answer:To show that 2^{4n+1} + 3^{4n+1} is divisible by 5 for any integer n, we can work modulo 5. This means we will consider the remainders when the expressions are divided by 5. 1. Start by expressing 2^{4n+1} + 3^{4n+1} as: 2^{4n+1} + 3^{4n+1} 2. Notice that 2^{4n+1} can be rewritten using exponent properties: 2^{4n+1} = 2^{4n} cdot 2 Similarly, rewrite 3^{4n+1}: 3^{4n+1} = 3^{4n} cdot 3 Thus: 2^{4n+1} + 3^{4n+1} = 2^{4n} cdot 2 + 3^{4n} cdot 3 3. Next, evaluate the terms 2^{4n} and 3^{4n} modulo 5. Notice the following periodic patterns because 4 is the order of 2 and 3 under modulo 5 arithmetic: - For 2^{4n} mod 5: A few powers of 2 modulo 5 are: [ begin{aligned} 2^1 &equiv 2 mod 5, 2^2 &equiv 4 mod 5, 2^3 &equiv 3 mod 5, 2^4 &equiv 1 mod 5 end{aligned} ] Hence, for any integer n, 2^{4n} equiv 1 mod 5. - For 3^{4n} mod 5: A few powers of 3 modulo 5 are: [ begin{aligned} 3^1 &equiv 3 mod 5, 3^2 &equiv 4 mod 5, 3^3 &equiv 2 mod 5, 3^4 &equiv 1 mod 5 end{aligned} ] Hence, for any integer n, 3^{4n} equiv 1 mod 5. 4. Substituting back, we get: 2^{4n} cdot 2 equiv 1 cdot 2 equiv 2 mod 5 3^{4n} cdot 3 equiv 1 cdot 3 equiv 3 mod 5 Thus: 2^{4n+1} + 3^{4n+1} equiv 2 + 3 equiv 5 equiv 0 mod 5 # Conclusion We have shown that 2^{4n+1} + 3^{4n+1} is divisible by 5 for any integer n. blacksquare

answer:Given the student population in the first year (Grade 10), second year (Grade 11), and third year (Grade 12) are 900, 900, and 1200 respectively, the ratio of student populations among the three grades is 3:3:4. Since stratified sampling is being used, we need to retain this proportion in our sample. Therefore, to determine the number of third-year students to be sampled, we calculate the fraction of the total student population that they represent and apply this fraction to the sample size of 50. The population proportion for third-year students is their population divided by the total population: frac{1200}{900 + 900 + 1200} = frac{1200}{3000} = frac{4}{10} = 0.4 Next, we apply this proportion to the sample size of 50 students: 50 times 0.4 = 20 Therefore, 20 students should be sampled from the third year class. [ boxed{20 text{ third-year students should be sampled.}} ]

answer:Let's denote the rate at which pipe A fills the cistern as A, the rate at which pipe B fills the cistern as B, and the rate at which the third pipe (let's call it C) empties the cistern as C. Since pipe A can fill the cistern in 60 minutes, its rate is 1/60 of the cistern per minute. Similarly, since pipe B can fill the cistern in 75 minutes, its rate is 1/75 of the cistern per minute. When all three pipes are open, the combined rate at which the cistern fills is the sum of the rates of pipes A and B minus the rate of pipe C. We know that with all three pipes open, the cistern fills in 50 minutes, so the combined rate is 1/50 of the cistern per minute. Therefore, we can write the following equation: A + B - C = 1/50 Substituting the rates of A and B, we get: (1/60) + (1/75) - C = 1/50 To solve for C, we need to find a common denominator for the fractions. The least common multiple of 60, 75, and 50 is 300. So we can rewrite the equation as: (5/300) + (4/300) - C = 6/300 Combining the fractions on the left side, we get: (5 + 4)/300 - C = 6/300 9/300 - C = 6/300 Now, we isolate C: -C = 6/300 - 9/300 -C = -3/300 C = 3/300 C = 1/100 This means that the third pipe empties the cistern at a rate of 1/100 of the cistern per minute. To find out how long it would take for the third pipe to empty the cistern by itself, we take the reciprocal of the rate: Time = 1 / (1/100) Time = 100 minutes Therefore, the third pipe alone can empty the cistern in boxed{100} minutes.

answer:Let's use variables to represent the number of each type of pet Frankie has: Let C = the number of cats Let S = the number of snakes Let P = the number of parrots Let D = the number of dogs From the information given, we can set up the following equations: 1. S = C + 6 (Frankie has six more snakes than cats) 2. P = C - 1 (Frankie has one less parrot than cats) 3. C + D = 6 (Six of his pets have four legs, which would be the cats and dogs) We also know that the total number of pets is 19, so: 4. C + S + P + D = 19 Now we can substitute equations 1 and 2 into equation 4: C + (C + 6) + (C - 1) + D = 19 Combine like terms: 3C + 6 - 1 + D = 19 3C + 5 + D = 19 Now we can substitute equation 3 into the above equation: 3C + 5 + (6 - C) = 19 Combine like terms: 3C - C + 5 + 6 = 19 2C + 11 = 19 Subtract 11 from both sides: 2C = 8 Divide by 2: C = 4 Now we know Frankie has 4 cats. We can use this information to find the number of other pets: S = C + 6 = 4 + 6 = 10 snakes P = C - 1 = 4 - 1 = 3 parrots D = 6 - C = 6 - 4 = 2 dogs Now we can add up all the pets: C + S + P + D = 4 + 10 + 3 + 2 = 19 pets So Frankie has a total of boxed{19} pets, which is consistent with the information given.