There are \(28 \times 365 + 7=10227\) days in those years.

Usually, a year contains 365 days, but there are some which have 366 days. Between 1990 and 2018, there are 7 years having 366 days (1992, 1996, 2000, 2004, 2008, 2012. 2016), so we have the plus 7 in the sentence.

The total number of drops in a bottle is: \(20 \times 30=600 \) 

The total number of drops used in each time David properly clean his glasses is: \(3\times2\times2=12\).

So the maximum number of times is: \(300\div12=25\).

The answer is \(25\).

From the question, we have:\(\displaystyle \left [ \begin{array}{l} N= p_1\times p_2^2\times p_3^2\quad (1) \\ N=p_4\times p_5^8\qquad (2) \end{array} \right.\)

with \(p_1, p_2, p_3,p_4,p_5\)are prime numbers.

From\((2)\), we cannot have \(p_5\leq3 \) because \(3^8=6561\). If \(p_5=2\), \(N=256 \times p_4\), thus,\(p_4 \leq3\) and\(p_4=3\) (beacause \(p_4 \not = p_5\)). So, \(N= 2^8\times3=768\).

From \((1)\):

If \(p_2=2\) and \(p_3=3\) (and vice versa), \(3\leq p_1\leq 27\), therefore \(p_1 \in \{5,7,11,13,17,19,23\}\). We have 7 numbers for this case.

If  \(p_2=2\) and \(p_3=5\)  (and vice versa), \(0\leq p_1 \leq 10\), therefore \(p_1 \in \{3, 7\}\) . We have 2 numbers for this case.

If  \(p_2=2\) and \(p_3=7\)  (and vice versa), \(0\leq p_1\leq5\) , therefore\(p_1 = 3\)  . We have 1 number for this case.

If  \(p_2=2\)  and \(p_3 \geq11\)   (and vice versa),  \(n \geq 2^2 \times 11^2 \times 3 =1452\)  , therefore we don't have any numbers for this case.

If  \(p_2=3\)  and \(p_3=5\)  (and vice versa),  \(p_1 \leq 4\) , therefore  \(p_1=2\) . We have 1 number for this case.

If  \(p_2= 3\)  and \(p_3=7\)   (and vice versa),  \(p_1 \leq 2\)  , therefore ​ ​\(p_1=2\)  . We have 1 number for this case.

​ If   \(p_2=3\)  and   \(p_3 > 7\)  (and vice versa),  \(N > 3^2\times11^2= 1089\), therefore we don't have any numbers for this case.

If \(p_2 >3 \) (or \(p_3>3\)), \(N\geq 5^2\times7^2= 1225\), therefore we don't have any numbers for this case.

Thus, we have 13 numbers satisfied the question.

Note that \(p_1 \not =p_2 \not = p_3\) and\(p_4 \not= p_5\).

We have:

\(\overline{ab}=10a+b=b^2 \Rightarrow 10a=b(b-1)\\ \Rightarrow b(b-1) \mathrel{\vdots} 10 \Rightarrow \left [ \begin{array}{l} b \mathrel{\vdots}5\\(b-1)\mathrel{\vdots}5 \end{array} \right.\\ \Rightarrow b\in \{ 5,6\} (we \hskip 0.1cm cannot\hskip 0.1cm have\hskip 0.1cm b=0 \hskip 0.1cm or \hskip 0.1cm b=1 \hskip 0.1cm because \hskip 0.1cm it \hskip 0.1cm leads \hskip 0.1cm to \hskip 0.1cm a=0) \quad (1) \)

We also have:

\(\overline{acbc}=1000a+101c+10b=(\overline{ba})^2=(10b+a)^2=100b^2+20ab+a^2\\ \Rightarrow 1000a +10b(1-10b-2a) - 100c=(a^2-c) \Rightarrow (a^2-c)\mathrel{\vdots}10 \Rightarrow \left [ \begin{array}{l} a^2=c\\ a^2 \mod 10=c \end{array} \right.\\ \Rightarrow (a,c)\in \{(1,1), (2,4), (3,9), (4,6), (5,5),(6,6),(7,9),(8,4),(9,1)\} \quad (2) \)

From\((1)\), we have \(a=2\) if \(b=5\), or \(a=3\) if \(b=6\). Combine with \((2)\), we have: \(\overline{abc} \in \{ 254,369\}\).

As we can see in this problem, we have 3 situations that may happen:

Assume that you choose door 1, you can see that the probability to win the car is one-third. If the MC shows door 2 or 3, you may think that now the chance is fifty-fifty, but in fact not. If you stay with door 1, the probability to win is still one-third; but if you change the door, the fact is that now the probability to win is two-thirds, so if you change, you are more ly to win. 

Note: You can see the same problem on Google, 'The Monty Hall Problem', for an illustrated explanation.

From the question, we have 3 kinds of meat and 4 kinds of vegatables.

To get a two-distinct-meat pizza, we need to choose the first meat and the second meat.

From 3 kinds of meat, to choose the first meat, we have 3 options. After choosing the first meat, we have 2 kinds left (because the first and the second meat need to be distinct), so we have 2 options for the second meat.

