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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 < x < m. 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 == x^2 mod{5} would consist of the points (0,0), (1,1), (2,4), (3,4), and (4,1). The graph of 3x == 4y-1 mod{35} has a single x-intercept (x_0,0) and a single y-intercept (0,y_0), where 0 < x_0, y_0 < 35. What is the value of x_0+y_0?

Guest Aug 2, 2018
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heureka  Aug 2, 2018
edited by heureka  Aug 2, 2018
 #2
avatar+20033 
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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 < x < m.
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 == x^2 mod{5} would consist of the points (0,0), (1,1), (2,4), (3,4), and (4,1).
The graph of 3x == 4y-1 mod{35} has a single x-intercept (x_0,0) and a single y-intercept (0,y_0),
where 0 < x_0, y_0 < 35. What is the value of x_0+y_0?

 

\(\begin{array}{lrcl} &3x &\equiv& 4y-1 \pmod{35} \\ \text{or } & 3x &=& 4y-1 + 35n,\ \quad n\in Z \\\\ &&& \boxed{\mathbf{3x-4y-35n} \mathbf{=} \mathbf{-1} } \\ \end{array} \)

 

Solve:

\(\begin{array}{|rclrcl|} \hline 3x-4y-35n &=& -1 \quad | \quad x \text{ has the smallest coefficient} \\\\ x &=& \dfrac{-1+4y+35n}{3} \\\\ x &=& \dfrac{-1+3y+y+33n+2n}{3} \\\\ x &=& \dfrac{3y+33n+(-1+y+2n)}{3} \\\\ x &=& y+11n+\underbrace{\dfrac{ -1+y+2n }{3}}_{=a} \qquad a \in Z \\\\ && a = \dfrac{ -1+y+2n }{3} \\\\ && 3a = -1+y+2n \quad | \quad y \text{ has the smallest coefficient} \\\\ && y = 1 - 2n + 3a\\ && \qquad \qquad | \qquad \text{Since the variable n only occurs in the}\\ && \qquad \qquad | \qquad \text{integer part, set the parameter b = n. }\quad b\in Z \\\\ && \mathbf{y = 1+3a-2b} \\ && \quad \quad | \quad \text{Now on the right side of the equation there is no}\\ && \quad \quad | \quad \text{break and none of the variables any more is included.} \\ && \quad \quad | \quad \text{By inserting in reverse order, now in all equations,} \\ && \quad \quad | \quad \text{in which one variable has been isolated,} \\ && \quad \quad | \quad \text{eliminates the other variables.} \\\\ x &=& \dfrac{-1+4y+35n}{3} \quad | \quad y = 1+3a-2b,\ n=b \quad \\\\ x &=& \dfrac{-1+4(1+3a-2b)+35b}{3} \\\\ x &=& \dfrac{-1+4+12a-8b+35b}{3} \\\\ x &=& \dfrac{3+12a+27b}{3} \\\\ \mathbf{x }&\mathbf{=}& \mathbf{1+4a+9b} \\ \hline \end{array}\)

 

\(\begin{array}{|r|r|r|l|} \hline a & b & x = 1+4a+9b & y = 1+3a-2b & 3x \equiv 4y-1 \pmod{35} \\ \hline 1 & 2 & x_0 = 23 & 0 & 3\cdot 23 = 4\cdot0-1 \pmod{35} \\ & & 0\lt 23 \lt 35 \ \checkmark & & 69 = -1 \pmod{35} \ \checkmark \\ \hline 2 & -1 & 0 & y_0 = 9 & 3\cdot0 = 4\cdot 9 -1 \pmod{35} \\ & & & 0\lt 9 \lt 35 \ \checkmark & 0 = 35 \pmod{35} \\ & & & & 0 = 35-35 \pmod{35} \\ & & & & 0 = 0 \pmod{35} \ \checkmark \\ \hline \end{array}\)

 

\(x_0+y_0 = 23+9=32\)

 

laugh

heureka  Aug 3, 2018
edited by heureka  Aug 3, 2018

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