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 #4
avatar+18936 
+1
heureka Feb 16, 2018
 #1
avatar+18936 
+1

What is the smallest positive integer that will satisfy the following congruences:

N mod 105 = 104,

N mod 111 = 110,

N mod 121 = 111,

N mod 122 = 111

Thanks for any help.

 

\(\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{104 \pmod{105}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{110 \pmod{111}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -104 &=& 105x \\ & n -110 &=& 111y \\\\ \Rightarrow & n &=& 104+105x \\ & n &=& 110+111y \\\\ & n=104+105x&=& 110+111y \\ & 104+105x&=& 110+111y \\ & 105x-111y&=& 6 \quad & | \quad : 3 \\ & \mathbf{35x-37y}& \mathbf{=}& \mathbf{2} \qquad x,y \in Z \\ &\rightarrow \quad x &=& -1+37b^{^1)} \qquad b \in Z \\\\ & n &=& 104+105x \\ & n &=& 104+105(-1+37b) \\ & n &=& 104-105+105\cdot 37b \\ & n &=& -1+ 3885b \\ & \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad (1)\\ \end{array}\\ \hline \end{array}\)

\(\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{121}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{122}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -111 &=& 121x \\ & n -111 &=& 122y \\\\ \Rightarrow & n &=& 111+121x \\ & n &=& 111+122y \\\\ & n=111x+121y&=& 111+122y \\ & 111x+121y&=& 111+122y \\\\ & \mathbf{121x-122y}&\mathbf{=}& \mathbf{0} \qquad x,y \in Z \\ &\rightarrow \quad x &=& 122a^{^2)} \qquad a \in Z \\\\ & n &=& 111+121x \\ & n &=& 111+121(122a) \\ & n &=& 111+121\cdot 122a \\ & n &=& 111+14762a \\ & \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad (2)\\ \end{array} \\ \hline \end{array} \)

 

After reducing, we have two formulas:

\( \begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad &(1)\\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad &(2)\\ \hline \end{array} \)

 

\( \begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n +1 &=& 3885x \\ & n -111 &=& 14762y \\\\ \Rightarrow & n &=& -1+3885x \\ & n &=& 111+14762y \\\\ & n=-1+3885x&=& 111+14762y \\ & -1+3885x&=& 111+14762y \\ & \mathbf{3885x-14762y}& \mathbf{=}& \mathbf{112} \qquad x,y \in Z \\ &\rightarrow \quad x &=&3378+14762g^{^3)} \qquad g \in Z \\\\ & n &=& -1+3885x \\ & n &=& -1+3885\cdot (3378+14762g) \\ & n &=& -1+3885\cdot 3378+ 3885\cdot 14762g \\ & \mathbf{n} &\mathbf{=}& \mathbf{13123529 + 57350370g \qquad g \in Z } \\ \end{array}\)

 

The smallest positive integer is  \(\mathbf{13\ 123\ 529}\)

 

Proof:

\(\begin{array}{|rcll|} \hline \mathbf{13\ 123\ 529} &\text{mod } 105 =& 104\\ \mathbf{13\ 123\ 529} &\text{mod } 111 =& 110\\ \mathbf{13\ 123\ 529} &\text{mod } 121 =& 111 \\ \mathbf{13\ 123\ 529} &\text{mod } 122 =& 111 \\ \hline \end{array} \)

 

\(\begin{array}{lcll} \hline ^1) \\ \text{Solve of the diophantine equation $35x - 37y = 2$ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 35x &=& 2 + 37y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \\ &=& \dfrac{ 2 +35y+2y } {35} \\ &=& \dfrac{ 35y + 2 +2y } {35} \\ &=& \dfrac{35y}{35} + \dfrac{ 2 +2y } {35} \\ x &=& y + \dfrac{ 2 +2y} {35} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 + 2y } {35} \\ & 35a &=& 2 + 2y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 2y &=& -2 + 35a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \\ &=& \dfrac{ -2 + 34a+a } {2} \\ &=& \dfrac{ 34a - 2 +a } {2} \\ &=& -\dfrac{-2}{2} +\dfrac{34a}{2} + \dfrac{ a } {2} \\ y &=& -1+17a+ \dfrac{ a } {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{a} {2} \\ & 2b &=& a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2b } \\ \end{array} \end{array} \)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \quad & | \quad \mathbf{a = 2b }\\ & = & \dfrac{-2 + 35\cdot (2b) } {2} \\ \mathbf{y} & \mathbf{=} & \mathbf{-1+35b} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \quad & | \quad \mathbf{y = -1+35b }\\ & = & \dfrac{ 2 + 37\cdot (-1+35b ) } {11} \\ \mathbf{x} & \mathbf{=} & \mathbf{-1 + 37b } \\ \end{array} \)

