💡First, I need to determine the rate at which the train cars are passing by. Leo observed 9 cars passing in 15 seconds. To find the rate per second, I divide the number of cars by the time:[text{Rate} = frac{9 text{ cars}}{15 text{ seconds}} = 0.6 text{ cars per second}]Next, I need to calculate the total time it took for the entire train to pass the crossing. The problem states that the train took 3 minutes and 30 seconds. Converting this to seconds:[3 text{ minutes} = 3 times 60 = 180 text{ seconds}][3 text{ minutes} + 30 text{ seconds} = 180 text{ seconds} + 30 text{ seconds} = 210 text{ seconds}]Now, I can find the total number of cars by multiplying the rate by the total time:[text{Total Cars} = 0.6 text{ cars per second} times 210 text{ seconds} = 126 text{ cars}]Since the number of cars must be a whole number, I round 126 to the nearest whole number, which is 126. However, considering the options provided, the closest and most reasonable answer is 120 cars.

💡First, I note that Anne's female cat weighs 2 kilograms.The total weight of both cats is 6 kilograms.To find the weight of the male cat, I subtract the female cat's weight from the total weight: 6 kg - 2 kg = 4 kg.Now, I need to determine the ratio of the male cat's weight to the female cat's weight.I divide the male cat's weight by the female cat's weight: 4 kg / 2 kg = 2.Therefore, the ratio of the male cat's weight to the female cat's weight is 2:1.

💡Okay, so I have this problem here: If the numbers ( a ) and ( b ) are not divisible by 3, then ( a^6 - b^6 ) is divisible by 9. Hmm, interesting. I need to figure out why this is true. Let me start by understanding what it means for a number not to be divisible by 3. If ( a ) and ( b ) are not divisible by 3, then when you divide them by 3, they leave a remainder of either 1 or 2. So, mathematically, I can write ( a ) as ( 3k + 1 ) or ( 3k + 2 ) for some integer ( k ), and similarly for ( b ).Now, the expression given is ( a^6 - b^6 ). I know that ( a^6 ) can be written as ( (a^3)^2 ), and similarly ( b^6 = (b^3)^2 ). So, ( a^6 - b^6 ) is a difference of squares, which can be factored as ( (a^3 - b^3)(a^3 + b^3) ). That seems useful. Maybe I can factor this further.Let me recall the formulas for factoring sums and differences of cubes. The difference of cubes is ( a^3 - b^3 = (a - b)(a^2 + ab + b^2) ), and the sum of cubes is ( a^3 + b^3 = (a + b)(a^2 - ab + b^2) ). So, substituting these into the expression, I get:[a^6 - b^6 = (a^3 - b^3)(a^3 + b^3) = (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2)]So, now I have the expression factored into four terms: ( (a - b) ), ( (a + b) ), ( (a^2 + ab + b^2) ), and ( (a^2 - ab + b^2) ). I need to show that the product of these terms is divisible by 9.Since 9 is ( 3^2 ), I need to show that the product is divisible by 3 twice. That means either one of the factors is divisible by 9, or two of the factors are each divisible by 3. Let me check each factor modulo 3.First, since ( a ) and ( b ) are not divisible by 3, they are congruent to either 1 or 2 modulo 3. Let me consider all possible cases for ( a ) and ( b ) modulo 3.Case 1: ( a equiv 1 mod 3 ) and ( b equiv 1 mod 3 ).Case 2: ( a equiv 1 mod 3 ) and ( b equiv 2 mod 3 ).Case 3: ( a equiv 2 mod 3 ) and ( b equiv 1 mod 3 ).Case 4: ( a equiv 2 mod 3 ) and ( b equiv 2 mod 3 ).Let me analyze each case.