answer:Let's calculate the total cost of the items Candice buys each visit, excluding the sourdough bread. She buys: - 2 loaves of white bread at 3.50 each, so 2 * 3.50 = 7.00 - 1 baguette at 1.50 - 1 almond croissant at 2.00 The total cost per visit for these items is 7.00 + 1.50 + 2.00 = 10.50. Since Candice spends 78 over 4 weeks, we can calculate the total amount she spends on sourdough bread over 4 weeks by subtracting the cost of the other items from the total amount spent. Total spent on sourdough bread over 4 weeks = Total spent over 4 weeks - (Cost per visit for other items * Number of visits) Total spent on sourdough bread over 4 weeks = 78 - (10.50 * 4) Now, let's calculate the total spent on sourdough bread over 4 weeks: Total spent on sourdough bread over 4 weeks = 78 - 42 Total spent on sourdough bread over 4 weeks = 36 If the 36 is the total amount spent on sourdough bread over 4 weeks, and Candice buys the sourdough bread each week, we can find the cost of each loaf of sourdough bread by dividing the total amount spent by the number of weeks. Cost per loaf of sourdough bread = Total spent on sourdough bread over 4 weeks / Number of weeks Cost per loaf of sourdough bread = 36 / 4 Cost per loaf of sourdough bread = 9 Therefore, each loaf of sourdough bread costs boxed{9} .

answer:1. **Consider the space of real regular Borel measures on (X)**: - Let (M(X)) be the space of real regular Borel measures on (X). This space can be viewed as the dual space of (C(X)), the space of continuous real-valued functions on (X), with the weak-* topology. 2. **Define the action of (T) on measures**: - Define the action of (T) on a measure (mu in M(X)) by (Tmu(A) = mu(T^{-1}(A))) for all Borel subsets (A) of (X). This defines a continuous map on the space of probability measures on (X). 3. **Use fixed point theory**: - The set of probability measures on (X) is a convex compact set in the weak-* topology. By the Banach-Alaoglu theorem, this set is compact. By the Schauder fixed-point theorem, there exists a probability measure (mu) on (X) such that (Tmu = mu). This means (T) is measure-preserving with respect to (mu). 4. **Apply the Recurrence Theorem**: - By the Recurrence Theorem, for every Borel subset (A) of (X), the set (S_A) of those (x in A) for which only finitely many of the points (T^n(x)), (n geq 1), belong to (A) has measure zero. 5. **Construct a countable basis of open sets**: - Let ((U_n)_{n}) be a countable basis of open sets for (X). Define (S = bigcup S_{U_n}). Since (S) is a countable union of sets of measure zero, (S) itself has measure zero. 6. **Identify returning points**: - Every point (x in X setminus S) has the property that whenever (x in U) (where (U) is an open subset of (X)), infinitely many of the points (T^n(x)) belong to (U). This implies that for such (x), there exists a strictly increasing sequence (n_i) such that (lim_{k to infty} T^{n_k}(x) = x). 7. **Conclusion**: - Therefore, any point (x in X setminus S) is a returning point, proving the existence of a returning point in (X). (blacksquare)