 

\(\begin{array}{lcll} \hline ^2) \\ \text{Solve of the diophantine equation $121x - 122y = 0 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 121x &=& 122y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \\ &=& \dfrac{ 121y+y } {121} \\ &=& \dfrac{121y}{121} + \dfrac{y} {121} \\ x &=& y + \dfrac{ y} {121} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{y} {121} \\ & 121a &=& y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{y} &\mathbf{=}& \mathbf{ 121a } \\ \end{array} \end{array} \)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \quad & | \quad \mathbf{y=121a }\\ & = & \dfrac{122\cdot (121a) } {121} \\ \mathbf{x} & \mathbf{=} & \mathbf{122a} \\ \end{array}\)

 

\( \begin{array}{lcll} \hline ^3) \\ \text{Solve of the diophantine equation $3885x - 14762y = 112 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 3885x &=& 112 + 14762y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y} {3885}} \\ &=& \dfrac{ 112 + 11655y + 3107y } {3885} \\ &=& \dfrac{11655y + 112 + 3107y }{3885} \\ &=& \dfrac{11655y }{3885} + \dfrac{112 + 3107y} {3885} \\ x &=& 3y + \dfrac{ 112 + 3107y } {3885} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{112 + 3107y } {3885} \\ & 3885a &=& 112 + 3107y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 3107y &=& -112 + 3885a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a} {3107}} \\ &=& \dfrac{ -112 + 3107a + 778a } {3107} \\ &=& \dfrac{3107a - 112 + 778a }{3107} \\ &=& \dfrac{3107a }{3107} + \dfrac{- 112 + 778a} {3107} \\ y &=& 3a + \dfrac{ 112 + 3107a } {3107} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{-112 + 778a } {3107} \\ & 3107b &=& -112 + 778a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{rcll} 778a &=& 112 + 3107b \\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b} {778}} \\ &=& \dfrac{ 112 + 2334b + 773b } {778} \\ &=& \dfrac{2334b +112 + 773b }{778} \\ &=& \dfrac{2334b }{3107} + \dfrac{112 + 773b} {778} \\ a &=& 3b + \dfrac{ 112 + 773b} {778} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &c &=& \dfrac{112 + 773b } {778} \\ & 778c &=& 112 + 773b \\ \end{array} \\ \text{The variable with the smallest coefficient is $b$. The equation is transformed after $b$: }\\ \begin{array}{rcll} 773b &=& -112 + 778c\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c} {773}} \\ &=& \dfrac{ -112 + 773c + 5c } {773} \\ &=& \dfrac{773c -112 + 5c }{773} \\ &=& \dfrac{773c }{773} + \dfrac{-112 + 5c } {773} \\ b &=& c + \dfrac{ -112 + 5c} {773} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &d &=& \dfrac{-112 + 5c } {773} \\ & 773d &=& -112 + 5c \\ \end{array} \\ \text{The variable with the smallest coefficient is $c$. The equation is transformed after $c$: }\\ \begin{array}{rcll} 5c &=& 112 + 773d\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \\ &=& \dfrac{ 110+2 + 770d + 3d } {5} \\ &=& \dfrac{110+770d+2 + 5d }{5} \\ &=& \dfrac{110 }{5}+\dfrac{770d }{5} + \dfrac{2 + 3d } {5} \\ c &=& 22 + 154d + \dfrac{2 + 3d } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &e &=& \dfrac{ 2 + 3d } {5} \\ & 5e &=& 2 + 3d \\ \end{array} \\ \text{The variable