answer:To find the greatest number of kits Veronica can make without any items left over, we need to find the greatest common divisor (GCD) of the quantities of the items she has, which are 20 bottles of water, 12 cans of food, 30 flashlights, and 18 blankets. The GCD will tell us the maximum number of kits she can make such that each kit has the same number of each item and there are no items left over. Let's find the GCD of 20, 12, 30, and 18: - The factors of 20 are 1, 2, 4, 5, 10, 20. - The factors of 12 are 1, 2, 3, 4, 6, 12. - The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. - The factors of 18 are 1, 2, 3, 6, 9, 18. The common factors of all four numbers are 1 and 2. Since Veronica must make at least 5 kits, the number 1 is not a viable option. Therefore, the GCD is 2. Now, let's see if we can make kits with 2 of each item: - 20 bottles of water / 2 = 10 kits - 12 cans of food / 2 = 6 kits - 30 flashlights / 2 = 15 kits - 18 blankets / 2 = 9 kits The limiting factor here is the cans of food, which only allow for 6 kits. However, we need to ensure that each kit has no more than 10 items. If we make 6 kits, each kit would have: - 2 bottles of water - 2 cans of food - 2 flashlights - 2 blankets That's a total of 8 items per kit, which is within the constraint of no more than 10 items per kit. Therefore, Veronica can make boxed{6} kits, which is the greatest number of kits she can make without any item left over, while also meeting the additional constraints.

answer:Given the problem involves an infinite geometric sequence {a_{n}}, we are particularly interested in the relationship between a_{3} and a_{1} and the implications for the sum S_{n} of the first n terms. Let's analyze the provided solution step by step, adhering to the guidelines for clarity and precision. 1. **Identify the common ratio**: Let the common ratio of the geometric sequence {a_{n}} be denoted as q. The relationship between consecutive terms in a geometric sequence is defined by this ratio. 2. **Examine the condition a_{3} > a_{1}**: Given a_{1} > 0 and a_{3} > a_{1}, we can express a_{3} as a_{1}q^{2}. This leads to the inequality a_{1}q^{2} > a_{1}. 3. **Infer about q**: From a_{1}q^{2} > a_{1} and knowing a_{1} > 0, we deduce q^{2} > 1. This implies two possible intervals for q: either q > 1 or q < -1. 4. **Analyze the sum S_{n} for different values of q**: - When q < -1, for odd n, we find that S_{n} = frac{a_{1}(1-q^{n})}{1-q} = frac{a_{1}}{1-q}(-q)^{n} + frac{a_{1}}{1-q}. Given frac{a_{1}}{1-q} > 0 and (-q) > 1, we conclude that S_{n} increases monotonically for odd n, which means it has no maximum value. - When q > 1, the expression for S_{n} simplifies to frac{a_{1}(1-q^{n})}{1-q} = frac{a_{1}}{q-1}(q^{n}-1). Since frac{a_{1}}{q-1} > 0 and q > 1, it's evident that S_{n} also increases monotonically in this case, indicating no maximum value for S_{n}. 5. **Consider q = 1**: In this scenario, S_{n} = ncdot a_{1}. With a_{1} > 0, it's clear that S_{n} grows indefinitely as n increases, signifying no maximum value for S_{n}. Through the analysis above, we establish that the condition "a_{3} > a_{1}" is indeed sufficient for the sum S_{n} to have no maximum value. This sufficiency is deduced from the fact that for all cases considered (q < -1, q > 1, and q = 1), S_{n} lacks a maximum value. However, the necessity of this condition is not demonstrated, meaning there may be other conditions under which S_{n} also has no maximum value. Consequently, the conclusion is that "a_{3} > a_{1}" is a **Sufficient but not Necessary condition** for "S_{n} has no maximum value". Therefore, the correct choice is: boxed{A}

answer:Evie collects 10 shells each day for 6 days, so the total number of shells she collects is: 10 shells/day * 6 days = 60 shells After giving some shells to her brother, she has 58 shells left. To find out how many shells she gave to her brother, we subtract the number of shells she has left from the total number she collected: 60 shells - 58 shells = 2 shells Evie gave boxed{2} shells to her brother.

answer:If a group of 10 people can hold 20 boxes in total, that means each person can hold 2 boxes (since 20 boxes ÷ 10 people = 2 boxes per person). Assuming that each person is holding an equal number of boxes in each hand, a man can lift 1 box in each hand (since 2 boxes per person ÷ 2 hands = boxed{1} box per hand).