answer:1. **Define the dimensions of the rectangle**: Let the width be w. The length is then 3w given it is three times the width. 2. **Apply the Pythagorean Theorem**: The diagonal forms a right triangle with the rectangle's length and width. Using the Pythagorean theorem: [ w^2 + (3w)^2 = 16^2 ] Simplifying the equation: [ w^2 + 9w^2 = 256 implies 10w^2 = 256 ] 3. **Solve for w**: [ w^2 = frac{256}{10} implies w = sqrt{frac{256}{10}} = frac{16}{sqrt{10}} = frac{16sqrt{10}}{10} ] 4. **Calculate the area of the rectangle**: The area A is given by width times length: [ A = w times 3w = 3w^2 ] Substitute the expression for w^2: [ A = 3 left(frac{256}{10}right) = frac{768}{10} = 76.8 ] The area of the rectangle, in terms of its dimensions, is 76.8 square units. The final answer is boxed{76.8}

answer:If Jack has 198 books in the classics section and each author has 33 books, we can find out how many classic authors he has by dividing the total number of books by the number of books per author. 198 books ÷ 33 books/author = 6 authors Jack has boxed{6} classic authors in his collection.

answer:1. Calculate the total number of rearrangements for the name "Anna". The name has four letters, 'A', 'N', 'N', 'A', with 'A' and 'N' each repeated twice. The formula for rearrangements when there are repeated elements is: [ frac{n!}{p_1! p_2! dots p_k!} ] where (n) is the total number of letters, and (p_1, p_2, dots, p_k) are the numbers of times each distinct element is repeated. Thus for "Anna" it is: [ frac{4!}{2! times 2!} = frac{24}{4} = 6 text{ arrangements.} ] 2. Calculate the time taken to write all the arrangements. - Anna writes 8 rearrangements per minute. [ text{Time in minutes} = frac{6 text{ arrangements}}{8 text{ arrangements/minute}} = 0.75 text{ minutes.} ] 3. Convert minutes to hours. [ text{Time in hours} = frac{0.75 text{ minutes}}{60 text{ minutes/hour}} = frac{0.75}{60} text{ hours} = boxed{0.0125 text{ hours}.} ] Conclusion: The calculation shows that it will take Anna (0.0125) hours, or (45) seconds, to write down all possible rearrangements of her name "Anna".

answer:1. Define the sequence (a_1, a_2, a_3, ldots) by (a_1 = 1) and, for (n > 1), [ a_n = a_{lfloor n/2 rfloor} + a_{lfloor n/3 rfloor} + ldots + a_{lfloor n/n rfloor} + 1. ] We need to prove that there are infinitely many (n) such that (a_n equiv n pmod{2^{2010}}). 2. Let (M = 2010). 3. Define (d_n = a_n - a_{n-1}) for (n > 1) and (d_1 = 1). Note that for (n > 1), [ d_n = a_n - a_{n-1} = (a_{lfloor n/2 rfloor} - a_{lfloor (n-1)/2 rfloor}) + (a_{lfloor n/3 rfloor} - a_{lfloor (n-1)/3 rfloor}) + ldots + (a_{lfloor n/(n-1) rfloor} - a_{lfloor (n-1)/(n-1) rfloor}) + a_{lfloor n/n rfloor}. ] 4. For all integers (k not| n), (a_{lfloor n/k rfloor} = a_{lfloor (n-1)/k rfloor}) and thus cancel out in the sum. For all (k | n), [ a_{lfloor n/k rfloor} - a_{lfloor (n-1)/k rfloor} = a_{n/k} - a_{n/k-1} = d_{n/k}. ] Thus, [ d_n = sum_{k|n, k<n} d_k. ] 5. **Lemma 1:** Suppose that (p^k | m) for a prime (p) and positive integer (k). Then, (2^{k-1} | d_m). 6. **Proof of Lemma 1:** The proof will be by induction on (m). - For (m = 2), the statement follows from the fact that the only prime factor is (2) and thus, (2^{1-1} = 1 | d_2). - Suppose the statement holds for all (m < n). Now, let (m = n). If (k = 1), then the statement follows trivially. Suppose that (k ge 2). Note that [ d_n = sum_{i|n, i<n} d_i = d_{n/p} + sum_{i|(n/p), i<n/p} d_i + sum_{i|(n/p^k), i<n/p^k} d_{i cdot p^k} = d_{n/p} + d_{n/p} + sum_{i|(n/p^k), i<n/p^k} d_{i cdot p^k} = 2d_{n/p} + sum_{i|(n/p^k), i<n/p^k} d_{i cdot p^k}. ] Note that every value in the summation (if any exist) [ sum_{i|(n/p^k), i<n/p^k} d_{i cdot p^k} ] is divisible by (2^{k-1}) by the inductive assumption. Thus, [ d_n equiv 2d_{n/p} pmod{2^{k-1}}. ] Note that since (p^{k-1} | (n/p)), by the inductive assumption, (2^{k-2} | d_{n/p}). Thus, (2^{k-1} | 2d_{n/p}) and therefore (2^{k-1} | d_n), which completes the induction. (blacksquare) 7. Let (p_1, p_2, ldots, p_{2^M}) be (2^M) distinct odd primes greater than (2^M). By the Chinese Remainder Theorem, there exist infinitely many integers (n) greater than (2^M) such that [ n equiv i - 1 pmod{p_i^{M+1}} ] for all (1 le i le 2^M). Note that by Lemma 1, since (p_i^{M+1} | (n - i + 1)), (d_{n - i + 1} equiv 0 pmod{2^M}) for all (1 le i le 2^M). Thus, [ a_n equiv a_{n-1} equiv ldots equiv a_{n-2^M+1} equiv c pmod{2^M} ] where (c) is some integer. Note that among the integers (n, n-1, ldots, n-2^M+1), there exists exactly one that is congruent to (c) modulo (2^M). Let this integer be (l). Then, [ a_l equiv c equiv l pmod{2^M}. ] Thus, there exist infinitely many (n) such that (a_n equiv n pmod{2^M}).