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question:Consider the sequence a_{n}=cos (underbrace{100 ldots 0^{circ}}_{n1}) For example, a_{1}=cos 1^{circ}, a_{6}=cos 100000^{circ}. How many of the numbers a_{1}, a_{2}, ldots, a_{100} are positive?
answer:1. We start by examining the sequence given by [ a_n = cosleft(underbrace{100 ldots 0^circ}_{n-1 text{ zeros}}right). ] For example, [ a_1 = cos(1^circ), quad a_6 = cos(100000^circ). ] 2. Notice that for an integer ( x ) divisible by 40, the following holds: [ cos(x^circ) = cos((10x)^circ). ] This is because the difference between ( x^circ ) and ( 10x^circ ), which is ( 9x^circ ), is an integer multiple of ( 360^circ ). Therefore, ( cos(9x^circ) = cos(0^circ) = 1 ) when divided by ( 360^circ ). 3. Consequently, [ cosleft(left(10^kright)^circright) = cosleft(left(10^{k+1}right)^circright) quad text{for all} quad k geq 3. ] 4. By examining the cosine values, we observe: [ cosleft(1^circright) > 0, ] [ cosleft(10^circright) > 0, ] [ cosleft(100^circright) < 0. ] 5. For values ( k geq 3 ): [ 0 < cosleft(1000^circright) = cosleft(10000^circright) = cdots = cosleft(left(10^{99}right)^circright). ] 6. Therefore, out of the 100 terms in our sequence, only ( cosleft(100^circright) ) is negative. The rest 99 values are positive. Conclusion: [ boxed{99} ]
question:Twelve people sit around a circular table. Some of them are knights who always tell the truth, and the rest are liars who always lie. Each of the seated people declared the person sitting to their left to be a liar. Can it be definitively determined how many knights and how many liars are at the table?
answer:: 1. Let us denote the people sitting around the table by ( P_1, P_2, ldots, P_{12} ) in a clockwise manner. 2. Each person ( P_i ) claims that the person ( P_{i-1} ) (to their left) is a liar. 3. This implies that if ( P_i ) is a knight (truth-teller), then ( P_{i-1} ) must indeed be a liar. 4. Conversely, if ( P_i ) is a liar, then ( P_{i-1} ) must be a knight (since the liar always lies). Given the above logic, we must consider two adjacent people and deduce their roles based on the statements made. For instance: - If ( P_1 ) is a knight, then ( P_{12} ) must be a liar. - If ( P_1 ) is a liar, then ( P_{12} ) must be a knight. Hence, following this pattern: - If ( P_1 ) is a knight (truth-teller), then ( P_2 ) must be a liar since ( P_2 ) would call ( P_1 ) a liar inaccurately. - If ( P_1 ) is a liar, then ( P_2 ) must be a knight, since ( P_1 )'s statement would be false. We continue this pattern around the entire table: - If ( P_1 ) is a knight, this implies an alternating sequence of knights and liars around the table. - If ( P_1 ) is a liar, the pattern still necessitates an alternating sequence of knights and liars in order for everyone's statements to remain consistent. Now, since there are 12 people sitting around the table, and our observations require that knights and liars alternate positions around the table, there are only two scenarios: - 6 knights and 6 liars. Observing the above patterns: - If ( P_1 ) is a knight, then ( P_2, P_4, P_6, P_8, P_{10}, P_{12} ) are all liars, and ( P_3, P_5, P_7, P_9, P_{11} ) are knights. - If ( P_1 ) is a liar, then ( P_2, P_4, P_6, P_8, P_{10}, P_{12} ) are all knights, and ( P_3, P_5, P_7, P_9, P_{11} ) are liars. In either case, we can conclude that the number of knights and liars is: [ boxed{text{6 knights and 6 liars}}.
question:A grocer has 400 pounds of coffee in stock, 40 percent of which is decaffeinated. If the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated, what percent, by weight, of the grocer's stock of coffee is decaffeinated?
answer:First, let's calculate the amount of decaffeinated coffee in the initial stock: 40% of 400 pounds = 0.40 * 400 = 160 pounds of decaffeinated coffee. Now, let's calculate the amount of decaffeinated coffee in the additional 100 pounds: 60% of 100 pounds = 0.60 * 100 = 60 pounds of decaffeinated coffee. Adding the decaffeinated coffee from both stocks: 160 pounds + 60 pounds = 220 pounds of decaffeinated coffee. Now, let's calculate the total weight of the grocer's stock after the additional purchase: 400 pounds + 100 pounds = 500 pounds of total coffee. Finally, let's calculate the percentage of decaffeinated coffee in the total stock: (220 pounds decaffeinated / 500 pounds total) * 100 = 44%. So, boxed{44%} of the grocer's stock of coffee is decaffeinated by weight.
question:When Derek was 7 years old, he had four times as many dogs as cars. Fifteen years later, after selling some of his dogs and buying 350 more cars, the number of cars became three times the number of dogs. How many dogs does Derek have now if he had 120 dogs when he was seven years old?
answer:When Derek was 7 years old, he had 120 dogs. At that time, he had four times as many dogs as cars. So, the number of cars he had when he was 7 years old can be calculated as: Number of cars = Number of dogs / 4 Number of cars = 120 / 4 Number of cars = 30 Fifteen years later, Derek sold some of his dogs and bought 350 more cars. Let's denote the number of dogs he sold as "D" and the new number of dogs he has as "Dogs_now". So, the new number of dogs he has is: Dogs_now = 120 - D The new number of cars he has is: Cars_now = 30 + 350 Cars_now = 380 At this point, the number of cars is three times the number of dogs: Cars_now = 3 * Dogs_now 380 = 3 * (120 - D) Now, let's solve for D: 380 = 360 - 3D 3D = 360 - 380 3D = -20 D = -20 / 3 D = -6.666... However, it's not possible to have a negative number of dogs sold, so there must be a mistake in the interpretation of the problem. Let's re-evaluate the information given: Fifteen years later, the number of cars became three times the number of dogs. This means that after buying 350 more cars, Derek had 380 cars, and this number is three times the number of dogs he has now. So, we can write: 380 cars = 3 * Dogs_now Now, let's solve for Dogs_now: Dogs_now = 380 / 3 Dogs_now = 126.666... Since Derek cannot have a fraction of a dog, we round down to the nearest whole number: Dogs_now = 126 So, Derek has boxed{126} dogs now.