heureka

delete

Find the least positive integer x that satisfies x+4609 == 2104 mod(12).

\(\begin{array}{|rcll|} \hline x+4609 &\equiv& 2104 \pmod{12} \quad & | \quad 4609 \equiv 1 \pmod{12} \quad 2104 \equiv 4 \pmod{12} \\ x+1 &\equiv& 4 \pmod{12} \quad & | \quad -1 \\ \mathbf{x} & \mathbf{\equiv} & \mathbf{3 \pmod{12}} \\ \hline \end{array}\)

Determine the smallest non-negative integer a that satisfies the congruences:  

a == 2 mod 3,

a == 4 mod 5,

a == 6 mod 7 

a == 8 mod 9

1.

\(\begin{array}{|rcll|} \hline & \mathbf{ a } & \mathbf{\equiv}& \mathbf{8 \pmod{9}} \\\\ \text{or} & a &=& 8+9n \\ & a &=& 8 + 3\cdot 3n \\ \text{or} & a &\equiv & 8 \pmod{3} \quad & | \quad 8 &\equiv & 2 \pmod{3}\\\\ & \mathbf{ a } & \mathbf{\equiv}& \mathbf{2 \pmod{3}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{ a } & \mathbf{\equiv}& \mathbf{4 \pmod{5}} \\ (2) & \mathbf{ a } & \mathbf{\equiv}& \mathbf{6 \pmod{7}} \\ (3) & \mathbf{ a } & \mathbf{\equiv}& \mathbf{8 \pmod{9}} \quad & | \quad \text{implicit } a \equiv 2 \pmod{3} \\ \hline \end{array}\)

Solve:

\(\begin{array}{|rcll|} \hline a &=& 4\cdot 7 \cdot 9 \cdot \frac{1}{7 \cdot 9}\pmod{5} \\ &+& 6\cdot 5 \cdot 9 \cdot \frac{1}{5 \cdot 9}\pmod{7} \\ &+& 8\cdot 5 \cdot 7 \cdot \frac{1}{5 \cdot 7}\pmod{9} \\ &+& 5\cdot 7 \cdot 9 n \qquad n\in Z \\\\ a &=& 252 \cdot \left( 63^{-1} \pmod{5} \right) \\ &+& 270 \cdot \left( 45^{-1} \pmod{7} \right) \\ &+& 280 \cdot \left( 35^{-1} \pmod{9} \right) \\ &+& 315 n \\\\ a &=& 252 \cdot \left( 63^{\phi(5)-1} \pmod{5} \right) \quad &|\quad \gcd(63,5) = 1,\ \phi(5) = 5-1=4 \\ &+& 270 \cdot \left( 45^{\phi(7)-1} \pmod{7} \right) \quad &|\quad \gcd(45,7) = 1,\ \phi(7) = 7-1=6 \\ &+& 280 \cdot \left( 35^{\phi(9)-1} \pmod{9} \right) \quad &|\quad \gcd(35,9) = 1,\ \phi(9) = 9(1-\frac13)=6 \\ &+& 315 n \\\\ a &=& 252 \cdot \left( 63^{4-1} \pmod{5} \right) \quad &|\quad \gcd(63,5) = 1,\ \phi(5) = 5-1=4 \\ &+& 270 \cdot \left( 45^{6-1} \pmod{7} \right) \quad &|\quad \gcd(45,7) = 1,\ \phi(7) = 7-1=6 \\ &+& 280 \cdot \left( 35^{6-1} \pmod{9} \right) \quad &|\quad \gcd(35,9) = 1,\ \phi(9) = 9(1-\frac13)=6 \\ &+& 315 n \\\\ a &=& 252 \cdot \left( 63^{3} \pmod{5} \right) \quad &|\quad 63^{3} \pmod{5} = 2 \pmod{5} \\ &+& 270 \cdot \left( 45^{5} \pmod{7} \right) \quad &|\quad 45^{5} \pmod{7} = 5 \pmod{7} \\ &+& 280 \cdot \left( 35^{5} \pmod{9} \right) \quad &|\quad 35^{5} \pmod{9} = 8 \pmod{9} \\ &+& 315 n \\\\ a &=& 252 \cdot 2 + 270 \cdot 5 + 280 \cdot 8 + 315 n \\ a &=& 4096 + 315 n \quad &|\quad 4096 \equiv 314 \pmod{315} \\ \mathbf{a} & \mathbf{=}&\mathbf{ 314 + 315 n \qquad n\in Z }\\ \hline \end{array}\)

 The smallest non-negative integer a is 314

The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2 + 14t - 0.4 at time t (in seconds).

As an improper fraction, for how long is the cannonball above a height of 6 meters?