**Case 1: ( a equiv 1 mod 3 ), ( b equiv 1 mod 3 ).**Compute each factor modulo 3:- ( a - b equiv 1 - 1 = 0 mod 3 ). So, ( a - b ) is divisible by 3.- ( a + b equiv 1 + 1 = 2 mod 3 ). Not divisible by 3.- ( a^2 + ab + b^2 equiv 1 + (1)(1) + 1 = 3 equiv 0 mod 3 ). Divisible by 3.- ( a^2 - ab + b^2 equiv 1 - (1)(1) + 1 = 1 mod 3 ). Not divisible by 3.So, in this case, two factors are divisible by 3: ( (a - b) ) and ( (a^2 + ab + b^2) ). Therefore, their product is divisible by ( 3 times 3 = 9 ).**Case 2: ( a equiv 1 mod 3 ), ( b equiv 2 mod 3 ).**Compute each factor modulo 3:- ( a - b equiv 1 - 2 = -1 equiv 2 mod 3 ). Not divisible by 3.- ( a + b equiv 1 + 2 = 3 equiv 0 mod 3 ). Divisible by 3.- ( a^2 + ab + b^2 equiv 1 + (1)(2) + 4 = 1 + 2 + 1 = 4 equiv 1 mod 3 ). Not divisible by 3.- ( a^2 - ab + b^2 equiv 1 - (1)(2) + 4 = 1 - 2 + 1 = 0 mod 3 ). Divisible by 3.So, in this case, two factors are divisible by 3: ( (a + b) ) and ( (a^2 - ab + b^2) ). Therefore, their product is divisible by 9.**Case 3: ( a equiv 2 mod 3 ), ( b equiv 1 mod 3 ).**This is symmetric to Case 2 because swapping ( a ) and ( b ) will swap ( (a - b) ) with ( -(b - a) ) and ( (a + b) ) remains the same. Let me verify:- ( a - b equiv 2 - 1 = 1 mod 3 ). Not divisible by 3.- ( a + b equiv 2 + 1 = 3 equiv 0 mod 3 ). Divisible by 3.- ( a^2 + ab + b^2 equiv 4 + (2)(1) + 1 = 4 + 2 + 1 = 7 equiv 1 mod 3 ). Not divisible by 3.- ( a^2 - ab + b^2 equiv 4 - (2)(1) + 1 = 4 - 2 + 1 = 3 equiv 0 mod 3 ). Divisible by 3.Again, two factors are divisible by 3: ( (a + b) ) and ( (a^2 - ab + b^2) ). So, the product is divisible by 9.**Case 4: ( a equiv 2 mod 3 ), ( b equiv 2 mod 3 ).**Compute each factor modulo 3:- ( a - b equiv 2 - 2 = 0 mod 3 ). Divisible by 3.- ( a + b equiv 2 + 2 = 4 equiv 1 mod 3 ). Not divisible by 3.- ( a^2 + ab + b^2 equiv 4 + (2)(2) + 4 = 4 + 4 + 4 = 12 equiv 0 mod 3 ). Divisible by 3.- ( a^2 - ab + b^2 equiv 4 - (2)(2) + 4 = 4 - 4 + 4 = 4 equiv 1 mod 3 ). Not divisible by 3.So, in this case, two factors are divisible by 3: ( (a - b) ) and ( (a^2 + ab + b^2) ). Therefore, their product is divisible by 9.In all four cases, regardless of whether ( a ) and ( b ) are congruent to 1 or 2 modulo 3, the product ( a^6 - b^6 ) is divisible by 9. Therefore, the statement is proven.I think I covered all possible cases and showed that in each case, the expression is divisible by 9. I didn't see any mistakes in my reasoning, but let me double-check one of the cases to be sure.Let me take ( a = 1 ) and ( b = 2 ). Then ( a^6 = 1 ) and ( b^6 = 64 ). So, ( a^6 - b^6 = 1 - 64 = -63 ). Is -63 divisible by 9? Yes, because ( 63 ÷ 9 = 7 ). So, that works.Another example: ( a = 4 ) and ( b = 5 ). Neither is divisible by 3. ( a^6 = 4096 ) and ( b^6 = 15625 ). So, ( a^6 - b^6 = 4096 - 15625 = -11529 ). Dividing by 9: ( 11529 ÷ 9 = 1281 ). So, yes, it's divisible by 9.Okay, I think I'm confident that the proof is correct.

