(Solution 1)We have: \(\sqrt{1+1\cdot2\cdot3\cdot4}=1+1\cdot4=5\)

\(\sqrt{1+2\cdot3\cdot4\cdot5}=1+2\cdot5=11\)

.........

\(\sqrt{1+204\cdot205\cdot206\cdot207}=1+204\cdot207=42229\)

(Solution 2) We have: \(\sqrt{1+1\cdot2\cdot3\cdot4}=2\cdot3-1=5\)

\(\sqrt{1+2\cdot3\cdot4\cdot5}=3\cdot4-1=11\)

..............

\(\sqrt{1+204\cdot205\cdot206\cdot207}=205\cdot206-1=42229\)

Alone has done solutions 3+4.

Read the P/s: It's 7, not 8

We have: 1 + 5 = 2 + 4 = 6 (there are no 2 number-3 cards)

=> A has (1;5) or (2;4) cards [1]

B said the difference between two cards is 5

=> B has (1;6), (2;7), (3;8) or (4;9) cards [2]

We have: 2*9 = 3*6 = 18

=> C has (2;9) or (3;6) cards [3]

We have: 2 = 2*1; 4 = 2*2; 6 = 2*3; 8 = 2*4

=> D has (1;2), (2;4), (3;6) or (4;8) cards [4]

If A has (2;4) cards => Unsatisfy

If A has (2;4) cards -> B has (1;6) or (3;8) cards

 - If B has (1;6) cards => Unsatisfy for C

 - If B has (3;8) cards => Unsatisfy for C, too

=> A has (1;5) cards

=> B has (2;7), (3;8) or (4;9) cards

- If B has (2;7) cards -> C has (3;6) cards -> D has (4;8) cards -> The number of the remaining card is 9.

- If B has (3;8) cards -> C has (2;9) cards -> Unsatisfy for D

- If B has (4;9) cards -> C has (3;6) cards -> Unsatisfy for D

So the number of the remaining card is 9

Sorry, wrong answer. Why does 21 = 13 + 7?

We'll fix the avatar soon in the future.

Calculate(v), not "calculator"(n) or "calculating"(v-ing)

We have: \(\left|7x-5y\right|\ge0\); \(\left|2z-3x\right|\ge0\) and \(\left|xy+yz+zx-2000\right|\ge0\)

=> \(P\ge0+0+0\ge0\)

\(P=0\) only when \(7x-5y=2z-3x=xy+yz+zx-2000=0\)

=> \(7x=5y;2z=3x;xy+yz+zx=2000\)

=> \(\dfrac{x}{5}=\dfrac{y}{7}\); \(\dfrac{z}{3}=\dfrac{x}{2}\) and \(xy+yz+zx=2000\)

=> \(\dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{15}\) and \(xy+yz+zx=2000\)

Put \(\dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{15}=k\)

We have: \(k^2=\dfrac{x}{10}\cdot\dfrac{y}{14}=\dfrac{y}{14}\cdot\dfrac{z}{15}=\dfrac{z}{15}\cdot\dfrac{x}{10}=\dfrac{xy+yz+zx}{10\cdot14+14\cdot15+15\cdot10}=\dfrac{2000}{140+210+150}\)

\(=\dfrac{2000}{500}=4\)

=> \(k=\pm2\)

=> \(\left(x;y;z\right)=\left(20;28;30\right)\) and  \(\left(-20;-28;-30\right)\)

So \(minP=0\) only when (x;y;z) = (20;28;30) and (-20;-28;-30)

I think the title is wrong. It is: \(\dfrac{3}{4}\left(AB+AC+BC\right)< AD+BE+CF< AB+AC+BC\)

The numbers satisfy is: 202;211;220

So there are 3 numbers satisfy.

A dozen pens =12 pens

So each pen cost: \(\dfrac{24.6}{12}=2.05\left(dollars\right)\)

Beause Le Anh Duy copyed the solution from the source (same solution 100% with the source, even the words) so I won't give Duy any points, also the questions next time I'll change. 

Link: https://www.pbvusd.k12.ca.us/site/handlers/filedownload.ashx?moduleinstanceid=10900&dataid=12282&FileName=1718AnswersIndexSolutions.pdf

17) \(G=\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

\(G=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

\(G=1-\dfrac{1}{\left(n+1\right)^2}< 1\) (proved)

AUTO ANSWER (AFTER SEVERAL DAYS):

The probability is: \(\dfrac{6}{6}.\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{2}{6}.\dfrac{1}{6}\approx1,5\%\)

AUTO ANSWER (AFTER SEVERAL DAYS):

This is a >146 IQ question. If you've not found the solution than don't mind about it because this question has no suitable answer.

Sam Loyd - the greatest person in history on IQ questions has found this. But he gave the price $1000 for someone who can solve this. It was a mathematic shock in the 1880s, and over many years later people knew that the question has no solutions. This is 1 of billions of similar questions can be written down.

AUTO ANSWER (AFTER SEVERAL DAYS): I don't khow why the moderator selected this answer. The answer is wrong.

If Amos and Butch shoot first, they will shoot each other (always on target). So Cody has 50% chance to win, while others has \(\dfrac{50\%}{2}=25\%\) chance to win because their probability at shooting is all 100% target.

If Cody shoots first, he'll shoot miss (because if he shoots on target, the other person will defeat him), let the others shoot each other. Then he'll have 50% chance to win, while others has only 25% chance to win.

So the probability that Cody is alive is 50% and the two others are 25%/ person.

16) We have: 

SOLUTION 1

8 + 1 + 3 = 12; 1 + 2 = 3

9 + 0 + 9 = 18; 1 + 8 = 9

4 + 5 + 3 = 12; 1 + 2 = 3

So 1 + 4 + 6 + 2 = 13; 1 + 3 = 4.

8 + 1 + 3 = 12; 12 - 9 = 3

9 + 0 + 9 = 18; 18 - 9 = 9

4 + 5 + 3 = 12; 12 - 9 = 3

So 1 + 4 + 6 + 2 = 13; 13 - 9 = 4

ANSWER: 4

15) If John took the chocolate => Only Wendy told the truth => SATISFY

      If Wendy took the chocolate => John, Charlies and Sally told the truth => UNSATISFY

      If Charlies took the chocolate => John and Charlies told the truth => UNSATISFY

      If Sally took the chocolate => John and Charlies told the truth => UNSATISFY

So John took the chocolate.

What? The source's answer is wrong. Pls check again.

It is correct (friendly numbers) but can you explain the rules of the numbers?

This is a 145-150 IQ question. Very few people knew that in flipping the coin 200 times, you must have at least 6-consecutive DOWN/UP. So in the report, he didn't have this, so the expert knew exactly it was a fake report.