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Let a , b , and c be values chosen from {0,1,2,3,…,100} . They need not be distinct, but a≥b . How many ordered triples (a,b,c) are there satisfying both |a−b|=c and |b−c|=a ? 

 Mar 3, 2020
 #1
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The answer is 201, but i don't know why

 Mar 3, 2020
 #2
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Let \(a\) , \(b\) , and \(c\) be values chosen from \(\{~0,~1,~2,~3,~\ldots ~,100~\}\) .
They need not be distinct, but \(a\geq b\) .
How many ordered triples \((~a,~b,~c~)\) are there satisfying both \(|~a-b~|=c\) and \(|~b-c~|=a\) ?

 

\(\small{ \begin{array}{|lrcll|} \hline & |~a-b~| &=& c \qquad \text{square both sides} \\ (1) & (a-b)^2 &=& c^2 \\\\ & |~b-c~| &=& a \qquad \text{square both sides} \\ (2) & (b-c)^2 &=& a^2 \\ \hline \\ (1)+(2): & (a-b)^2 + (b-c)^2 &=& c^2 + a^2 \\ & a^2-2ab+b^2+b^2-2bc+c^2 &=& c^2 + a^2 \\ & -2ab+b^2+b^2-2bc &=& 0 \\ & 2b^2-2ab-2bc &=& 0 \quad | \quad : 2 \\ & b^2-ab-bc &=& 0 \\ &\mathbf{ b(b-a-c) } &=& \mathbf{0} \\ \hline \\ 1) & \mathbf{b} &=& \mathbf{0} \\ & |~a-b~| &=& c \quad | \quad b = 0 \\ & |~a-0~| &=& c \\ & |~a ~| &=& c \\ & \mathbf{a} &=& \mathbf{c} \\ \quad \text{List}~ 1: & \mathbf{(~a,~0,~a~)} && \begin{array}{|lcll|} \hline (~0,~0,~0~),\quad (~1,~0,~1~),\quad (~2,~0,~2~),\ \ldots (~99,~0,~99~),\ (~100,~0,~100~) \\ \mathbf{101} ~\text{ordered triples} \\ \hline \end{array} \\ \hline \\ 2) & \mathbf{b-a-c} &=& \mathbf{0} \\ & a+c &=& b \\ & a &\geq& b \quad | \quad b = a+c \\ & a&\geq & a+c \quad \text{possible if $c=0$} \\ & \mathbf{c} &=& \mathbf{0} \\ & |~b-c~| &=& a \quad | \quad c = 0 \\ & |~b-0~| &=& a \\ & |~b ~| &=& a \\ & \mathbf{b} &=& \mathbf{a} \\ \quad \text{List}~ 2: & \mathbf{(~a,~a,~0~)} && \begin{array}{|lcll|} \hline (~1,~1,~0~),\quad (~2,~2,~0~),\quad (~3,~3,~0~),\ \ldots (~99,~99,~0~),\ (~100,~100,~0~) \\ \mathbf{100} ~\text{ordered triples}\qquad (~0,~0,~0~)~\text{ is already included in the first list } \\ \hline \end{array} \\ \hline \end{array} }\)

 

\(\text{There are $101+100 = \mathbf{201}$ ordered triples}\)

 

laugh

 Mar 3, 2020
edited by heureka  Mar 3, 2020

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