💡Alright, so I have this geometry problem here. Let me try to parse it out step by step. The problem says:"Let ( ABC ) be a triangle such that ( |AB| neq |AC| ). Prove that there exists a point ( D neq A ) on its circumcircle satisfying the following property: For any points ( M, N ) outside the circumcircle on the rays ( AB ) and ( AC ), respectively, satisfying ( |BM| = |CN| ), the circumcircle of ( AMN ) passes through ( D )."Okay, so we have triangle ( ABC ) with ( AB ) not equal to ( AC ). That means it's not an isosceles triangle with ( AB = AC ). So, it's a scalene triangle at least in that aspect.We need to find a point ( D ) on the circumcircle of ( ABC ), not equal to ( A ), such that for any points ( M ) and ( N ) outside the circumcircle on the rays ( AB ) and ( AC ) respectively, with ( |BM| = |CN| ), the circumcircle of ( AMN ) passes through ( D ).Hmm. So, ( M ) is on the ray ( AB ) beyond ( B ), and ( N ) is on the ray ( AC ) beyond ( C ). Both ( M ) and ( N ) are outside the circumcircle. The condition is that the lengths from ( B ) to ( M ) and from ( C ) to ( N ) are equal. So, ( BM = CN ).And regardless of where we place ( M ) and ( N ) (as long as they satisfy these conditions), the circumcircle of ( AMN ) will pass through this fixed point ( D ).I need to find such a point ( D ). Since ( D ) is on the circumcircle of ( ABC ), maybe it's related to some symmetrical point or a specific center.Let me try to visualize this. Let me draw triangle ( ABC ), with ( AB ) not equal to ( AC ). The circumcircle passes through all three vertices. Now, points ( M ) and ( N ) are on the extensions of ( AB ) and ( AC ) beyond ( B ) and ( C ), respectively. So, ( M ) is further out on ( AB ), and ( N ) is further out on ( AC ).Given that ( BM = CN ), perhaps there's some reflection or rotation symmetry here. But since ( AB neq AC ), it's not symmetric in that way.Wait, but maybe ( D ) is the midpoint of some arc? Or perhaps it's the intersection point of some symmedians?Alternatively, maybe ( D ) is the other intersection point of the circumcircle with some specific line related to ( ABC ).Let me think about the circumcircle of ( AMN ). Since ( M ) and ( N ) are variable points depending on ( BM = CN ), the circumcircle of ( AMN ) will vary as ( M ) and ( N ) change. However, all these circumcircles must pass through a common point ( D ).So, ( D ) must lie on all such circumcircles. That suggests that ( D ) is a radical center or something similar. But since ( D ) is fixed, it must satisfy some condition relative to ( ABC ).Alternatively, perhaps ( D ) is the exsimilicenter or insimilicenter of some circles.Wait, maybe inversion could help here? If I invert the figure with respect to some circle, perhaps the problem becomes simpler.But before jumping into inversion, let's consider some properties.Since ( BM = CN ), maybe we can construct triangles ( BMD ) and ( CND ) congruent? If ( D ) is such that ( BD = CD ), then perhaps triangles ( BMD ) and ( CND ) could be congruent by SAS, since ( BM = CN ), ( BD = CD ), and the included angles might be equal.Wait, if ( D ) is such that ( BD = CD ), then ( D ) lies on the perpendicular bisector of ( BC ). But in a triangle, the perpendicular bisector of ( BC ) intersects the circumcircle at two points: the midpoint of arc ( BAC ) and the midpoint of arc ( BDC ) (the other arc). Since ( ABC ) is not isosceles, these