answer:Let's denote the number of male students as M and the number of female students as F. According to the information given: F = 658 (number of female students) M = F - 38 (since there are 38 more female students than male students) Now we can find the number of male students: M = 658 - 38 M = 620 The total number of students who came to school would be the sum of male and female students minus the number of students who were absent. Let's denote the total number of students who came to school as T and the number of absent students as A. A = 17 (number of absent students) T = M + F - A T = 620 + 658 - 17 T = 1278 - 17 T = 1261 So, the total number of students who came to school is boxed{1261} .

answer:1. **Claim**: There exist infinitely many ( n ) such that both intervals ( (na, na+a] ) and ( (nb, nb+b] ) contain only composite numbers. **Proof**: For the sake of contradiction, suppose there exists ( N in mathbb{N} ), and for all ( n ge N ) there is a prime either in ( (na, na+a] ) or in ( (nb, nb+b] ) (or in both). Then for any ( k in mathbb{N} ), in some of the intervals ( (Na, Na+ka] ) or ( (Nb, Nb+kb] ) there exist at least ( k/2 ) prime numbers. This implies that the prime numbers have a positive density, which contradicts the prime number theorem (PNT). (blacksquare) 2. For those ( n ) satisfying the above claim, we have ( pi((n+1)a) = pi(na) ) and ( pi((n+1)b) = pi(nb) ). Hence, [ frac{P(n)}{Q(n)} = frac{P(n+1)}{Q(n+1)} ] for infinitely many natural numbers ( n ). This yields [ P(z)Q(z+1) = P(z+1)Q(z), quad forall z in mathbb{C} quad quad quad (1) ] 3. If ( z_0 in mathbb{C} ) is a root of ( P(z) ) with maximal ( mathrm{Re}(z) ), the above identity implies ( z_0 ) is also a root of ( Q(z) ). Note that ( P_1(z) := frac{P(z)}{z - z_0} ) and ( Q_1(z) := frac{Q(z)}{z - z_0} ) also satisfy ( (1) ). Thus, applying that operation several times, we get ( P(z) = cQ(z) ) for some constant ( c in mathbb{C} ). 4. By the prime number theorem (PNT), ( pi(an) sim frac{an}{ln(an)} ), so it easily follows that ( c = frac{a}{b} ). Hence, [ frac{pi(an)}{pi(bn)} = frac{a}{b} ] 5. It gives [ frac{pi(an+a) - pi(an)}{pi(bn+b) - pi(bn)} = frac{a}{b} ] 6. Let ( n = kb! ). Then ( bn+2, bn+3, dots, bn+b ) are all composite numbers. Consider the sequence ( kb cdot b! + 1, k = 1, 2, dots ). By Dirichlet's theorem, there exists ( k in mathbb{N} ) such that ( kb cdot b! + 1 ) is a prime. For that ( k ), we get ( pi(bn+b) - pi(bn) = 1 ). It yields ( frac{a}{b} ) is an integer. With the same trick, we can find ( n ) such that ( pi(an+a) - pi(an) = 1 ), which yields ( frac{b}{a} ) is an integer. 7. Hence, the only possibility is ( a = b ). In this case, the condition of the statement obviously holds. The final answer is ( boxed{ (a, b) = (k, k) } ) for any ( k in mathbb{N} ).

answer:First, let's calculate the cost of the items Mary needs for each class: - Folders: 12 classes * 3.50 = 42 - Notebooks: 12 classes * 2 notebooks * 3 = 72 - Binders: 12 classes * 5 = 60 - Pencils: 12 classes * 3 pencils * 1 = 36 - Erasers: For every 6 pencils, she needs 2 erasers. She has 36 pencils, so she needs 36/6 * 2 erasers = 12 erasers * 0.75 = 9 - Highlighters: 1 pack * 3.25 = 3.25 - Markers: 1 pack * 3.50 = 3.50 - Sticky notes: 1 pack * 2.50 = 2.50 Now, let's add up the cost of these items: 42 + 72 + 60 + 36 + 9 + 3.25 + 3.50 + 2.50 = 228.25 Since Mary spent 210, we need to subtract the cost of the items we've already calculated from the total amount she spent to find out how much she spent on the set of paints, the pack of color pencils, the calculator, and the sketchbook: 210 - 228.25 = -18.25 This means our calculations show that Mary would be short by 18.25 if she bought all the items listed with the given prices. However, since we need to find out the cost of the set of paints, the pack of color pencils, the calculator, and the sketchbook, let's calculate those separately: - Set of paints: 18 - Pack of color pencils: 7 - Calculator: 10.50 - Sketchbook: 4.50 Adding these up gives us: 18 + 7 + 10.50 + 4.50 = 40 So, the set of paints, the pack of color pencils, the calculator, and the sketchbook cost boxed{40} altogether.

answer:1. **Sequences and Relationship** For the geometric progression (GP) with terms a, ar, ar^2, ldots, ar^{n-1}: [ P = a^n r^{frac{n(n-1)}{2}} ] Sum of the GP, S: [ S = a frac{1-r^n}{1-r} ] Sum of the squares of the GP, S'': [ S'' = a^2 + (ar)^2 + (ar^2)^2 + ldots + (ar^{n-1})^2 = a^2(1 + r^2 + r^4 + ldots + r^{2(n-1)}) ] Using the sum formula for a geometric series: [ S'' = a^2 frac{1 - r^{2n}}{1 - r^2} ] 2. **Relate P, S, and S''** We need to derive a relationship combining these: [ P^2 = (a^n r^{frac{n(n-1)}{2}})^2 = a^{2n} r^{n(n-1)} ] Compare this to re-expressions of S and S'': [ S^2 = left(a frac{1-r^n}{1-r}right)^2 quad text{and} quad S'' = a^2 frac{1 - r^{2n}}{1 - r^2} ] Notice [ frac{S^2}{S''} = frac{left(a frac{1-r^n}{1-r}right)^2}{a^2 frac{1-r^{2n}}{1-r^2}} = frac{(1-r^n)^2}{(1-r)^2} frac{1-r^2}{1-r^{2n}} = frac{(1-r^n)^2}{1-r^{2n}} ] Simplifying and considering for all cases (n = 2 as special case): [ (S^2/S'') = r^{n(n-1)} quad Rightarrow quad sqrt{S^2/S''} = r^{frac{n(n-1)}{2}} ] 3. **Conclusion** Thus, P can be expressed as: [ sqrt{a^{2n frac{S^2}{S''}}} ] The final answer is **A)** boxed{sqrt{a^{2n} frac{S^2}{S''}}}