with the smallest coefficient is $d$. The equation is transformed after $d$: }\\ \begin{array}{rcll} 3d &=& -2 + 5e\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \\ &=& \dfrac{ -2 + 3e + 2e } {3} \\ &=& \dfrac{3e -2 + 2e}{3} \\ &=& \dfrac{3e }{3} + \dfrac{-2 + 2e} {3} \\ d &=& e + \dfrac{-2 + 2e } {3} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &f &=& \dfrac{ -2 + 2e } {3} \\ & 3f &=& -2 + 2e \\ \end{array} \\ \text{The variable with the smallest coefficient is $e$. The equation is transformed after $e$: }\\ \begin{array}{rcll} 2e &=& 2 + 3f\\ \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{2 + 3f } {2}} \\ &=& \dfrac{ 2+2f+f } {2} \\ &=& \dfrac{2 }{2} + \dfrac{2f }{2} + \dfrac{f} {2} \\ e &=& 1 + f + \dfrac{f} {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &g &=& \dfrac{ f} {2} \\ & 2g &=& f \\ \end{array} \\ \text{The variable with the smallest coefficient is $f$. The equation is transformed after $f$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{f} &\mathbf{=}& \mathbf{ 2g } \\ \end{array} \end{array}\)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{ 2 + 3f } {2}} \quad & | \quad \mathbf{f=2g }\\ & = & \dfrac{2 + 3\cdot 2g } {2} \\ \mathbf{e} & \mathbf{=} & \mathbf{1 + 3g } \\\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \quad & | \quad \mathbf{e=1 + 3g }\\ & = & \dfrac{-2 + 5\cdot (1 + 3g) } {2} \\ \mathbf{d} & \mathbf{=} & \mathbf{1 + 5g } \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \quad & | \quad \mathbf{d=1 + 5g }\\ & = & \dfrac{112 + 773\cdot (1 + 5g) } {2} \\ \mathbf{c} & \mathbf{=} & \mathbf{177 + 773g } \\\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c } {773}} \quad & | \quad \mathbf{c=177 + 773g }\\ & = & \dfrac{-112 + 778\cdot (177 + 773g) } {773} \\ \mathbf{b} & \mathbf{=} & \mathbf{ 178 + 778g } \\\\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b } {778}} \quad & | \quad \mathbf{b=178 + 778g }\\ & = & \dfrac{112 + 3107\cdot (178 + 778g) } {778} \\ \mathbf{a} & \mathbf{=} & \mathbf{ 711 + 3107g } \\\\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a } {3107}} \quad & | \quad \mathbf{a=711 + 3107g }\\ & = & \dfrac{-112 + 3885\cdot (711 + 3107g) } {3107} \\ \mathbf{y} & \mathbf{=} & \mathbf{ 889 + 3885g } \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y } {3885}} \quad & | \quad \mathbf{y=889 + 3885g }\\ & = & \dfrac{112 + 14762\cdot (889 + 3885g) } {3885} \\ \mathbf{x} & \mathbf{=} & \mathbf{ 3378 + 14762g } \\ \end{array}\)

 

 

laugh

heureka Feb 15, 2018
 #3
avatar+18936 
+2

Triangle AHI is equilateral. We know BC, DE, and FG are all parallel to HI and AB = BD = DF = FH.
What is the ratio of the area of the trapezoid FGIH to the area of triangle AHI?
Express your answer as a common fraction.

 

 

\(\text{Let $FG = s $} \\ \text{Let $HI = c $} \\ \text{Area of the triangle $AHI = A_{AHI}$ } \\ \text{Area of the triangle $AGF = A_{AGF}$ } \\ \text{Area of the trapezoid $FGIH = A_{FGIH}$ } \\ \text{Let $AK = h$ (height of the triangle$_{AGF}$) } \\ \text{Let $AL = H$ (height of the triangle$_{AHI}$) } \\ \text{Let $AF = \frac34 c $}\)

 

1.

