only ( c=9, d=9 ) gives ( c times d = 81 geq 75 ).So, triplet ( (9,9,9) ) gives ( bcd=729 ).Let me check this case:[3a times 9 times 9 times 9 = 2000a + 200 times 9 + 20 times 9 + 2 times 9][3a times 729 = 2000a + 1800 + 180 + 18][2187a = 2000a + 1998][187a = 1998][a = frac{1998}{187} approx 10.68]Not an integer, and ( a ) must be a single digit, so invalid.Next, let's try ( b=8 ):Then ( c times d geq frac{667}{8} approx 83.375 ). So ( c times d geq 84 ).Possible pairs ( (c,d) ):- ( c=9 ): ( d geq lceil 84/9 rceil = 10 ). Not possible- ( c=8 ): ( d geq lceil 84/8 rceil = 11 ). Not possible- Similarly, no valid pairs since ( d ) can't exceed 9.So, no solutions for ( b=8 ).Next, ( b=7 ):( c times d geq frac{667}{7} approx 95.285 ). So ( c times d geq 96 ).But the maximum ( c times d ) is 81 (9x9), so no solutions here.Similarly, for ( b leq 6 ), ( c times d ) would have to be even larger, which isn't possible.So, no four-digit numbers satisfy the condition.Given that we've checked up to four-digit numbers and found only 48 as a solution, and considering that higher-digit numbers would likely have even larger gaps between the left and right sides of the equation, it's reasonable to conclude that 48 is the only solution.Just to be thorough, let me check if there are any numbers with more than four digits that could satisfy the condition. However, as the number of digits increases, the left side (the number itself) grows exponentially with each additional digit, while the right side (the product of digits) grows multiplicatively. Since the product of digits can't grow as fast as the number itself, it's unlikely that any number with more than four digits would satisfy the condition. Therefore, 48 is the only solution.

answer：Okay, so I have this problem here: the sum of the binomial coefficients of the expansion of (2x - 1/√x)^n is 64, and I need to find the constant term in its expansion. Hmm, let me try to figure this out step by step.First, I remember that the sum of the binomial coefficients for any binomial expression (a + b)^n is found by setting a = 1 and b = 1. So, in this case, the expression is (2x - 1/√x)^n. To find the sum of the binomial coefficients, I should set x to a value that makes both terms coefficients. Wait, actually, if I set x = 1, then 2x becomes 2 and 1/√x becomes 1. So, substituting x = 1, the expression becomes (2 - 1)^n, which simplifies to (1)^n = 1. But the problem says the sum is 64, not 1. Hmm, that doesn't make sense. Maybe I'm misunderstanding something.Wait, no. The sum of the binomial coefficients is actually the sum of the coefficients when you expand the expression. So, if I set x = 1, that should give me the sum of all the coefficients. Let me try that again. If x = 1, then 2x is 2 and 1/√x is 1, so the expression becomes (2 - 1)^n = 1^n = 1. But the problem states that the sum is 64. That means I must have made a mistake.Oh, I think I get it now. The sum of the binomial coefficients is when you consider only the coefficients, not the variables. So, in the expression (2x - 1/√x)^n, the coefficients are 2 and -1. So, to find the sum of the coefficients, I should set x = 1, which gives me (2*1 - 1/√1)^n = (2 - 1)^n = 1^n = 1. But the problem says the sum is 64, so that can't be right. Maybe I need to consider the absolute values of the coefficients or something else?Wait, no. The sum of the binomial coefficients is actually the sum of the coefficients when you expand the expression, regardless of the variables. So, if I have (a + b)^n, the sum of the coefficients is (1 + 1)^n = 2^n. But in this case, the expression is (2x - 1/√x)^n. So, to find the sum of the coefficients, I set x = 