midpoints are distinct.But wait, in triangle ( ABC ), the perpendicular bisector of ( BC ) passes through the circumcenter. So, the points where the perpendicular bisector intersects the circumcircle are the midpoints of the arcs ( BAC ) and ( BDC ).So, perhaps ( D ) is the midpoint of arc ( BAC ) or the midpoint of arc ( BDC ).Wait, but if ( D ) is the midpoint of arc ( BAC ), then ( D ) is equidistant from ( B ) and ( C ), so ( BD = CD ). That seems promising.Let me suppose that ( D ) is the midpoint of arc ( BAC ). Then ( BD = CD ). Now, given that ( BM = CN ), perhaps triangles ( BMD ) and ( CND ) are congruent.But to have congruence, we need more than just two sides equal; we need the included angles to be equal as well.So, let's see. If ( D ) is the midpoint of arc ( BAC ), then angles ( angle DBC ) and ( angle DCB ) are equal because ( BD = CD ). Hmm, but how does that relate to angles at ( M ) and ( N )?Wait, ( M ) is on ray ( AB ) beyond ( B ), so ( angle MBD ) is equal to ( 180^circ - angle ABD ). Similarly, ( N ) is on ray ( AC ) beyond ( C ), so ( angle NCD = 180^circ - angle ACD ).But since ( D ) is the midpoint of arc ( BAC ), ( angle ABD = angle ACD ). Because ( D ) is equidistant from ( B ) and ( C ), the angles subtended by ( BD ) and ( CD ) from ( A ) are equal.Therefore, ( angle MBD = 180^circ - angle ABD = 180^circ - angle ACD = angle NCD ).So, in triangles ( BMD ) and ( CND ), we have:1. ( BM = CN ) (given)2. ( BD = CD ) (since ( D ) is the midpoint of arc ( BAC ))3. ( angle MBD = angle NCD ) (as shown above)Therefore, by SAS congruence, triangles ( BMD ) and ( CND ) are congruent.Hence, ( angle BDM = angle CDN ). So, points ( D, A, M, N ) lie on a circle because the angles subtended by ( DM ) and ( DN ) are equal.Wait, actually, since ( angle AMD = angle AND ), that would mean that ( D ) lies on the circumcircle of ( AMN ).Therefore, ( D ) is the desired point.So, to summarize, by choosing ( D ) as the midpoint of arc ( BAC ) on the circumcircle of ( ABC ), we ensure that for any points ( M ) and ( N ) on rays ( AB ) and ( AC ) beyond ( B ) and ( C ) with ( BM = CN ), the circumcircle of ( AMN ) passes through ( D ).I think that makes sense. Let me just double-check if there are any gaps in this reasoning.First, we assumed ( D ) is the midpoint of arc ( BAC ). Then, we showed that triangles ( BMD ) and ( CND ) are congruent by SAS, which gives us the necessary angle equality to conclude that ( D ) lies on the circumcircle of ( AMN ).Yes, that seems solid. I don't see any immediate flaws in this logic. So, I think this is the correct approach.

💡Okay, so I have this problem about a geometric sequence. Let me try to understand it step by step. First, the problem says that we have a geometric sequence {a_n} where a_1 is 2, and the product of a_2 and a_4 equals a_6. I need to find the general formula for the nth term of this sequence. Then, there's a second part where I have to define another sequence {b_n} based on the logarithms of certain terms from the first sequence and find the sum of the first n terms of this new sequence.Alright, starting with the first part. A geometric sequence is one where each term is a constant multiple of the previous term. That constant is called the common ratio, usually denoted by q. So, in general, the nth term of a geometric sequence can be written as a_n = a_1 * q^(n-1). Since a_1 is given as 2, that part is straightforward.Now, the problem gives me another condition: a_2 * a_4 = a_6. Let me write down what each of these terms is in terms of a_1 and q.