\(\begin{array}{|rcll|} \hline -4.9t^2 + 14t - 0.4 &=& 6 \\ -4.9t^2 + 14t - 0.4 - 6 &=& 0 \quad & | \quad \\ -4.9t^2 + 14t - 6.4 &=& 0 \quad & | \quad \cdot (-1) \\ 4.9t^2 - 14t + 6.4 &=& 0 \\\\ t &=& \dfrac{14\pm \sqrt{14^2-4\cdot 4.9\cdot 6.4 } } { 2\cdot 4.9} \\\\ &=& \dfrac{14\pm \sqrt{196-125.44 } } {9.8} \\\\ &=& \dfrac{14\pm \sqrt{70.56} } {9.8} \\\\ &=& \dfrac{14\pm 8.4 } {9.8} \\\\ t_1 &=& \dfrac{14- 8.4 } {9.8} \\\\ \mathbf{t_1} & \mathbf{=} & \mathbf{ \dfrac{5.6 } {9.8} } \\\\ t_2 &=& \dfrac{14+ 8.4 } {9.8} \\\\ \mathbf{t_2} & \mathbf{=} & \mathbf{\dfrac{22.4} {9.8}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t_2-t_1 &=& \dfrac{22.4} {9.8} - \dfrac{5.6 } {9.8} \\\\ &=& \dfrac{22.4-5.6} {9.8} \\\\ &=& \dfrac{16.8} {9.8} \\\\ &=& 1.71428571429\ \text{seconds} \\ \hline \end{array} \)

The cannonball is \(\approx 1.7\ \text{seconds} = (\dfrac{16.8} {9.8}\ \text{seconds})\)above a height of 6 meters?

delete

The mean and range of ten consecutive terms in an arithmetic progression are all 82.
What is the first term in the sequence?

\(\text{Let $a$ is the first term} \\ \text{Let $b$ is the last term}\)

\(\begin{array}{|lrcll|} \hline \text{mean:} & \dfrac{a+b}{2} &=& 82 \quad & | \quad \cdot 2 \\\\ & a+b &=& 2\cdot 82 \qquad (1) \\ \\ \text{range:} & b-a &=& 82 \qquad (2) \\ \hline \\ (1)-(2):& (a+b) - (b-a) &=& 2\cdot 82 - 82 \\ & a+b - b+a &=& 82 \\ & a+a &=& 82 \\ & 2a &=& 82 \quad & | \quad : 2 \\ & \mathbf{ a } & \mathbf{=} & \mathbf{41} \\ \hline \end{array}\)

The first term in the sequence 41

The solution of 8x+1 == 5 mod 12 is x == a mod m for some positive integers m >= 2 and a

\(\begin{array}{lrcl} &8x+1 &\equiv& 5 \pmod{12} \\ \text{or } & 8x+1 &=& 5 + 12n,\ \quad n\in Z \\\\ & 8x+1 &=& 5 + 12n \quad & | \quad -1 \\ & 8x &=& 4 + 12n \quad & | \quad :4 \\ & 2x &=& 1 + 3n \\ &&& \boxed{\mathbf{2x-3n} \mathbf{=} \mathbf{1} } \\ \end{array} \)

\(\begin{array}{|rclrcl|} \hline 2x-3n &=& 1 \quad | \quad x \text{ has the smallest coefficient} \\\\ 2x &=& 1+3n \\\\ \mathbf{x} & \mathbf{=}& \mathbf{\dfrac{1+3n}{2}} \\\\ x &=& \dfrac{1+2n+n}{2} \\\\ x &=& \dfrac{2n+(1+n)}{2} \\\\ x &=& n+\underbrace{\dfrac{ 1+n }{2}}_{=b} \qquad a \in Z \\\\ && b = \dfrac{ 1+n }{2} \\\\ && 2b = 1+n \quad | \quad n \text{ has the smallest coefficient} \\\\ && n = -1+2b \\ && \quad \quad | \quad \text{Now on the right side of the equation there is no}\\ && \quad \quad | \quad \text{break and none of the variables any more is included.} \\ && \quad \quad | \quad \text{By inserting in reverse order, now in all equations,} \\ && \quad \quad | \quad \text{in which one variable has been isolated,} \\ && \quad \quad | \quad \text{eliminates the other variables.} \\\\ x &=& \dfrac{1+3n}{2} \quad | \quad n = -1+2b \\\\ x &=& \dfrac{1+3(-1+2b)}{2} \\\\ x &=& \dfrac{1-3+6b }{2} \\\\ x &=& \dfrac{-2+6b}{2} \\\\ \mathbf{x }&\mathbf{=}& \mathbf{-1+3b} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x & = & \underbrace{ -1 }_{=a} + \underbrace{3b}_{=m} \\\\ && \text{The positive integer $ m\ge 2$ , is $3$ or multiples of $3$}. \\ && \mathbf{m= 3,6,9,12,\ldots} \text{ and } \mathbf{a=-1} \\ \hline \end{array} \)

\(\begin{array}{|r|c|c|c|c|c|} \hline b & m=3b & m\ge 2 & x = -1+m & 8x+1 \equiv 5 \pmod {12} & x \equiv -1 \pmod{m} \\ \hline 1 & 3 & \checkmark & 2 & 17 \equiv 5 \pmod {12}\ \checkmark & 2 \equiv -1 \pmod {3}\ \checkmark \\ \hline 2 & 6 & \checkmark & 5 & 41 \equiv 5 \pmod {12}\ \checkmark & 5 \equiv -1 \pmod {3}\ \checkmark \\ \hline 3 & 9 & \checkmark & 8 & 65 \equiv 5 \pmod {12}\ \checkmark & 8 \equiv -1 \pmod {3}\ \checkmark \\ \hline \ldots & \ldots &\ldots & \\ \hline \end{array}\)

"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:

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

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

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

delete

delete