\(\begin{array}{|rcll|} \hline A_{AHI} &=& A_{AGF} + A_{FGIH} \\\\ \dfrac{cH}{2} &=& \dfrac{sh}{2} + \left( \dfrac{s+c}{2}\right)(H-h) \quad & | \quad \times 2 \\\\ cH &=& sh + (s+c)(H-h) \\\\ \not{cH} &=& \not{sh} + sH-\not{sh}+\not{cH}-ch \\\\ \mathbf{ch} &\mathbf{=}& \mathbf{ sH } \qquad (1) \\\\ &\text{or}& \\\\ \mathbf{\dfrac{s}{c}} &\mathbf{=}& \mathbf{\dfrac{h}{H}} \qquad (2) \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline \text{ratio} &=& \dfrac{ A_{FGIH} } {A_{AHI}} \\\\ &=& \dfrac{ \left( \dfrac{s+c}{2}\right)(H-h) } {\dfrac{cH}{2}} \\\\ &=& \dfrac{(s+c)(H-h)}{cH} \\\\ &=& \dfrac{sH-sh+cH-ch}{cH} \quad & | \quad ch=sH \qquad (1) \\\\ &=& \dfrac{cH-sh}{cH} \\\\ &=& 1-\dfrac{sh}{cH} \quad & | \quad \dfrac{s}{c} = \dfrac{h}{H} \qquad (2) \\\\ &=& 1-\dfrac{h^2}{H^2} \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\dfrac{h}{H}\right)^2} \\\\ && \boxed{\mathbf{3.}\\ \dfrac{h}{\frac34 c} = \dfrac{H}{c} \\ h = \frac34 c \cdot \dfrac{H}{c} \\ h = \frac34 \cdot H \\ \mathbf{\dfrac{h}{H} = \frac34} } \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\frac34 \right)^2} \\\\ &=& 1- \frac{9}{16} \\\\ &=& \frac{16-9}{16} \\\\ &\mathbf{=}& \mathbf{\dfrac{7}{16}} \\ \hline \end{array}\)

 

So  the ratio  of areas is \(\mathbf{\dfrac{7}{16}}\)

 

laugh

heureka Feb 13, 2018
 #1
avatar+18936 
+1

I assume the question is:

\(\mathbf{\text{A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed $\\$ so that the first three terms are in geometric progression, $\\$ the second, third, and fourth terms are in arithmetic progression, $\\$ and, in general, for all $n\ge1,$ $\\$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, $\\$ and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. $\\$ Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $.}}\)

 

Let AP = arithmetic progression,

Let GP = geometric progression

 

\(\text{AP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i(j-k) + t_j(k-i) + t_k(i-j) = 0 \\\\ & i = 2n & t_i = a_{2n} \\ & j = 2n+1 & t_j = a_{2n+1} \\ & k = 2n+2 & t_k = a_{2n+2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t_i(j-k) + t_j(k-i) + t_k(i-j) &=& 0 \\ a_{2n}(2n+1-(2n+2)) + a_{2n+1}(2n+2-2n) + a_{2n+2}(2n-(2n+1)) &=& 0 \\ \mathbf{-a_{2n} + 2a_{2n+1} - a_{2n+2}} &\mathbf{=}& \mathbf{0} \\ \hline \end{array} \)

 

\(\text{GP Formula:}\\ \begin{array}{|llcll|} \hline \text{ $t =$ term }\\ \text{ $i,j,k = $ indices}\\\\ t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ & i = 2n-1 & t_i = a_{2n-1} \\ & j = 2n & t_j = a_{2n} \\ & k = 2n+1 & t_k = a_{2n+1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline t_i^{j-k} \times t_j^{k-i} \times t_k^{i-j} = 1 \\\\ a_{2n-1}^{2n-(2n+1)} \times a_{2n}^{2n+1-(2n-1)} \times a_{2n+1}^{2n-1-(2n)} = 1 \\\\ \mathbf{a_{2n-1}^{-1} \times 2a_{2n}^{2} \times a_{2n+1}^{-1} } &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

 

So we have:

