answer:To solve the problem, we need to prove three parts: a) 2001 is *good*, but 3001 isn't *good*. b) The product of two *good* numbers is a *good* number. c) If the product of two numbers is *good*, then at least one of the numbers is *good*. Let's start by understanding the definition of a *good* number. A number ( X ) is *good* if it can be written as the sum of two consecutive positive integers and as the sum of three consecutive positive integers. 1. **Sum of Two Consecutive Positive Integers:** Let ( X ) be the sum of two consecutive positive integers. Then we can write: [ X = n + (n+1) = 2n + 1 ] This implies that ( X ) is an odd number. 2. **Sum of Three Consecutive Positive Integers:** Let ( X ) be the sum of three consecutive positive integers. Then we can write: [ X = m + (m+1) + (m+2) = 3m + 3 = 3(m+1) ] This implies that ( X ) is a multiple of 3. Combining these two conditions, we see that a *good* number must be both odd and a multiple of 3. Therefore, a *good* number must be of the form: [ X equiv 3 pmod{6} ] # Part (a) **2001 is *good*, but 3001 isn't *good*.** 1. **Check 2001:** [ 2001 div 6 = 333 text{ remainder } 3 implies 2001 equiv 3 pmod{6} ] Since 2001 satisfies ( 2001 equiv 3 pmod{6} ), it is a *good* number. 2. **Check 3001:** [ 3001 div 6 = 500 text{ remainder } 1 implies 3001 equiv 1 pmod{6} ] Since 3001 does not satisfy ( 3001 equiv 3 pmod{6} ), it is not a *good* number. # Part (b) **The product of two *good* numbers is a *good* number.** Let ( X ) and ( Y ) be two *good* numbers. Then: [ X equiv 3 pmod{6} quad text{and} quad Y equiv 3 pmod{6} ] We need to show that ( XY ) is also *good*: [ XY equiv 3 cdot 3 = 9 equiv 3 pmod{6} ] Thus, ( XY ) is also of the form ( 3 pmod{6} ), and hence it is a *good* number. # Part (c) **If the product of two numbers is *good*, then at least one of the numbers is *good*.** Let ( X ) and ( Y ) be two numbers such that ( XY ) is *good*. Then: [ XY equiv 3 pmod{6} ] We need to show that at least one of ( X ) or ( Y ) is *good*. Assume for contradiction that neither ( X ) nor ( Y ) is *good*. Then: [ X notequiv 3 pmod{6} quad text{and} quad Y notequiv 3 pmod{6} ] The possible residues modulo 6 for ( X ) and ( Y ) are 0, 1, 2, 4, and 5. We check the product of each pair: - ( 0 times text{anything} equiv 0 pmod{6} ) - ( 1 times 1 equiv 1 pmod{6} ) - ( 1 times 2 equiv 2 pmod{6} ) - ( 1 times 4 equiv 4 pmod{6} ) - ( 1 times 5 equiv 5 pmod{6} ) - ( 2 times 2 equiv 4 pmod{6} ) - ( 2 times 4 equiv 8 equiv 2 pmod{6} ) - ( 2 times 5 equiv 10 equiv 4 pmod{6} ) - ( 4 times 4 equiv 16 equiv 4 pmod{6} ) - ( 4 times 5 equiv 20 equiv 2 pmod{6} ) - ( 5 times 5 equiv 25 equiv 1 pmod{6} ) None of these products are congruent to 3 modulo 6. Therefore, our assumption is false, and at least one of ( X ) or ( Y ) must be *good*. (blacksquare)

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answer:The issue in fact is hidden in this mapping: ```csharp HasMany(x => x.TitleSetTitles) ... .Inverse(); ``` This mapping is instructing NHibernate: The child will care about this relation. But that means, that the child must know about its parent, which is not true in this code: ```csharp var titleSet = new TitleSetDataEntity { Name = Somename, ... TitleSetTitles = new List<TitleSetTitleData> { new TitleSetTitleData... }, }; session.Save(titleSet); ``` In this code, the child instance created by `new TitleSetTitleData` is missing the parent reference! And that's the problem. NOTE - I guess that the property `TitleSetTitles` is of type `IList<ChildDataClass1> TitleSetTitles` but in the code we use `new List<TitleSetTitleData>`... I guess it is a typo. The correct syntax would be: ```csharp var titleSet = new TitleSetDataEntity { Name = Somename, ... TitleSetTitles = new List<TitleSetTitleData>() }; var child = new TitleSetTitleData { ... ParentDataClassData = titleSet }; titleSet.TitleSetTitles.Add(child); session.Save(titleSet); ``` And now, NHibernate will have enough information. `Parent` has a `Child` added in the list, `Child` knows about the `Parent`, and the `.Inverse()` mapping will work because the ID of the `Parent` won't be null anymore.

answer:Some plant-based foods that are high in both iron and protein include lentils, chickpeas, quinoa, tofu, tempeh, black beans, spinach, pumpkin seeds, hemp seeds, and chia seeds. These foods do not contain calcium, but they offer a range of other nutrients essential for good health. One healthier alternative that offers all three essential nutrients in one is a vegan smoothie made with spinach, tofu or soy milk, and chia seeds. This smoothie is a great source of iron, protein, and calcium, and it is a delicious and easy way to get all three in one meal. For adults, the recommended daily intake of iron is 18mg for women and 8mg for men. The recommended daily intake of protein is around 0.36g per pound of body weight, or roughly 56g per day for the average sedentary man and 46g per day for the average sedentary woman. The recommended daily intake of calcium is 1000-1200mg per day for adults, depending on age and sex.