answer:# Solution: Part (1): Given vectors overrightarrow{m}=(2sin(x-frac{pi}{6}),1) and overrightarrow{n}=(2cos x,1), and since overrightarrow{m} parallel overrightarrow{n}, we have the condition for parallel vectors that their components are proportional. Thus, we can set up the equation from their x-components: [ 2sin(x-frac{pi}{6})=2cos x ] Dividing both sides by 2 and using the sine subtraction formula, we get: [ sin(x-frac{pi}{6})=cos x implies sin x cos frac{pi}{6} - cos x sin frac{pi}{6} = cos x ] Substituting cos frac{pi}{6} = frac{sqrt{3}}{2} and sin frac{pi}{6} = frac{1}{2}, we have: [ frac{sqrt{3}}{2}sin x - frac{1}{2}cos x = cos x ] Rearranging and combining like terms, we find: [ frac{sqrt{3}}{2}sin x = frac{3}{2}cos x implies tan x = sqrt{3} ] Given x in [0,2pi], the solutions for tan x = sqrt{3} are: [ x=frac{pi}{3} quad text{or} quad x=frac{4pi}{3} ] Therefore, the values of x are boxed{x=frac{pi}{3} text{ or } x=frac{4pi}{3}}. Part (2): Given vectors overrightarrow{m}=(2sin(x-frac{pi}{6}),1) and overrightarrow{n}=(2cos x,1), the dot product function f(x)=overrightarrow{m} cdot overrightarrow{n} is: [ f(x) = 4sin(x-frac{pi}{6})cos x + 1 ] Using trigonometric identities, we simplify f(x): [ f(x) = 2sqrt{3}sin x cos x - 2cos^2 x + 1 = sqrt{3}sin 2x - cos 2x ] Using the sine addition formula, we further simplify: [ f(x) = 2sin(2x-frac{pi}{6}) ] Given x in [-frac{pi}{4},frac{pi}{4}], the range of 2x-frac{pi}{6} is [-frac{2pi}{3},frac{pi}{3}]. Thus, the range of f(x) is [-2,sqrt{3}]. Therefore, the maximum value of f(x) on the interval [-frac{pi}{4},frac{pi}{4}] is boxed{sqrt{3}}, and the minimum value is boxed{-2}.

answer:As with the three-digit case, each of the four digits in these integers has three choices: either 2, 5, or 7. Since the choices are independent for each digit, the total number of such four-digit integers can be computed by multiplying the choices together for each of the four positions: [ 3 times 3 times 3 times 3 = 3^4 = 81. ] Thus, there are boxed{81} different four-digit integers composed of the digits 2, 5, and 7.

answer:1. **Problem Restatement:** We are given that in city (N) there are exactly three monuments. A group of 42 tourists visited the city, and each tourist took at most one photo of each of the three monuments. It was found that any two tourists together had photos of all three monuments. We need to find the minimum total number of photos taken by all the tourists. 2. **Notation and Setup:** Let's number the three monuments as (A), (B), and (C). Denote the number of photos taken of monument (A), (B), and (C) respectively by (f_A), (f_B), and (f_C). 3. **Analysis of Conditions:** - If a tourist did not take a photo of monument (A), then they must have taken photos of monuments (B) and (C). - Similarly, if a tourist did not take a photo of monument (B), they must have taken photos of monuments (A) and (C). - And if a tourist did not take a photo of monument (C), they must have taken photos of monuments (A) and (B). 4. **Derivation of Minimum Photos:** - Consider a monument, say (A). If any more than one tourist did not take a photo of (A), then the combined photos of this group would lack photos of monument (A). Hence, at most one tourist can skip taking a photo of (A). - Therefore, if at most one tourist did not take a photo of each monument, then for each monument, there are at least (42 - 1 = 41) photos taken. Hence, (f_A, f_B, f_C geq 41). 5. **Total Minimum Photos Computation:** - Summing up the minimum photos for all three monuments, we have: [ f_A + f_B + f_C geq 41 + 41 + 41 = 123 ] 6. **Verification of the Minimum Count:** - A scenario that achieves this minimum is as follows: Suppose 1 tourist took no photos at all, and the remaining 41 tourists each took a single photo of each of the three monuments. In this case, the total number of photos is: [ 3 times 41 = 123 ] - In this scenario, every pair of tourists would still have photos covering all three monuments, satisfying the problem's condition. Conclusion. The minimum total number of photos taken by all tourists is therefore ( boxed{123} ).

answer:1. **Understanding the Problem:** - The grandmother has three grandchildren. - The first grandchild calls every 3 days. - The second grandchild calls every 4 days. - The third grandchild calls every 5 days. - All three called on December 31, 2016. - We need to determine how many days in the following year (2017) did she receive no calls. 2. **Calculating the Total Days in the Year:** - 2017 is not a leap year, so it has 365 days. 3. **Applying the Principle of Inclusion-Exclusion:** Let's ( N ) be the total number of days in 2017 (i.e., 365). We will use the Inclusion-Exclusion Principle to find the number of days on which at least one grandchild calls. 4. **Finding Days Each Grandchild Calls:** - Days the first grandchild calls in 2017: ( leftlfloor frac{365}{3} rightrfloor = 121 ) - Days the second grandchild calls in 2017: ( leftlfloor frac{365}{4} rightrfloor = 91 ) - Days the third grandchild calls in 2017: ( leftlfloor frac{365}{5} rightrfloor = 73 ) 5. **Finding Days Two Grandchildren Call Together:** - Days both the first and second grandchildren call: ( leftlfloor frac{365}{3 cdot 4} rightrfloor = leftlfloor frac{365}{12} rightrfloor = 30 ) - Days both the second and third grandchildren call: ( leftlfloor frac{365}{4 cdot 5} rightrfloor = leftlfloor frac{365}{20} rightrfloor = 18 ) - Days both the first and third grandchildren call: ( leftlfloor frac{365}{3 cdot 5} rightrfloor = leftlfloor frac{365}{15} rightrfloor = 24 ) 6. **Finding Days All Three Grandchildren Call Together:** - Days all three call together: ( leftlfloor frac{365}{3 cdot 4 cdot 5} rightrfloor = leftlfloor frac{365}{60} rightrfloor = 6 ) 7. **Applying the Inclusion-Exclusion Formula:** The number of days on which at least one of the grandchildren calls is: [ |A cup B cup C| = |A| + |B| + |C| - |A cap B| - |A cap C| - |B cap C| + |A cap B cap C| ] Plugging in the values: [ |A cup B cup C| = 121 + 91 + 73 - 30 - 18 - 24 + 6 ] Simplifying: [ = 285 - 72 + 6 = 219 ] 8. **Calculating the Number of Days without Calls:** The number of days on which none of the grandchildren call: [ 365 - |A cup B cup C| = 365 - 219 = 146 ] **Conclusion:** [ boxed{D} ]