1, which gives me (2*1 - 1/√1)^n = (2 - 1)^n = 1^n = 1. But the problem says the sum is 64, so that must mean that 2^n = 64 because the coefficients are 2 and -1, but when you set x = 1, it's (2 - 1)^n = 1^n = 1, which doesn't help. Wait, maybe I need to consider the coefficients differently.Hold on, maybe the sum of the binomial coefficients is actually the sum of the absolute values of the coefficients. So, if I take the absolute values, it would be (2 + 1)^n = 3^n. But 3^n = 64? Let me check: 3^4 = 81, which is more than 64, and 3^3 = 27, which is less. So that's not it either.Wait, perhaps I'm overcomplicating this. The sum of the binomial coefficients is when you set x = 1, regardless of the coefficients. So, in this case, (2x - 1/√x)^n becomes (2 - 1)^n = 1^n = 1. But the problem says it's 64. That doesn't add up. Maybe the question is referring to the sum of the coefficients in the expansion, not the binomial coefficients. Hmm.Wait, binomial coefficients are the C(n, k) terms, right? So, the sum of the binomial coefficients is C(n, 0) + C(n, 1) + ... + C(n, n) = 2^n. So, if the sum is 64, then 2^n = 64, which means n = 6 because 2^6 = 64. Okay, that makes sense. So, n is 6.Now, I need to find the constant term in the expansion of (2x - 1/√x)^6. To find the constant term, I need to look for the term where the exponent of x is zero. Let's recall the binomial expansion formula: (a + b)^n = Σ (from k=0 to n) C(n, k) * a^(n-k) * b^k.In this case, a = 2x and b = -1/√x. So, each term in the expansion will be C(6, k) * (2x)^(6 - k) * (-1/√x)^k. Let's simplify this term.First, let's write out the general term:T_k = C(6, k) * (2x)^(6 - k) * (-1/√x)^k.Let's simplify each part:(2x)^(6 - k) = 2^(6 - k) * x^(6 - k).(-1/√x)^k = (-1)^k * (1/√x)^k = (-1)^k * x^(-k/2).So, combining these, the term becomes:T_k = C(6, k) * 2^(6 - k) * (-1)^k * x^(6 - k - k/2).Simplify the exponent of x:6 - k - k/2 = 6 - (3k)/2.We need the exponent to be zero for the term to be constant. So, set 6 - (3k)/2 = 0.Solving for k:6 = (3k)/2 => 12 = 3k => k = 4.So, the constant term occurs when k = 4. Now, let's compute that term.T_4 = C(6, 4) * 2^(6 - 4) * (-1)^4.Compute each part:C(6, 4) = C(6, 2) because C(n, k) = C(n, n - k). C(6, 2) = (6*5)/(2*1) = 15.2^(6 - 4) = 2^2 = 4.(-1)^4 = 1.So, T_4 = 15 * 4 * 1 = 60.Therefore, the constant term is 60.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, n = 6 because 2^6 = 64. That seems correct.Then, the general term: C(6, k) * (2x)^(6 - k) * (-1/√x)^k. Simplifying gives C(6, k) * 2^(6 - k) * (-1)^k * x^(6 - (3k)/2). Setting the exponent to zero: 6 - (3k)/2 = 0 => k = 4. That seems right.Calculating T_4: C(6, 4) = 15, 2^(2) = 4, (-1)^4 = 1. So, 15 * 4 = 60. Yep, that looks correct.I think I've got it. The constant term is 60.

answer：First, I'll define the variables: let ( h ) represent the cost of one hotdog and ( b ) represent the cost of one burger.From the problem, I have two equations based on the given sales:[4h + 5b = 3.75][5h + 3b = 3.45]To eliminate one of the variables, I'll multiply the first equation by 3 and the second equation by 5:[12h + 15b = 11.25][25h + 15b = 17.25]Next, I'll subtract the first modified equation from the second to eliminate ( b ):[13h = 6.00][h = frac{6.00}{13} = 0.4615]Now that I have the value of ( h ), I'll substitute it back into the first original equation to find ( b ):[4(0.4615) + 5b = 3.75][1.846 + 5b = 3.75][5b = 1.904][b = frac{1.904}{5} = 0.3808]Finally, I'll calculate the cost for six hotdogs and seven burgers using the values of ( h ) and ( b ):[6h + 7b = 6(0.4615) + 7(0.3808)][2.769 + 2.6656 = 5.4346]Rounding to the nearest hundredth, the total cost is 5.43.