answer:1. Let ( n = 1990! ). The positive divisors of ( n ) are denoted by ( d_1, d_2, ldots, d_k ). 2. Consider a divisor ( d_i ) of ( n ). By the definition of divisors, there exists another divisor ( d_j ) such that ( d_j = frac{n}{d_i} ). 3. Notice that for each ( d_i ), there is a corresponding ( d_j ) such that ( d_i cdot d_j = n ). This implies ( d_j = frac{n}{d_i} ). 4. We need to show that: [ sum_{i=1}^k frac{d_i}{sqrt{n}} = sum_{i=1}^k frac{sqrt{n}}{d_i} ] 5. Consider the left-hand side of the equation: [ sum_{i=1}^k frac{d_i}{sqrt{n}} ] Since ( d_i cdot d_j = n ), we can rewrite ( d_j ) as ( frac{n}{d_i} ). Therefore, the left-hand side becomes: [ sum_{i=1}^k frac{d_i}{sqrt{n}} = sum_{i=1}^k frac{d_i}{sqrt{n}} ] 6. Now consider the right-hand side of the equation: [ sum_{i=1}^k frac{sqrt{n}}{d_i} ] Using the fact that ( d_j = frac{n}{d_i} ), we can rewrite the right-hand side as: [ sum_{i=1}^k frac{sqrt{n}}{d_i} = sum_{i=1}^k frac{sqrt{n}}{frac{n}{d_i}} = sum_{i=1}^k frac{d_i}{sqrt{n}} ] 7. Since both sides of the equation are equal, we have: [ sum_{i=1}^k frac{d_i}{sqrt{n}} = sum_{i=1}^k frac{sqrt{n}}{d_i} ] (blacksquare)

answer:First, let's find out the fodder consumption ratio for each type of animal based on the given information: 3 buffaloes = 4 cows = 2 oxen = 6 goats This means: 1 buffalo = 4/3 cows 1 buffalo = 2/3 oxen 1 buffalo = 6/3 goats = 2 goats Now, let's calculate the total fodder consumption for the current animals on the farm: 15 buffaloes = 15 * (4/3 cows) = 20 cows worth of fodder 8 oxen = 8 * (3/2 buffaloes) = 12 buffaloes worth of fodder = 12 * (4/3 cows) = 16 cows worth of fodder 24 cows = 24 cows worth of fodder 12 goats = 12 * (1/2 buffalo) = 6 buffaloes worth of fodder = 6 * (4/3 cows) = 8 cows worth of fodder Total fodder consumption in terms of "cow units" is: 20 (buffaloes) + 16 (oxen) + 24 (cows) + 8 (goats) = 68 cow units The fodder available is enough for 48 days for these animals. So, the total amount of fodder available is: 68 cow units * 48 days = 3264 cow unit-days Now, let's calculate the total fodder consumption for the additional animals: 80 more cows = 80 cows worth of fodder 50 more buffaloes = 50 * (4/3 cows) = 200/3 cows worth of fodder ≈ 66.67 cows worth of fodder 20 more oxen = 20 * (3/2 buffaloes) = 30 buffaloes worth of fodder = 30 * (4/3 cows) = 40 cows worth of fodder 30 more goats = 30 * (1/2 buffalo) = 15 buffaloes worth of fodder = 15 * (4/3 cows) = 20 cows worth of fodder Total additional fodder consumption in terms of "cow units" is: 80 (cows) + 66.67 (buffaloes) + 40 (oxen) + 20 (goats) = 206.67 cow units Now, let's add the additional animals to the current animals to get the new total fodder consumption: 68 (current cow units) + 206.67 (additional cow units) = 274.67 cow units Finally, let's find out how many days the fodder will last for all the animals: 3264 cow unit-days / 274.67 cow units = 11.88 days Therefore, the fodder will last for approximately boxed{11.88} days after the additional animals are brought in.

answer:First, let's find out how many notebooks Amanda had after ordering more but before losing any: Initial notebooks: 65 Ordered more: 23 Total before loss: 65 + 23 = 88 notebooks Now, let's calculate how many notebooks she lost: 15% of 88 notebooks = 0.15 * 88 = 13.2 notebooks Since Amanda can't lose a fraction of a notebook, we'll round this to the nearest whole number. Depending on the context, you might round down to 13 notebooks (since you typically can't lose part of a physical notebook), or round up to 14 notebooks (if the percentage represents a statistical average over many instances). Let's assume we round down to 13 notebooks for this calculation. Now, let's find out how many notebooks Amanda has after the loss: Total after loss: 88 - 13 = 75 notebooks So, Amanda has boxed{75} notebooks now.

answer:Given: - overrightarrow{a}=(3,1) - overrightarrow{b}=(1,-1) - overrightarrow{c}=overrightarrow{a}+koverrightarrow{b} Since vectors overrightarrow{a} and overrightarrow{c} are perpendicular, their dot product equals zero. This can be formulated as overrightarrow{a} cdot overrightarrow{c} = 0. We expand overrightarrow{c} based on its definition: overrightarrow{c} = overrightarrow{a} + koverrightarrow{b} = (3,1) + k(1,-1) = (3+k, 1-k) Now, applying the dot product of overrightarrow{a} and overrightarrow{c}: overrightarrow{a} cdot overrightarrow{c} = (3,1) cdot (3+k, 1-k) = 3(3+k) + 1(1-k) Simplifying the dot product expression gives: 9 + 3k + 1 - k = 10 + 2k Setting the equation equal to zero (because overrightarrow{a} and overrightarrow{c} are perpendicular): 10 + 2k = 0 Solving for k involves isolating k: 2k = -10 k = frac{-10}{2} k = -5 Therefore, the value of k that satisfies the condition is k = boxed{-5}.