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{a_{2n+2}} &\mathbf{=}& \mathbf{2a_{2n+1} -a_{2n} } \\ (2) & \mathbf{a_{2n+1} } &\mathbf{=}& \mathbf{\dfrac{ a_{2n}^{2} }{a_{2n-1} } } \\ \hline \end{array}\)

 

\(\mathbf{a_1=1 }\\ \begin{array}{|rcll|} \hline a_3 &=& \dfrac{a_2^2}{a_1} \quad a_1 = 1 & | \quad \text{ Formula $(2)$} \\ \mathbf{a_3} & \mathbf{=} & \mathbf{ a_2^2 } \\ \hline a_4 &=& 2a_3-a_2 \quad a_3 = a_2^2 & | \quad \text{ Formula $(1)$} \\ &=& 2a_2^2-a_2 \\ \mathbf{a_4} & \mathbf{=} & \mathbf{ a_2(2a_2-1) } \\ \hline a_5 &=& \dfrac{a_4^2}{a_3} & | \quad \text{ Formula $(2)$} \\ a_5 &=& \dfrac{a_2^2(2a_2-1)^2}{a_2^2} \\ \mathbf{a_5} & \mathbf{=} & \mathbf{ (2a_2-1)^2 } \\ \hline a_6 &=& 2a_5-a_4 & | \quad \text{ Formula $(1)$} \\ &=& 2(2a_2-1)^2-a_2(2a_2-1) \\ \mathbf{a_6} & \mathbf{=} & \mathbf{ (2a_2-1)(3a_2-2) } \\ \hline a_7 &=& \dfrac{a_6^2}{a_5} & | \quad \text{ Formula $(2)$} \\ a_7 &=& \dfrac{(2a_2-1)^2(3a_2-2)^2}{(2a_2-1)^2} \\ \mathbf{a_7} & \mathbf{=} & \mathbf{ (3a_2-2)^2 } \\ \hline a_8 &=& 2a_7-a_6 & | \quad \text{ Formula $(1)$} \\ &=& 2(3a_2-2)^2-(2a_2-1)(3a_2-2) \\ \mathbf{a_8} & \mathbf{=} & \mathbf{ (3a_2-2)(4a_2-3) } \\ \hline a_9 &=& \dfrac{a_8^2}{a_7} & | \quad \text{ Formula $(2)$} \\ a_9 &=& \dfrac{(3a_2-2)^2(4a_2-3)^2}{(3a_2-2)^2} \\ \mathbf{a_9} & \mathbf{=} & \mathbf{ (4a_2-3)^2 } \\ \hline a_{10} &=& 2a_9-a_8 & | \quad \text{ Formula $(1)$} \\ &=& 2(4a_2-3)^2-(3a_2-2)(4a_2-3) \\ \mathbf{a_{10}} & \mathbf{=} & \mathbf{ (4a_2-3)(5a_2-4) } \\ \hline \end{array}\)

 

\(\mathbf{a_9+a_{10}=646}\\ \begin{array}{|rcll|} \hline \mathbf{a_9+a_{10}} & \mathbf{=}& \mathbf{646} \\ (4a_2-3)^2+(4a_2-3)(5a_2-4) &=& 646 \\ (4a_2-3) (4a_2-3+5a_2-4) &=& 646 \\ (4a_2-3) (9a_2-7) &=& 646 \\ \ldots \\ 36a_2^2-55a_2-625 &=& 0 \\\\ a_2 &=& \dfrac{55\pm\sqrt{55^2-4\cdot36\cdot(-625)} }{2\cdot 36} \\\\ &=& \dfrac{55\pm\sqrt{93025} }{72} \\\\ &=& \dfrac{55\pm305 }{72} \\\\ a_2 &=& \dfrac{55+305 }{72} \\\\ \mathbf{a_2 } & \mathbf{=} & \mathbf{5} \\\\ && \text{or} \\\\ a_2 &=& \dfrac{55-305 }{72} \\\\ a_2 & = & -3.47222222222 \\ && \text{no solution, because $a_n$ is a sequence}\\ && \text{of positive integers!}\\ \hline \end{array}\)

 