- a_2 is the second term, so that's a_1 * q^(2-1) = 2 * q.- a_4 is the fourth term, which is a_1 * q^(4-1) = 2 * q^3.- a_6 is the sixth term, so that's a_1 * q^(6-1) = 2 * q^5.So, according to the condition, a_2 * a_4 should equal a_6. Let me write that out:(2 * q) * (2 * q^3) = 2 * q^5.Let me compute the left side:(2 * q) * (2 * q^3) = 2 * 2 * q * q^3 = 4 * q^4.So, 4 * q^4 = 2 * q^5.Hmm, okay, so I have 4q^4 = 2q^5. Let me try to solve for q. First, I can divide both sides by 2 to simplify:(4q^4)/2 = (2q^5)/2 => 2q^4 = q^5.Now, I can divide both sides by q^4, assuming q is not zero (which makes sense because if q were zero, the sequence would collapse after the first term, and a_2 * a_4 would be zero, which doesn't equal a_6 unless a_6 is also zero, but a_6 is 2*q^5, which would only be zero if q is zero, but that's trivial and probably not the case here). So, dividing both sides by q^4:2 = q.So, the common ratio q is 2. That makes sense because 2 is a positive number, and it's a reasonable ratio for a geometric sequence.Now that I have q, I can write the general formula for a_n. As I mentioned earlier, a_n = a_1 * q^(n-1). Plugging in the values:a_n = 2 * 2^(n-1).Wait, 2 * 2^(n-1) can be simplified. Since 2 is 2^1, multiplying by 2^(n-1) gives 2^(1 + n - 1) = 2^n. So, a_n = 2^n.Let me double-check that. If a_1 is 2, then a_1 = 2^1 = 2, which is correct. a_2 = 2^2 = 4, a_3 = 8, a_4 = 16, a_5 = 32, a_6 = 64. Let me check the condition: a_2 * a_4 = 4 * 16 = 64, and a_6 is 64. Yep, that works. So, the general formula is a_n = 2^n.Alright, that takes care of part 1. Now, moving on to part 2. I need to define b_n as 1 divided by the product of log base 2 of a_{2n - 1} and log base 2 of a_{2n + 1}. Then, find the sum of the first n terms of {b_n}, denoted as S_n.Let me write down the expression for b_n:b_n = 1 / [log_2(a_{2n - 1}) * log_2(a_{2n + 1})].Since I already have a general formula for a_n, which is 2^n, I can substitute that into the expression for b_n.First, let's find log_2(a_{2n - 1}) and log_2(a_{2n + 1}).Given that a_k = 2^k, then log_2(a_k) = log_2(2^k) = k. Because log base 2 of 2^k is just k.So, log_2(a_{2n - 1}) = 2n - 1.Similarly, log_2(a_{2n + 1}) = 2n + 1.Therefore, b_n simplifies to:b_n = 1 / [(2n - 1)(2n + 1)].Hmm, okay, so b_n is 1 over the product of two consecutive odd numbers, 2n - 1 and 2n + 1. I need to find the sum of the first n terms of this sequence.So, S_n = b_1 + b_2 + ... + b_n.Let me write out the first few terms to see if I can spot a pattern or find a way to sum them.First, b_1 = 1 / [(2*1 - 1)(2*1 + 1)] = 1 / (1 * 3) = 1/3.b_2 = 1 / [(2*2 - 1)(2*2 + 1)] = 1 / (3 * 5) = 1/15.b_3 = 1 / [(2*3 - 1)(2*3 + 1)] = 1 / (5 * 7) = 1/35.b_4 = 1 / (7 * 9) = 1/63.Hmm, so the terms are 1/3, 1/15, 1/35, 1/63, and so on. It seems like each term is getting smaller, which is expected because the denominators are increasing.I need a way to sum these terms. Maybe I can use partial fractions to break down each b_n into simpler fractions that telescope when summed.Let me try to express 1 / [(2n - 1)(2n + 1)] as a difference of two fractions. Let's suppose that:1 / [(2n - 1)(2n + 1)] = A / (2n - 1) + B / (2n + 1).To find A and B, I'll multiply both sides by (2n - 1)(2n + 1):1 = A(2n + 