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"Modulo m graph paper" consists of a grid of m^2 points, representing all pairs of integer residues (x,y) where \(0\le x . To graph a congruence on modulo m graph paper, we mark every point (x,y) that satisfies the congruence. For example, a graph of  \(y\equiv x^2\pmod 5\) would consist of the points (0,0), (1,1), (2,4), (3,4), and (4,1).

The graph of 

\(3x\equiv 4y-1 \pmod{35}\)

has a single x-intercept \((x_0,0)\) and a single y-intercept \((0, y_0)\), where \(0\le x_0,y_0<35\).

What is the value of \(x_0+y_0\)?

 Jul 24, 2019
 #1
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"Modulo \(m\) graph paper" consists of a grid of \(m^2\) points, representing all pairs of integer residues \((x,y)\) where \((0\le x)\) .
To graph a congruence on modulo m graph paper, we mark every point \((x,y)\) that satisfies the congruence.
For example, a graph of \(y\equiv x^2\pmod 5\)  would consist of the points \((0,0),\ (1,1),\ (2,4),\ (3,4),\ \text{and}\ (4,1)\).

 

The graph of
\(3x\equiv 4y-1 \pmod{35}\)
has a single x-intercept \((x_0,0)\)  and a single y-intercept \((0, y_0)\), where \(0\le x_0,y_0<35\).

 

What is the value of \(x_0+y_0\)?

 

\(\begin{array}{|lrclrcl|} \hline P(x_0,\ 0): & 3x &\equiv& 4y-1 \pmod{35} \\ & 3x_0 &\equiv& 4\cdot 0 -1 \pmod{35} \\ & 3x_0 &\equiv& -1 \pmod{35} \\\\ & 3x_0 &=& -1 +35n \quad n\in \mathbb{Z} \\ & x_0 &=& \dfrac{-1 +35n} {3} \\ & x_0 &=& \dfrac{-1 +36n-n} {3} \\ & x_0 &=& 12n-\underbrace{\dfrac{1+n} {3}}_{=a}\quad a\in \mathbb{Z} \\ & x_0 &=& 12n-a & a&=& \dfrac{1+n} {3} \\ & & & & 3a&=& 1+n \\ & & & &\mathbf{ n}&=&\mathbf{ 3a-1} \\ & x_0 &=& 12(3a-1)-a \\ & x_0 &=& 36a-12-a \\ & x_0 &=& -12+35a \\\\ & x_0&\equiv& -12 \pmod{35} \\ & x_0&\equiv& 35 -12 \pmod{35} \\ & x_0&\equiv& 23 \pmod{35} \\ & \mathbf{x_0}&=& \mathbf{23} \qquad 0\leq x_0 < 35\\ \hline \end{array}\)

\(\begin{array}{|lrclrcl|} \hline P(0,\ y_0): & 3x &\equiv& 4y-1 \pmod{35} \\ & 0 &\equiv& 4y_0 -1 \pmod{35} \quad | \quad \cdot (-1) \\ & 0 &\equiv& -4y_0 +1 \pmod{35} \\ \\ & 0 &=& -4y_0 +1 +35n \quad n\in \mathbb{Z} \quad | \quad +4y_0 \\ & 4y_0 &=& 1 +35n \quad | \quad : 4 \\ & y_0 &=& \dfrac{1 +35n} {4} \\ & y_0 &=& \dfrac{1 +36n-n} {4} \\ & y_0 &=& 9n+\underbrace{\dfrac{1-n} {4}}_{=a}\quad a\in \mathbb{Z} \\ & y_0 &=& 9n+a & a&=& \dfrac{1-n} {4} \\ & & & & 4a&=& 1-n \\ & & & &\mathbf{ n}&=&\mathbf{ 1-4a} \\ & y_0 &=& 9(1-4a)+a \\ & y_0 &=& 9-36a+a \\ & y_0 &=& 9-35a \\\\ & y_0&\equiv& 9 \pmod{35} \\ & \mathbf{y_0}&=& \mathbf{9} \qquad 0\leq y_0 < 35\\ \hline \end{array}\)

 

\(\mathbf{x_0 + y_0} = 23+9 \mathbf{=32}\)

 

laugh

 Jul 25, 2019

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