So, we have  \(3\times2=6\) two-distinct-meat pizzas.

To get a two-distinct-vegatable pizza, we need to choose the first vegatable and the second vegatable.

From 4 kinds of vegatables, to choose the first vegatable, we have 4 options. After choosing the first vegatable, we have 3 kinds left (because the first and the second vegatable need to be distinct), so we have 3 options for the second vegatable.

So, we have \(4\times3=12\) two-distinct-vegatable pizzas.

To get a one-meat-one-vegatable pizza, we need to choose the meat and the vegatable.

From 3 kinds of meat, we have 3 options. From 4 kinds of vegatables, we have 4 options.

So, we have  \(3\times4=12\) one-meat-one-vegatable pizzas.

After all, we have \(6+12+12=30\) combinations of pizza.

From the question, we have some equations:

\(\left \{ \begin{array}{l} A+B=150\times 2=300\quad(1)\\ B+C=127\times 2=254\quad (2)\\ C+D=168\times 2 =336\quad(3) \end{array}\right.\)

We do the following operation \((1)+(3)-(2)\). after that, we have \(A+D=300+336-254=382\).

Therefore, the average weight of sheep A and sheep D is: \(382\div 2= 191\)(pounds)

Note that \(A,B,C,D\) is the weight of each sheep A, B, C, D, respectively.

\(\widehat{ABC}=90^o \Rightarrow O\in AC \\\Rightarrow The\hskip0.1cmlength\hskip0.1cm of\hskip0.1cm arc \hskip0.1cmABC\hskip0.1cmis\\ l= \dfrac12\times2\pi\times3=3\pi \hskip0.1cm(meters)\)

Let \(A= 1+\dfrac1 4+ \dfrac{1}{16}+...=\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{4^n}\).

We calculate the sum of \(n\) first numbers of the series \(A\).

\(S_n=\displaystyle \sum_{i=0}^n \dfrac{1}{4^i}=\dfrac{1-\dfrac{1}{4^{n+1}}}{1-\dfrac1 4}=\dfrac4 3 \times \Bigg (1-\dfrac{1}{4^{n+1}} \Bigg )\)

Let \(n \rightarrow \infty\), we have \(S_n \rightarrow A\), and \(S_n \rightarrow \dfrac4 3\).

As the limit of \(S_n\) is unique, we have \(A=\dfrac4 3\).

Let \(A= 1+\dfrac1 2+ \dfrac1 4+...=\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{2^n}\).

We calculate the sum of \(n\) first numbers of the series \(A \).

\(S_n=\displaystyle \sum_{i=0}^n \dfrac{1}{2^i}=\dfrac{1-\dfrac{1}{2^{n+1}}}{1-\dfrac1 2}=2-\dfrac{1}{2^n}\).

Let \(n \rightarrow \infty\), we have \(S_n \rightarrow A\), and \(S_n \rightarrow 2\).

As the limit of \(S_n\) is unique, we have \(A=2\).

Note that the denotation of \(\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{2^n}\) is just a way we write for short, in fact we cannot find the exact sum of the infinite series, we just can find the limit of the sequence \(S_n\) and we define \(\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{2^n}= \lim_{n\rightarrow \infty}S_n\).

Let the average driving speed be \(v\) and the time it takes Alysha to drive to the market be \(t\).

So, we have the length of the route from home to the market is: \(v \times t\).

Because the average driving speed is 8 times the average walking speed, we have the average walking speed is: \(\dfrac{v}{8}\).

Therefore, the time it takes Alysha to walk to the market is:\(\dfrac{v \times t}{\dfrac{v}{8}}=\dfrac{8 \times v \times t}{v}=8t\).

We can see that walking takes her 21 minutes longer than when she drives to the market; thus, the difference between the walking time and the driving time is: \(8t-t=7t\).

\(\Rightarrow 7t=21 \Rightarrow t=3\).

Thus, it takes her 3 minutes to drive from home to market.

Let AB be the path that both Jana and Zhao follow.

After 5 minutes or \(\dfrac{1}{12}\)of an hour, Jana has jogged to C; therefore, the length of AC is \(6 \times \dfrac{1}{12}=\dfrac1 2 \)  mile.

So, now Jana has to jog from C to B, and the length of BC is \(2 - \dfrac 1 2 =\dfrac3 2\) miles.

Jana jog on path CB for \(\dfrac 3 2 \div 2 = \dfrac 3 4\)hour, or 45 minutes.

While Jana jog on CB, Zhao ride on AB for \(2 \div 10 = \dfrac 15\) hour, or 20 minutes.

So, Zhao come to B before Jana or Jana come to B after Zhao a period of \(45 - 20 = 25\) minutes.

Thus, rhe answer is 25 minutes.

Because each person's food level does not change, we can assume that each person's rice level everyday is a.

Therefore, at the beginning, the army unit has 200 x 30 x a = 6000a rice level.

After 10 days, the unit remains 6000a - 200 x 10 x a =4000a rice level 

and has 200 + 50 = 250 people.

Thus, the number of days that the unit has rice to eat is 4000a /(250 x a) = 16.

So, the rice is left for the unit to eat in 16 days.