answer：First, I need to determine how many ways to select three distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8} such that their sum equals 9.I'll start by listing all possible combinations of three different numbers from this set.Next, I'll calculate the sum for each combination and check if it equals 9.After identifying all valid combinations, I'll count them to find the total number of ways to achieve the sum of 9 with three cards.

answer：First, I need to find two 2-digit numbers whose product is 3774.To start, I'll factorize 3774 to identify its prime components. Dividing by 2 gives me 1887. Then, dividing by 3 yields 629. Next, I recognize that 629 is the product of 17 and 37. So, the prime factors of 3774 are 2, 3, 17, and 37.Now, I'll combine these prime factors to form two 2-digit numbers. Multiplying 2 and 3 gives 6, and multiplying 17 and 37 gives 629. However, 629 is a 3-digit number, which doesn't fit the requirement.I'll try a different combination. Multiplying 2 and 17 results in 34, and multiplying 3 and 37 gives 111. Again, 111 is a 3-digit number, so this combination also doesn't work.Finally, I'll multiply 2 and 37 to get 74, and 3 and 17 to get 51. Both 74 and 51 are 2-digit numbers, and their product is 3774. Therefore, the smaller of the two numbers is 51.

answer：Alright, let's try to figure out this problem. So, Vasya walks from his house to school, and there's a traffic light right in the middle of the road. On Monday, he encountered a green light, which I guess means he didn't have to stop. On Tuesday, he walked at the same speed but had to stand at the traffic light for 5 minutes, and then he doubled his speed afterward. Despite this change, he spent the same amount of time traveling both days. We need to find out how long it took him each day.Hmm, okay. Let me break this down. First, let's define some variables to represent the different parts of the journey. Let's say the total distance from his house to school is D. Since the traffic light is in the middle, the distance from his house to the traffic light is D/2, and the same from the traffic light to school.On Monday, he didn't stop at the traffic light because it was green. So, he walked the entire distance D at his usual speed, let's call it v. Therefore, the time he took on Monday is just the total distance divided by his speed, which is D/v.On Tuesday, things are a bit different. He walked the first half of the journey, which is D/2, at his usual speed v. Then, he encountered the traffic light and had to wait for 5 minutes. After that, he doubled his speed, so his speed became 2v for the remaining half of the journey, which is another D/2.So, let's calculate the time he took on Tuesday. First, the time to walk the first half: (D/2)/v = D/(2v). Then, he waits for 5 minutes. After that, he walks the second half at double speed: (D/2)/(2v) = D/(4v). So, the total time on Tuesday is D/(2v) + 5 minutes + D/(4v).But wait, the problem says that the total time on both days was the same. So, the time on Monday, which is D/v, is equal to the time on Tuesday, which is D/(2v) + 5 + D/(4v). Let me write that equation down:D/v = D/(2v) + 5 + D/(4v)Hmm, let me simplify the right side. First, let's combine the terms with D. So, D/(2v) + D/(4v) can be combined by finding a common denominator, which is 4v. So, that's (2D)/(4v) + D/(4v) = (3D)/(4v). So, the equation becomes:D/v = (3D)/(4v) + 5Now, let's subtract (3D)/(4v) from both sides to get:D/v - (3D)/(4v) = 5Simplify the left side. D/v is the same as (4D)/(4v), so:(4D)/(4v) - (3D)/(4v) = (D)/(4v) = 5So, D/(4v) = 5. Therefore, D/v = 5 * 4 = 20.Wait, D/v is the time he took on Monday, which is equal to the total time on Tuesday. So, the time he spent each day is 20 minutes.Let me just verify that. So, on Monday, he walked the entire distance D at speed v, taking 20 minutes. On Tuesday, he walked the first half D/2 at speed v, which would take (D/2)/v = (D/v)/2 = 10 minutes. Then he waited 5 minutes, and then walked the second half at 2v, which would take (D/2)/(2v) = (D)/(4v) = 5 minutes. So, total time on Tuesday is 10 + 5 + 5 = 20 minutes. Yep, that checks out.So, the time he spent each day is 20 minutes.**Final Answer**The time Vasya spent traveling each day is boxed{20} minutes.

answer：First, I recognize that if vectors a and b are perpendicular, their dot product must be zero. This gives me the equation 4*1 + m*(-2) = 0, which simplifies to 4 - 2m = 0. Solving for m, I find that m equals 2.Next, I substitute m = 2 into vector a, making it (4, 2). Then, I add vector b, which is (1, -2), to vector a. This results in the vector a + b = (5, 0).Finally, to find the magnitude of the resulting vector, I use the formula for the magnitude of a vector, which is the square root of the sum of the squares of its components. Calculating this, I get √(5² + 0²) = √25 = 5.

answer：To determine how much Mr. John will have left after deducting his monthly transport fare, I'll start by calculating 5% of his income.First, I'll find 5% of 2000 by multiplying 2000 by 0.05, which equals 100. This is the amount he spends on public transport each month.Next, I'll subtract the transport fare from his total income: 2000 minus 100 equals 1900.Therefore, Mr. John will have 1900 left after paying for his monthly transport fare.

answer：First, I'll define the fraction as ( frac{x}{y} ), where ( x ) is the numerator and ( y ) is the denominator.When 1 is subtracted from the denominator, the fraction becomes ( frac{1}{3} ). This gives the equation:[frac{x}{y - 1} = frac{1}{3}]Cross-multiplying, I get:[3x = y - 1]So, ( y = 3x + 1 ).Next, when 4 is added to the numerator, the fraction becomes ( frac{1}{2} ). This gives the equation:[frac{x + 4}{y} = frac{1}{2}]Cross-multiplying, I get:[2(x + 4) = y]Substituting ( y = 3x + 1 ) into this equation:[2x + 8 = 3x + 1]Solving for ( x ), I find:[x = 7]Then, substituting ( x = 7 ) back into ( y = 3x + 1 ):[y = 22]Therefore, the original fraction is ( frac{7}{22} ).