Coinciding Comets

The periods of the comets can be expressed as a system of linear congruences.

$\begin{array}{ll} \text{2P/Encke:} & x \equiv 2017 \pmod{3} \\ \text{4P/Faye:} & x \equiv 2014 \pmod{8} \\ \text{8P/Tuttle:} & x \equiv 2008 \pmod{13}. \end{array}$

This reduces to the following system of congruences:

$\begin{aligned} x &\equiv 1 \pmod{3} \\ x &\equiv 6 \pmod{8} \\ x &\equiv 6 \pmod{13}. \end{aligned}$

Since the last two congruences have the same value, they can be simplified to the following congruence with modulus $8\times 13=104:$

$x \equiv 6 \pmod{104}.$

Then, express this as an equation, and substitute it into the first congruence:

$\begin{aligned} x &= 104j+6 \\ 104j+6 &\equiv 1 \pmod{3} \\ j &\equiv 2 \pmod{3} \end{aligned}$

Express this as an equation, and substitute it into the expression for $x:$

$\begin{aligned} j &= 3k+2 \\ x &=104(3k+2)+6 \\ x &= 312k+214 \\ x &\equiv 214 \pmod{312} \end{aligned}$

The next year that satisfies this congruence is $312(6)+214=\boxed{2086}.$

We have $2008+13x=2014+8y=2017+3z$ or $13x=6+8y=9+3z$ .

Since the middle term is multiple of $2$ , and the last term a multiple of $3$ , $x$ is a multiple of $6$ .

Let $x=6$ for minimum year. $78=6+8y, y=9, 78=9+3z, z=23$ .

Thus $x,y,z$ all are integers. This is our required solution. Thus year = $2008+13x=2008+78=\boxed {2086}$ .

Suppose that the three comets reach perihelion together in the year $2008+x$ . The data in the question then gives

$\begin{aligned} x &\equiv 0 \pmod{3}\\x &\equiv 6 \pmod{8}\\ x &\equiv 0 \pmod{13}\end{aligned}$

The first and last of these equations allow us to write

$x=39s \dots(1)$

Substituting this into the middle equation gives

$39s \equiv 6 \pmod{8}$

Subtract $40s$ (a multiple of 8) from both sides to get

$-s \equiv 6 \pmod{8}\\ \implies s \equiv 2 \pmod{8}$

Substitute $s=2$ in equation (1) to get $x=78$

And so the required year is $2008+78=\boxed{2086}$

We solve the following equations for $a, c$

$3a+2017=8b+2014$
$13c+2008=8b+2014$

which works out to

$a=\dfrac{1}{3} (8b-3)$
$c=\dfrac{2}{13} (4b+3)$

This means that $b$ must be a multiple of $3$ , or $b=3d$ , so that we have

$c=\dfrac{2}{13} 3(4d+1)$

or $d=3$ as the smallest possible value. This gets us $(a, b, c)=(23, 9, 6)$ , and the year when they next line up is $2086$