We see: row x has \(2\left(x-1\right)+1\) flowers

So row 25 has: \(2\left(25-1\right)+1=2.24+1=49\)

Apply the same sequence ratio, we have:

\(\dfrac{\overline{ab}}{b}=\dfrac{\overline{bc}}{c}=\dfrac{\overline{ca}}{a}=\dfrac{\overline{ab}+\overline{bc}+\overline{ca}}{a+b+c}=\dfrac{10a+b+10b+c+10c+a}{a+b+c}=\dfrac{11a+11b+11c}{a+b+c}=11\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\overline{ab}}{b}=11\\\dfrac{\overline{bc}}{c}=11\\\dfrac{\overline{ca}}{a}=11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\overline{ab}=11b\\\overline{bc}=11c\\\overline{ca}=11a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}10a+b=11b\\10b+c=11c\\10c+a=11a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}10a=10b\\10b=10c\\10c=10a\end{matrix}\right.\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a}{\overline{bc}}+\dfrac{b}{\overline{ab}}+\dfrac{c}{\overline{ab}}=\dfrac{a}{10a+a}+\dfrac{a}{10a+b}+\dfrac{a}{10a+a}=\dfrac{3a}{11a}=\dfrac{3}{11}\)

\(tan\left(a\right)=\dfrac{1}{2}\Leftrightarrow a=26,5605118\)

\(\Rightarrow tan\left(90-2a\right)=tan\left(90-2\cdot26,56505118\right)=\dfrac{3}{4}\)

We have:

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

Apply, we have:

\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2}\left(1-\dfrac{1}{2}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)< \dfrac{1}{2}\)

\(S=\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{1992}{2^{1991}}\)

\(\Rightarrow2S=2+\dfrac{2}{1}+\dfrac{3}{2}+...+\dfrac{1992}{2^{1990}}\)

\(\Rightarrow2S-S=\left(2+\dfrac{2}{1}+\dfrac{3}{2}+...+\dfrac{1992}{2^{1990}}\right)-\left(\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{1992}{2^{1991}}\right)\)

\(\Rightarrow S=2+\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{1990}}-\dfrac{1992}{2^{1991}}\)

• Alone:  The answer must be 21

Ngô Phương: x = 4, y= 2 => xy = 6 ?

x² + y² = 36 - 2xy

=> x² + 2xy + y² = 36

=> (x+y)² = 36

=> x + y = 6 or -6

x² - y² = 12 

=> (x+y)(x-y) = 12

=> \(\left[{}\begin{matrix}6\left(x-y\right)=12\\-6\left(x-y\right)=12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-y=2\\x-y=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=6\\x-y=2\end{matrix}\right.\Rightarrow x=4;y=2}\\\left\{{}\begin{matrix}x+y=-6\\x-y=-2\end{matrix}\right.\Rightarrow x=-4;y=-2\end{matrix}\right.\)

So xy = 8

We see: 99 gives 9 + 9 = 18 and 10 give 1 + 0 = 1 => two numbers give 19

The same: 98 with 11; 97 with 12;... to 49 with 51.

So we have: \(\dfrac{99-10+1}{2}=45\left(\text{pair}\right)\)

And their sum are: 19.45 = 855

-5 ≤ 2x - 7 ≤ 21

=> 2 ≤ 2x ≤ 28

=> 1 ≤ x ≤ 14

So there are: 14 - 1 + 1 = 14 numbers is true

Tell the praction to find is \(\dfrac{a}{b}\)

\(\Rightarrow\left\{{}\begin{matrix}b-a=8\\\dfrac{2a}{b-1}=1\end{matrix}\right.\)

=> a + a = b - 1

=> a + 1 = b - a  = 8

=> a = 7

=> b = 15

So the fraction to find is \(\dfrac{7}{15}\)

Tell x is the number to find

=> x(x+2) = 783

=> x = 27 or x = -29

But x is a positive => x = 27

ABC + DBBB = 2011

=> 100A + 10B + C + 1000D + 100B + 10B + B = 2011

=> 100A + 1121B + 1000D + C = 2011

Because A,B,C,D are digits

=> \(\left\{{}\begin{matrix}9\ge A,D>0\\9\ge B,C\ge0\end{matrix}\right.\)

If B = 2 => 100A + 1121B + 1000D + C = 100A + 2242 + 1000D + C > 2011 [unsatisfactory]