\(\text{Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $ a_n $} \)

\(\begin{array}{|lclcl|} \hline a_1 && &=& 1 \\ a_2 && &=& 5 \\ a_3 &=& 5^2 &=& 25 \\ a_4 &=& 5\cdot 9 &=& 45 \\ a_5 &=& 9^2 &=& 81 \\ a_6 &=& 9\cdot 13 &=& 117 \\ a_7 &=& 13^2 &=& 169 \\ a_8 &=& 13\cdot 17 &=& 221 \\ a_9 &=& 17^2 &=& 289 \\ a_{10} &=& 17\cdot 21 &=& 357 \\ a_{11} &=& 21^2 &=& 441 \\ a_{12} &=& 21\cdot 25 &=& 525 \\ a_{13} &=& 25^2 &=& 625 \\ a_{14} &=& 25\cdot 29 &=& 725 \\ a_{15} &=& 29^2 &=& 841 \\ \color{red}a_{16} &=& 29\cdot 33 &\color{red}=& \color{red}957 \qquad a_n < 1000\\ a_{17} &=& 33^2 &=& 1089 \qquad a_n > 1000 \\ \hline \end{array} \)

 

 

\(\mathbf{\text{$a_n$ is $ 957$}}\)

 

 

\(\text{GP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{ratio} & a_{2n-1} & a_{2n} & a_{2n+1} \\ \hline 1 &5 &1 & 5 & 25 \\ 2&1.8 &25 & 45 & 81 \\ 3&1.4\ldots &81 & 117 & 169 \\ 4&1.30769230769 &169 & 221 & 289 \\ 5&1.23529411765 &289 & 357 & 441 \\ 6&1.19047619048 &441 & 525 & 625 \\ 7&1.16 &625 & 725 & 841 \\ 8&1.13793103448 &841 & 957 & 1089 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}\)

 

\(\text{AP:}\\ \begin{array}{|r|r|rrrl|} \hline n & \text{common difference} & a_{2n} & a_{2n+1} & a_{2n+2} \\ \hline 1 &20 & 5 & 25 & 45 \\ 2 &36 &45 & 81 & 117 \\ 3 &52 &117 & 169 & 221 \\ 4 &68 & 221 & 289 & 357 \\ 5 &84 &357 & 441 & 525 \\ 6 &100& 525 & 625 & 725 \\ 7 &116& 725 & 841 & 957 \\ \ldots & \ldots& \ldots& \ldots& \ldots \\ \hline \end{array}\)

 

laugh

heureka Feb 9, 2018
 #1
avatar+18936 
+1

Find the sum of the terms from the (n + 1) th to the m th term inclusive of an arithmetical progression whose

first term is a and whose second term is b.

If m = 13, n =   3 and the sum is 12a, find the ratio b: a.

 

Answer (m-n){a+1/2(m+n-1)(b-a)} ; 77/75

 

\(\text{Formula arithmetical progression: }\\ \begin{array}{|rcll|} \hline a_1 &=& a \\ a_2 &=& a+d \\ \hline \end{array} \)

 

\(\text{second term is b: }\\ \begin{array}{|rcll|} \hline a_2 &=& a+d \\ a_2 &=& b \\ \hline a+d &=& b \\ \mathbf{d} & \mathbf{=} & \mathbf{b-a} \\ \hline \end{array}\)

 

The common difference is \(\mathbf{b-a}\)

 

\(\text{The sum of the member is: }\\ \begin{array}{|rcll|} \hline s_n = n\cdot a + \dfrac{n(n-1)}{2}\cdot d \\ \hline \end{array} \)

 

\(\text{The sum of the terms from the (n + 1) th to the m th term is: }\\ \begin{array}{|rcll|} \hline s_m-s_n &=& \underbrace{m\cdot a + \dfrac{m(m-1)}{2}\cdot (b-a)}_{=s_m} - \left[ \underbrace{n\cdot a + \dfrac{n(n-1)}{2}\cdot (b-a)}_{=s_n} \right] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m(m-1)-n(n-1) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-m-n^2+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-n^2-m+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m^2-n^2-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ (m-n)(m+n)-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)(m-n)( m+n-1 ) \\\\ \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\ \hline \end{array}\)

