answer:1. Let ( X_n ) represent the number of dice throws needed for the sum of the outcomes to reach ( n ). 2. Define the indicator random variable ( I_k ) such that: I_k = begin{cases} 1, & text{if the first throw results in } k text{ points,} 0, & text{otherwise}. end{cases} 3. The expected value of ( I_k ) is: mathrm{E}[I_k] = 0 cdot frac{5}{6} + 1 cdot frac{1}{6} = frac{1}{6}. 4. Denoting ( X_{n-k} ) as the number of throws from the second throw onward to achieve a sum of ( n-k ), we have: X_n = X_{n-1}I_1 + X_{n-2}I_2 + cdots + X_{n-6}I_6 + 1. 5. Since ( X_{n-k} ) is independent of ( I_k ), taking the expectation on both sides, we obtain: mathrm{E}[X_n] = sum_{k=1}^{6} mathrm{E}[X_{n-k}] mathrm{E}[I_k] + 1 = frac{1}{6} sum_{k=1}^{6} mathrm{E}[X_{n-k}] + 1. 6. Let ( mathrm{E}[X_k] = e_k ) for simplicity. The recurrence relation becomes: e_n = frac{1}{6} sum_{k=1}^{6} e_{n-k} + 1. 7. Recognize that for ( n ) where ( n leq 6 ), the sums ( sum_{k=1}^{6} e_{n-k} ) will include terms ( e_k = 0 ) for ( k leq 0 ). 8. Generate these values one at a time; starting from base cases like ( e_0 = 0 ), we incrementally determine: begin{aligned} e_1 & = 1, e_2 & = frac{1}{6} cdot (1 + 0 + 0 + 0 + 0 + 0) + 1 = frac{7}{6}, e_3 & = frac{1}{6} cdot (1 + frac{7}{6} + 0 + 0 + 0 + 0) + 1 = frac{49}{36}, e_4 & = frac{1}{6} cdot (1 + frac{7}{6} + frac{49}{36} + 0 + 0 + 0) + 1 = frac{343}{216}, & vdots end{aligned} 9. For larger values of ( n ), recognizing that the expected number of throws, ( e_n ), increases approximately linearly, by mathematizing: e_n approx k cdot n, where calculations or computational tools like Excel might be effective for exact large ( n ). 10. By iterating through values up to 2010, an approximate calculation yields: e_{2010} approx 574.7619047619048. 11. Therefore, the expected number of throws necessary to achieve a sum of 2010 is approximately: boxed{574.761904}

answer:To find out how much money Mutter paid as lagaan, we need to calculate the proportion of the total lagaan that corresponds to Mutter's share of the total taxable land. Given that the total lagaan collected is Rs. 3,44,000 and Mutter's percentage of the total taxable land is 0.23255813953488372%, we can calculate Mutter's lagaan payment as follows: Mutter's lagaan payment = (Mutter's percentage of total taxable land / 100) * Total lagaan collected Plugging in the numbers: Mutter's lagaan payment = (0.23255813953488372 / 100) * 3,44,000 Mutter's lagaan payment = 0.0023255813953488372 * 3,44,000 Mutter's lagaan payment = Rs. 800 So, Mutter paid Rs. boxed{800} as lagaan.

answer:First, we recognize that the numbers form a decreasing arithmetic sequence where each number is 3 less than the preceding number. To count the terms easily, we first reverse the list to get 42, 45, ldots, 156, 159, 162. Next, since the common difference is 3, we divide each number by 3 to simplify: frac{42}{3}, frac{45}{3}, ldots, frac{156}{3}, frac{159}{3}, frac{162}{3} This simplifies to: 14, 15, ldots, 52, 53, 54. The sequence 14, 15, ldots, 54 is easier to count. To find the number of terms, we use the formula for the number of terms in an arithmetic sequence, which is: n = frac{text{last term} - text{first term}}{text{common difference}} + 1 = frac{54 - 14}{1} + 1 = 41. Hence, there are boxed{41} numbers in the list.

answer:First, we list all three-digit prime numbers less than 200: 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, and 197. Next, we check which of these numbers are palindromic primes: - **101** reverses to **101**, both prime. - **103** reverses to **301**, not prime. - **107** reverses to **701**, prime. - **109** reverses to **901**, not prime. - **113** reverses to **311**, not prime. - **127** reverses to **721**, not prime. - **131** reverses to **131**, both prime. - **137** reverses to **731**, not prime. - **139** reverses to **931**, not prime. - **149** reverses to **941**, not prime. - **151** reverses to **151**, both prime. - **157** reverses to **751**, prime. - **163** reverses to **361**, not prime. - **167** reverses to **761**, prime. - **173** reverses to **371**, not prime. - **179** reverses to **971**, not prime. - **181** reverses to **181**, both prime. - **191** reverses to **191**, both prime. - **193** reverses to **391**, not prime. - **197** reverses to **791**, prime. Adding the valid three-digit palindromic primes: 101, 107, 131, 151, 157, 167, 181, 191, 197, we get: [101 + 107 + 131 + 151 + 157 + 167 + 181 + 191 + 197 = boxed{1383}]