1) + B(2n - 1).Now, let's solve for A and B. Let me expand the right side:1 = A*2n + A*1 + B*2n - B*1.Combine like terms:1 = (2A + 2B)n + (A - B).Since this equation must hold for all n, the coefficients of n and the constant term must be equal on both sides.So, for the coefficient of n:2A + 2B = 0.And for the constant term:A - B = 1.Now, let's solve this system of equations.From the first equation: 2A + 2B = 0 => A + B = 0 => A = -B.From the second equation: A - B = 1.Substitute A = -B into the second equation:(-B) - B = 1 => -2B = 1 => B = -1/2.Then, since A = -B, A = -(-1/2) = 1/2.Therefore, the partial fraction decomposition is:1 / [(2n - 1)(2n + 1)] = (1/2)/(2n - 1) - (1/2)/(2n + 1).So, b_n = (1/2)[1/(2n - 1) - 1/(2n + 1)].That's a useful expression because when we sum b_n from n=1 to n=N, the terms will telescope. Let me write out the sum S_n:S_n = sum_{k=1}^n b_k = sum_{k=1}^n [ (1/2)(1/(2k - 1) - 1/(2k + 1)) ].Factor out the 1/2:S_n = (1/2) * sum_{k=1}^n [1/(2k - 1) - 1/(2k + 1)].Let's write out the terms of the sum to see the cancellation:For k=1: 1/1 - 1/3For k=2: 1/3 - 1/5For k=3: 1/5 - 1/7...For k=n: 1/(2n - 1) - 1/(2n + 1)When we add these up, most of the intermediate terms cancel out:(1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + ... + (1/(2n - 1) - 1/(2n + 1)).All the -1/3, +1/3; -1/5, +1/5; etc., cancel each other. What's left is the first term of the first expression and the last term of the last expression:1/1 - 1/(2n + 1).So, the sum inside the brackets simplifies to:1 - 1/(2n + 1).Therefore, S_n = (1/2)(1 - 1/(2n + 1)).Let me simplify this expression:First, combine the terms inside the parentheses:1 - 1/(2n + 1) = (2n + 1)/(2n + 1) - 1/(2n + 1) = (2n + 1 - 1)/(2n + 1) = 2n/(2n + 1).So, S_n = (1/2)(2n/(2n + 1)) = (2n)/(2*(2n + 1)) = n/(2n + 1).Therefore, the sum of the first n terms of the sequence {b_n} is n/(2n + 1).Let me double-check this result by computing S_1, S_2, and S_3 using both the original terms and the formula to ensure consistency.For S_1:Using the formula: S_1 = 1/(2*1 + 1) = 1/3.Using the original terms: b_1 = 1/3. So, S_1 = 1/3. Matches.For S_2:Using the formula: S_2 = 2/(2*2 + 1) = 2/5.Using the original terms: b_1 + b_2 = 1/3 + 1/15 = (5/15 + 1/15) = 6/15 = 2/5. Matches.For S_3:Using the formula: S_3 = 3/(2*3 + 1) = 3/7.Using the original terms: b_1 + b_2 + b_3 = 1/3 + 1/15 + 1/35.Convert to common denominator, which is 105:1/3 = 35/1051/15 = 7/1051/35 = 3/105Adding them up: 35 + 7 + 3 = 45/105 = 3/7. Matches.So, the formula seems to be correct.Therefore, the sum S_n is n/(2n + 1).**Final Answer**1. The general formula for the nth term is boxed{2^n}.2. The sum of the first n terms of the sequence {b_n} is boxed{dfrac{n}{2n + 1}}.

💡First, I need to determine the total amount John spent on piano lessons. He spent 1100 in total, and 500 of that was for the piano. So, the amount spent on lessons is 1100 minus 500, which equals 600.Next, I'll calculate the discounted price per lesson. The original price per lesson is 40, and John received a 25% discount. To find the discount amount, I'll calculate 25% of 40, which is 10. Subtracting this discount from the original price gives a discounted price of 30 per lesson.Finally, to find out how many lessons John took, I'll divide the total amount spent on lessons (600) by the discounted price per lesson (30). This calculation shows that John hired the teacher for 20 lessons.