 

\(\small{ \text{If $m = 13, n = 3$ and the sum is $12a$, find the ratio $b: a$ }\\ \begin{array}{|rcll|} \hline \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\\\ 12a & = & (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b}{2}\right)( m+n-1 )- \left(\dfrac{a}{2}\right)( m+n-1 ) \\\\ a\left( \dfrac{12}{m-n} - 1+ \dfrac{m+n-1}{2} \right) & = & b\left( \dfrac{m+n-1}{2} \right) \qquad m = 13 \qquad n = 3 \\\\ a\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) & = & b\left( \dfrac{15}{2} \right) \\\\ \dfrac{b}{a} &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) \\\\ &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12-10+75}{10} \right) \\\\ &=& \dfrac{2\cdot 77}{150} \\\\ &=& \dfrac{2\cdot 77}{2\cdot 75} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{ =}& \mathbf{ \dfrac{77}{75}} \\ \hline \end{array} } \)

 

 

laugh

heureka Feb 8, 2018
 #9
avatar+18936 
+2
heureka Feb 7, 2018
 #7
avatar+18936 
+2

When the minute hand of a clock points exactly at a full minute, the hour hand is exactly two minutes away.

Give all possible times in a 12-hour period that satisfy condition.

 

\(\text{ angular velocity minute hand: $ \omega_m^{\circ} =\dfrac{360^{\circ}}{1~ h} $ }\\ \text{ angular velocity hour hand: $ \omega_h^{\circ} =\dfrac{360^{\circ}}{12~ h} $ }\\ \boxed{\text{angle = angular velocity }\times \text{ time}}\\ \text{ angle minute hand: $ \alpha_m^{\circ} = \omega_m^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle hour hand: $ \alpha_h^{\circ} = \omega_h^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle difference minute hand-hour hand: $ \alpha_m^{\circ}-\alpha_h^{\circ} = \Delta\alpha^{\circ}$ } \\ \Delta\alpha^{\circ} = \alpha_m^{\circ}-\alpha_h^{\circ} \\ \Delta\alpha^{\circ} = \omega_m^{\circ} \times t^h - \omega_h^{\circ} \times t^h \\ \Delta\alpha^{\circ} = \left(\omega_m^{\circ} - \omega_h^{\circ} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(\dfrac{360^{\circ}}{1~ h} - \dfrac{360^{\circ}}{12~ h} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(360\cdot \dfrac{11}{12} \right) \times t^h \\ \Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} \\\)

 

\(\mathbf{\boxed{\Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} }}\)

 

Example:

\(t^h = 6\ h \\ \Delta\alpha^{\circ} = 330 \times 6 \pmod{360^{\circ}} \\ \Delta\alpha^{\circ} = 1980^{\circ} \pmod{ 360^{\circ} } \\ \Delta\alpha^{\circ} = 1980^{\circ} - 5\cdot 360^{\circ} \\ \Delta\alpha^{\circ} = 180^{\circ} \)

 

1.

\(\text{The hour hand of a clock points exactly at a full minute:} \\ \begin{array}{rcll} \dfrac{360^{\circ}}{60~\min.} &=& \dfrac{x}{1~\min.} \\\\ x&=&\dfrac{1~ \min.}{60~ \min.}\cdot 360^{\circ} \\\\ \mathbf{x}&\mathbf{=}&\mathbf{6^{\circ}} \\ \end{array} \)

 

1 minute on the clock conforms 6 degrees.

 

\(\text{The hour hand of a clock points exactly at a full minute, so} \\ \text{ $\alpha_h^{\circ} = 6^{\circ} \cdot n \qquad | \qquad n$ is an integer} \)

 

2. \(\mathbf{t^h =\ ?}\)

\(\begin{array}{rcll} \alpha_h^{\circ} &=& \omega_h^{\circ} \times t^h \\\\ t^h &=& \dfrac{ \alpha_h^{\circ} } {\omega_h^{\circ}} \quad & | \quad \alpha_h^{\circ} = 6^{\circ} \cdot n \qquad \omega_h^{\circ} = \dfrac{360^{\circ}}{12~ h} \\\\ t^h &=& \dfrac{ 6^{\circ} \cdot n } { \dfrac{360^{\circ}}{12~ h}} \\\\ \mathbf{t^h} &\mathbf{=}& \mathbf{0.2n} \\ \end{array}\)

 

3.

