Because A and B are center of two circles respectively. Let M, N denote intersections of A, B and two circles respectively

=> AM = BN = 2cm

Again: Cover MN and NM so AN = BM = 2cm - MN

So AB = 2cm + 2cm - MN = 2cm + AN = 2cm + BM

Because \(\overline{3AA1}\) divisible by 9

=> 3 + 2A + 1 divisible by 9

But 3 + 1 = 4

=> 2A divide 9 redundancy 5

But 0 \(\le A\le9\)

\(\Rightarrow0\le2A\le18\)

=> 2A = 5 or 2A = 14

If 2A = 5 => A unsatisfactory

If 2A = 14 => A = 7

So A = 7 and \(\overline{3AA1}=3771\)

2P = 8 + 2³ + 2⁴ + ... + 2²¹

2P - P = \(\left(8+2^3+2^4+...+2^{21}\right)-\left(4+2^2+2^3+...+2^{20}\right)\)

P = 8 + 2²¹ - 4 - 2² = 8 - 4 - 4 + 2²¹ = 2²¹

So P can be written as a power of 2

Because 1010 \(\le n\le2010\)

\(\Rightarrow204\le a\le227\)

Use Casio computer, you press:

203 --> = --> SHIFT --> RCL --> D

After that:

\(D+1:A=\dfrac{D^2-20223}{21}\)

The ":" is: ALPHA + the button next CALC

= ... [from 203 to 226]

We'll have a_n = 210 and n = 1137

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{2y+1}{6}\)

\(\Rightarrow x\left(2y+1\right)=30\)

But 2y + 1 is a odd => x is a even number

=> \(x\in\left\{-30;-10;-6;-2;2;6;10;30\right\}\)

We have:

So \(\left(x,y\right)=\left(-30;-1\right);\left(-10;-2\right);\left(-6;-3\right);\left(-2;-8\right);\left(2;7\right);\left(6;2\right);\left(10;1\right);\left(30;0\right)\)

\(\Rightarrow\dfrac{x}{6}-\dfrac{1}{30}=\dfrac{2}{y}\)

\(\Rightarrow\dfrac{5x}{30}-\dfrac{1}{30}=\dfrac{2}{y}\)

\(\Rightarrow\dfrac{5x-1}{30}=\dfrac{2}{y}\)

\(\Rightarrow y\left(5x-1\right)=60\)

But \(5x-1\equiv4\left(mod5\right)\)

\(\Rightarrow5x-1\in\left\{-6;-1;4\right\}\)

If 5x - 1 = -6

=> 5x = -5 => x = -1

And y = 60 : -6 = -10

If 5x-1 = -1 

=> 5x = 0 => x = 0

And y = 60 : -1 = -60

If 5x-1 = 4 

=> 5x = 5 => x = 1

And y = 60 : 4 = 15

So \(\left(x,y\right)=\left(-1;-10\right);\left(0;-60\right);\left(1;15\right)\)

Put A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\)

\(A=\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)\)

\(A=\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+...+\dfrac{100}{49\cdot51}\)

\(A=100\left(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{49\cdot51}\right)\)

Put B = \(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{99\cdot1}\)

\(B=\left(\dfrac{1}{1\cdot99}+\dfrac{1}{99\cdot1}\right)+\left(\dfrac{1}{3\cdot97}+\dfrac{1}{97\cdot3}\right)+...+\left(\dfrac{1}{49\cdot51}+\dfrac{1}{51\cdot49}\right)\)

\(B=2\left(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{49\cdot51}\right)\)

So 

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{99\cdot1}}=\dfrac{100\left(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{49\cdot51}\right)}{2\left(\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{49\cdot51}\right)}=\dfrac{100}{2}=50\)

\(\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{99}\)

\(=\left(\dfrac{1}{50}+\dfrac{1}{99}\right)+\left(\dfrac{1}{51}+\dfrac{1}{98}\right)+...+\left(\dfrac{1}{74}+\dfrac{1}{75}\right)\)

\(=\dfrac{149}{4950}+\dfrac{149}{4998}+...+\dfrac{149}{5550}=\dfrac{a}{b}\)

\(=\dfrac{149}{1}\left(\dfrac{1}{4950}+\dfrac{1}{4998}+...+\dfrac{1}{5550}\right)\)

=> a\(⋮149\)

We have:

\(1\dfrac{1}{3}=1\dfrac{1}{1\cdot3};1\dfrac{1}{8}=1\dfrac{1}{2\cdot4};....\)

The first number of denominator of mixed numbers 98^th is:

\(98+1-1=98\) 

So their sum are:

\(1\dfrac{1}{1\cdot3}+1\dfrac{1}{2\cdot4}+...+1\dfrac{1}{98\cdot100}\)

\(=\left(1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\)

\(=98+\dfrac{2}{2}\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+...+\dfrac{1}{98\cdot100}\right)\)

