answer:First, we find how many positive integers less than 200 are multiples of both 16 and 28. - 16 = 2^4 - 28 = 2^2 cdot 7 - The least common multiple (LCM) of 16 and 28 is found by taking the maximum power of each prime factor: 2^4 cdot 7 = 112. Next, we find how many multiples of 112 there are under 200: - Dividing 200 by 112 gives quotient 1 (and remainder 88), indicating there is only 1 multiple of 112 less than 200. We then count the number of multiples of each number: - For 16, 200 div 16 = 12.5, which means there are 12 multiples of 16 less than 200. - For 28, 200 div 28 approx 7.14, so there are 7 multiples of 28 less than 200. Alice and Alan can therefore choose from 12 cdot 7 = 84 different two-number combinations. Only 1 of these combinations involves selecting the same number (the multiple of 112). Thus, the probability that they selected the same number is frac{1}{84}. boxed{frac{1}{84}} is the probability that Alice and Alan selected the same number.

answer:Carol spends 1/4 of her savings on a stereo. For the television, she spends 2/3 less than what she spent on the stereo. To find out how much she spends on the television, we need to calculate 2/3 of the amount she spent on the stereo. Since she spent 1/4 of her savings on the stereo, 2/3 of that amount would be: (2/3) * (1/4) = 2/12 = 1/6 Now, to find out how much less she spends on the television compared to the stereo, we subtract this fraction from the amount she spent on the stereo: (1/4) - (1/6) = (3/12) - (2/12) = 1/12 So, Carol spends 1/12 of her savings on the television. To find the total fraction of her savings that she spent on both the stereo and the television, we add the fractions together: (1/4) + (1/12) = (3/12) + (1/12) = 4/12 = 1/3 Therefore, Carol spent boxed{1/3} of her savings on the stereo and television combined.

answer:1. Let's express the relative strength of each character in terms of the strength of Mouse (M): - Cat (C) is 6 times as strong as Mouse (M): [ C = 6M ] - Juchka (J) is 5 times as strong as Cat (C): [ J = 5C = 5 cdot 6M = 30M ] - Granddaughter (G) is 4 times as strong as Juchka (J): [ G = 4J = 4 cdot 30M = 120M ] - Babka (B) is 3 times as strong as Granddaughter (G): [ B = 3G = 3 cdot 120M = 360M ] - Dedka (D) is 2 times as strong as Babka (B): [ D = 2B = 2 cdot 360M = 720M ] 2. Calculate the total strength of Dedka, Babka, Granddaughter, Juchka, Cat, and Mouse together: [ D + B + G + J + C + M = 720M + 360M + 120M + 30M + 6M + M = 1237M ] 3. According to the problem, when Mouse is not there, Roshka cannot be pulled out. This means that the combined strength without the Mouse must be less than the required strength to pull out Roshka. Thus: [ 720M + 360M + 120M + 30M + 6M < 1237M ] Simplifying this, we have: [ 1236M < 1237M ] 4. From the problem, we understand that the required strength to pull out the Roshka is 1237M. To find the number of mice required to have this strength: [ frac{1237M}{M} = 1237 ] # Conclusion: [ boxed{1237} ]

answer:If the original number of cats on the island was 1800 and 600 were relocated on the first mission, then the number of cats remaining on the island after the first mission would be: 1800 - 600 = 1200 cats On the second mission, half of the remaining cats were relocated. So, half of 1200 cats would be: 1200 / 2 = 600 cats Therefore, 600 cats were relocated on the second mission. To find out how many cats remained on the island after the second mission, we subtract the number of cats relocated on the second mission from the number of cats remaining after the first mission: 1200 - 600 = 600 cats So, boxed{600} cats remained on the island after the rescue mission.