\(\text{The hour hand is exactly two minutes away: $\Delta\alpha^{\circ} = \pm 12^{\circ}$ } \\ \begin{array}{rcll} \Delta\alpha^{\circ} &=& 330 \times t^h \pmod{360^{\circ}} \quad & | \quad \mathbf{t^h=0.2n} \qquad \Delta\alpha^{\circ} = \pm 12^{\circ} \\ \pm 12^{\circ} &=& 330 \times 0.2n \pmod{360^{\circ}} \\ \pm 12^{\circ} &=& 66n \pmod{360^{\circ}} \\ \pm 12^{\circ} -66n &=& 360^{\circ}m \qquad & n \in N,\ m \in N \\ \pm 12^{\circ} &=& 66n + 360^{\circ}m \quad & | \quad : 6 \\ \pm 2^{\circ} &=& 11n + 60^{\circ}m \\ &&\boxed{1. \text{ Diophantine equation: } 11n + 60^{\circ}m = 2} \\ &&\boxed{2. \text{ Diophantine equation: } 11n + 60^{\circ}m = -2} \\ \end{array} \)

 

4. Solution diophantine equation 11n + 60m = 2:

\(\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& 2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \\ &=& \dfrac{ 2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m + 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ 2 - 5m } {11} \\ n &=&-5m+ \dfrac{ 2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 - 5m } {11} \\ & 11a &=& 2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& 2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \\ &=& \dfrac{ 2 - 10a-a } {5} \\ &=& \dfrac{ - 10a + 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ 2 -a } {5} \\ m &=& -2a+ \dfrac{ 2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ 2 - a } {5} \\ & 5b &=& 2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2 - 5b } \\ \end{array} \)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \quad & | \quad \mathbf{a = 2 - 5b }\\ & = & \dfrac{ 2 - 11 (2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{-4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = -4 + 11b }\\ & = & \dfrac{ 2 - 60(-4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{22 - 60b } \\ \end{array} \)

 

\(\boxed{ \text{1. Diophantine equation: }\\ \mathbf{n = 22 - 60b} \\\mathbf{m=-4 + 11b} \qquad b \in Z } \)\(\begin{array}{|lrcll|} \hline b = 0: & n &=& 22 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 22 \\ & &=& 4.4\quad (4:24) \\ \hline \end{array}\)

 

5. Solution diophantine equation 11n + 60m = -2:

\(\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& -2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ -2 - 60m } {11}} \\ &=& \dfrac{ -2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m - 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ -2 - 5m } {11} \\ n &=&-5m+ \dfrac{ -2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ -2 - 5m } {11} \\ & 11a &=& -2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& -2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \\ &=& \dfrac{ -2 - 10a-a } {5} \\ &=& \dfrac{ - 10a - 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ -2 -a } {5} \\ m &=& -2a+ \dfrac{ -2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ -2 - a } {5} \\ & 5b &=& -2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ -2 - 5b } \\ \end{array}\)

 

\(\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \quad & | \quad \mathbf{a = -2 - 5b }\\ & = & \dfrac{ -2 - 11 (-2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = 4 + 11b }\\ & = & \dfrac{ -2 - 60(4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{-22 - 60b } \\ \end{array}\)

 

\(\boxed{ \text{2. Diophantine equation: }\\ \mathbf{n = -22 - 60b} \\\mathbf{m=4 + 11b} \qquad b \in Z } \)\(\begin{array}{|lrcll|} \hline b = -1: & n &=& -22+60 \\ & &=& 38 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 38 \\ & &=& 7.6\quad (7:36) \\ \hline \end{array} \)

 

 

The solutions are: 4:24 and 7:36

 

laugh

heureka Feb 7, 2018