\(=98+\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

\(=98+\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=98+\dfrac{1}{2}\left(1-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=98+\dfrac{1}{2}\cdot\left(\dfrac{98}{99}+\dfrac{49}{100}\right)=98+\dfrac{49}{99}+\dfrac{49}{200}=\dfrac{9751}{99}+\dfrac{49}{200}=\dfrac{1955050}{19800}\)

Put A = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)

\(A=\left(\dfrac{99}{1}-98\right)+\left(\dfrac{98}{2}+1\right)+\left(\dfrac{97}{3}+1\right)+...+\left(\dfrac{1}{99}+1\right)\)

\(A=\dfrac{100}{100}+\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\dfrac{100}{100}\)

\(A=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

So \(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{x}{3}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{5x}{15}-\dfrac{3}{15}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{5x-3}{15}\)

\(\Rightarrow y\left(5x-3\right)=60\)

Because \(5x-3\equiv2\left(mod5\right)\) and 5x-3 \(\in U\left(60\right)\)

\(\Rightarrow5x-3\in\left\{2;12\right\}\)

If 5x - 3 = 2

=> 5x = 5 => x = 1

And y = 60: 2 = 30

If 5x  - 3 = 12

=> 5x = 15 => x = 3

And y = 60 : 12 = 5

So \(\left(x,y\right)=\left(1;30\right),\left(3;5\right)\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^8}\)

\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\)

\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^8}\right)\)

\(2A=1-\dfrac{1}{3^8}=1-\dfrac{1}{6561}=\dfrac{6560}{6561}\)

\(A=\dfrac{6560}{6561}\cdot\dfrac{1}{2}=\dfrac{3280}{6561}\)

Put A = \(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{37\cdot38\cdot39}\)

\(\Rightarrow A=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{37\cdot38}-\dfrac{1}{38\cdot39}\right)\)

\(\Leftrightarrow A=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{38\cdot39}\right)=\dfrac{1}{2}\cdot\dfrac{370}{741}=\dfrac{185}{741}\)

\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{780}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{9}{10}\cdot\dfrac{14}{15}\cdot...\cdot\dfrac{779}{780}\)

\(=\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\cdot\dfrac{28}{30}\cdot...\cdot\dfrac{1558}{1560}\)

\(=\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot\dfrac{3\cdot6}{4\cdot5}\cdot\dfrac{4\cdot7}{5\cdot6}\cdot...\cdot\dfrac{38\cdot41}{39\cdot40}\)

\(=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot38\right)\left(4\cdot5\cdot6\cdot...\cdot41\right)}{\left(2\cdot3\cdot4\cdot...\cdot39\right)\left(3\cdot4\cdot5\cdot...\cdot40\right)}=\dfrac{1\cdot41}{39\cdot3}=\dfrac{41}{117}\)

I don't sure, what do you think about my answer???

Oh, sorry, I write mistake, sorry

\(\dfrac{100}{400}=\dfrac{1}{4}\)

We have:

\(\dfrac{1}{5};\dfrac{1}{45};\dfrac{1}{117};...\) <=> \(\dfrac{1}{1\cdot5};\dfrac{1}{5\cdot9};\dfrac{1}{9\cdot13};...\)

So the first numbers of denominator of  fraction 100^th of this series is:

\(\left(100-1\right)\cdot4+1=397\)

=> This fraction is \(\dfrac{1}{397\cdot\left(397+4\right)}=\dfrac{1}{397\cdot401}\)

So their sum are:

\(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{397\cdot401}\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)=\dfrac{1}{4}\cdot\dfrac{400}{401}=\dfrac{100}{401}< \dfrac{100}{400}< \dfrac{1}{4}\)

So the sum of 100 fraction in the series is smaller than \(\dfrac{1}{4}\)

\(\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}=\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\dfrac{4\cdot6}{5\cdot5}\cdot...\cdot\dfrac{49\cdot51}{50\cdot50}\)

\(=\dfrac{2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot...\cdot3\cdot5\cdot5\cdot7\cdot...\cdot49\cdot51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}=\dfrac{2\cdot51}{3\cdot50}=\dfrac{17}{25}\)

\(B=\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

\(=\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(199-1-1-...-1\right)\)[198 digits 1]

\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{200}=200\left(\dfrac{1}{199}+\dfrac{1}{198}+...+\dfrac{1}{2}\right)=200A\)

So B = 200A 

=> \(\dfrac{A}{B}=\dfrac{1}{200}\)

\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}\div\dfrac{1}{3}\)

\(\Rightarrow\dfrac{x+3}{5x+15}-\dfrac{5}{5x+15}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x+3-5}{5x+15}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x-2}{5x+15}=\dfrac{303}{1540}\)

=> 1540x - 3080 = 1515x + 4545

=> 1540x - 1515x = 4545 + 3080

=> 25x = 7625

=> x = 305