answer:1. Define an n-digit number overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}} where a_{i} in {1, 2, 3, 4, 5, 6} for all i=1, 2, 3, 4, 5, 6. 2. According to the problem, each prefix of length k must be divisible by k. This yields the requirements: [ 1 mid a_{1}, quad 2 mid overline{a_{1} a_{2}}, quad 3 mid overline{a_{1} a_{2} a_{3}}, quad 4 mid overline{a_{1} a_{2} a_{3} a_{4}}, quad 5 mid overline{a_{1} a_{2} a_{3} a_{4} a_{5}}, quad 6 mid overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}. ] 3. Since overline{a_{1} a_{2}} must be divisible by 2, a_2 must be even. Therefore, a_2 in {2, 4, 6}. 4. Similarly, overline{a_{1} a_{2} a_{3} a_{4}} must be divisible by 4. For this to be true, overline{a_{3} a_{4}} must be divisible by 4, meaning a_{4} must be even and itself satisfying overline{a_{3} a_{4}} equiv 0 pmod{4}, so a_{4} is one of {2, 4, 6} depending on the possible values of a_{3}. 5. Next, overline{a_{1} a_{2} a_{3} a_{4} a_{5}} must be divisible by 5, which means a_{5} = 5. 6. overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}} must be divisible by 6, so a_{6} must be even (since the last digit of a number divisible by 2 must be even) and the entire sequence must sum to a multiple of 3. 7. Therefore, possible values of a_{1}, a_{3} must account for these constraints. 8. Considering that a_1 must provide for divisibility by 1 and 3, a_1 can be {1, 3} and similarly for a_3 given that the sum a_1 + a_2 + a_3 must also be divisible by 3. 9. Let's enumerate possibilities: - Since overline{a_1 a_2 a_3} divisible by 3 must imply (a_1 + a_2 + a_3) equiv 0 pmod{3}: - **If a_1 = 1 and a_3 = 3:** - a_2 in {2,4,6} - Checking, a_2 = 2, then 1 + 2 + 3 = 6 implies 0 pmod{3} - **If a_1 = 3 and a_3 = 1:** - a_2 in {2,4,6} - Checking a_2 = 2, then 3 + 2 + 1 = 6 implies 0 pmod{3} 10. Verify six-digit "good" numbers: - overline{123456, 123654, 321456, 321654}: - These numbers meet all individual checks for divisibility: - e.g., 123456: meets all overline{1}, overline{12}, overline{123}, overline{1234}, overline{12345}, overline{123456}, each modulo condition 1 through 6. 11. Thus, the valid numbers meeting all the criteria are 123654 and 321654. Conclusion: Option C is the number of such 6-digit "good" numbers. [ boxed{C} ]

answer:(1) Find a point ( F ) on segment ( PD ) such that ( AF parallel ) plane ( PEC ). 1. **Identify Key Points and Properties**: Given the pyramid ( P-ABCD ), with base ( ABCD ) which is a rhombus with side length 2 and ( angle ABC = 60^circ ). ( E ) is the midpoint of ( AB ), and ( PA perp ) plane ( ABCD ). 2. **Determine Midpoints**: Let ( G ) be the midpoint of ( PC ) and ( F ) be the midpoint of ( PD ). 3. **Show Parallelogram Property**: Since both ( G ) and ( F ) are midpoints, we have ( GF cong frac{1}{2} CD ). Given ( CD cong 2 ), it follows that ( GF cong 1 ). 4. **Confirm Parallelogram Formation**: Since ( EA cong GF ), ( GA parallel AF parallel EG ), thus quadrilateral ( AFGE ) is a parallelogram, implying ( AF parallel EG ). 5. **Conclusion**: Hence, point ( F ) on ( PD ) makes ( AF parallel ) plane ( PEC ). (2) Calculate the cosine of the dihedral angle ( D - PEC - A ). 1. **Determine Midpoint and Plane Relationships**: Let ( M ) be the midpoint of ( AD ). Since ( ABCD ) is a rhombus and ( E ) is the midpoint of ( AB ), ( M ) and ( E ) help in defining the planes and angles. 2. **Triangle and Perpendicular Relationships**: Since ( ABCD ) forms such that ( triangle ACD ) is equilateral (all sides equal and angles all ( 60^circ )): [ CM = sqrt{3}, quad text{and} quad CM perp text{plane } PAD ] 3. **Calculate Tangent of ( angle MPC )**: [ tan angle MPC = frac{CM}{PM} = frac{sqrt{3}}{sqrt{5}} ] We have ( PM = sqrt{5} ) and ( PA = 2 ). 4. **Construct ( DH ) and Calculate Related Angle**: Construct ( DH perp AB ) and another perpendicular ( HN perp PE ), connecting ( DN ). Hence, ( DH perp text{plane } PAB ) making ( angle HND ) the dihedral angle we desire. [ tan angle PEA = 2 ] Therefore: [ HN = 2 sin angle PEA = frac{4}{sqrt{5}}, quad DH = sqrt{3} ] This evaluates: [ tan angle HND = frac{sqrt{15}}{4} quad Rightarrow quad cos angle HND = frac{4}{sqrt{31}} = frac{4sqrt{31}}{31} ] 5. **Conclusion**: Therefore, the cosine of the dihedral angle ( D-PEA ) is: [ boxed{frac{4sqrt{31}}{31}} ]