=> B = 0 or 1

If D = 2 => 100A + 1121B + 1000D + C = 100A + 1121B + 2000 + C = 2011

<=> 100A + 1121B + C = 11 but 100A \(\ge100\) [unsatisfactory]

=> D = 0 or 1

** B = 0, D = 0

=> 100A + 1121B + 1000D + C = 100A + 0 + 0 + C = 2011

But 101 \(\le\) 100A + C \(\le909\)

<=> unsatisfactory 

** B = 0; D = 1

=> 100A + 1121B + 1000D + C = 100A + 0 + 1000 + C = 2011

<=> 100A + C = 1011 [the same this case]

<=> unsatisfactory

** B = 1; D = 0

=> 100A + 1121B + 1000D + C = 100A + 1121 + 0 + C = 2011

=> 100A + C = 890

But C < 10 => A = 8 <=> 800 + C = 890 <=> C = 90 > 10 [unsatisfactory]

** B=1; D = 1

=> 100A + 1121B + 1000D + C = 100A + 1121 + 1000 + C = 2011

=> 100A + C = - 11 [ unsatisfactory ]

So there are no numbers are satisfactory 

So A + B + C + D \(\in\phi\)

1800 = 2³.3².5² = 1.1800 = 2.900 = 3.600 = 4.450 = 5.360 = 6.300 = 8.225 = 9.200 = 10.180 = 12.150 = 15.120 = 18.100 = 20.90 = 24.75 = 25.72 = 30.60 = 36.50 = 40.45

So we have 36 numbers

1,2,3,4,5,6,8,9,10,12,15 is smaller or equal 15.

So we have 36 - 11 = 25 numbers left

Answer: 25

Tell the books is a, the CD is b

=> \(3a+8b=2\left(a+5b\right)\) and a+b = $45

=> 3a + 8b = 2a + 10b

=> a = 2b

=> a+b = b+2b = 3b = $45 <=> b = $15

=> a = $15.2 = $30

=> 6a = $30.6 = $180

So the answer is $180

The length of this rectange is:

\(\left(\dfrac{62}{2}\right)\div\left(1+4\right)\cdot4=24,8\left(m\right)\)

The width of this rectange is:

24,8 : 4 = 6,2 [m]

So the area of this rectange is:

24,8 . 6,2 = 153,76 [m²]

Answer: 153,76m²

Because 245n is a perfect number => 245n \(\ge0\)

When n = 0

So n = 0 is smallest number satisfactory 

We have:

\(2x^2+3y^2+4x=19\)

\(\Leftrightarrow2x^2+4x+2=21-3y^2\)

\(\Leftrightarrow2\left(x+1\right)^2=3\left(7-y^2\right)\)

Because \(2\left(x+1\right)^2⋮2\Rightarrow3\left(7-y^2\right)⋮2\)

But \(3⋮̸2\)=> \(7-y^2⋮2\)

=> y² is a odd => y is a odd

We have:

\(\left(x+1\right)^2\ge0\Rightarrow7-y^2\ge0\Rightarrow0\le y^2\le7\)

And y is a integer => \(y^2\in\left\{0;1;4\right\}\)

But y² is a odd => y² = 1 <=> \(y=\pm1\)

\(\Rightarrow2\left(x+1\right)^2=21-3\cdot1=18\)

\(\Rightarrow\left(x+1\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

So we have \(\left(x,y\right)=\left(2;1\right);\left(2;-1\right);\left(-4;1\right);\left(-4;-1\right)\)

Put A = 1³ + 2³ + ... + n³

We have if n = 1 => A = \(\left(\dfrac{1\left(1+1\right)}{2}\right)^2\) - right

So assume A always right with n = k, we have:

\(1^3+2^3+...+k^3=\left(\dfrac{k\left(k+1\right)}{2}\right)^2\)

So we need prove that if n = k+1 so A = n²

We have:

\(1^3+2^3+...+k^3+\left(k+1\right)^3=\left(\dfrac{k\left(k+1\right)}{2}\right)^2+\left(k+1\right)^3=\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}=\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\dfrac{\left[\left(k+1\right)\left(k+2\right)\right]^2}{2^2}=\left(\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right)^2\)

- Right

So 1³ + 2³ + ... + 2016³ = \(\left[\dfrac{2016\left(2016+1\right)}{2}\right]^2=\left(\dfrac{2016\cdot2017}{2}\right)^2=2033136^2\)

So n = 2033136

We have: 

a³ + b³ = (a + b)(a² - ab + b²)

So if a+b \(⋮6\) then a³ + b³ \(⋮6\)