Foretelling The Next Cicada Swarm

First note that both broods concur in $1998$ , because $2011-13=1998$ . Then by the Chinese Remainder Theorem, the next solution to this system of modular equations, i.e. when their lifecycles will next intersect, is $1998+13\cdot 17 = 2219$ , which is $2219-2014=205$ years from now.

Note that $2011$ is $13$ years from $1998$ , so to find when the $17$ year cycle meets the $13$ year cycle we can start from $1998$ .

$\text{lcm}(17,13)=221$ , so in $221$ years from $1998$ the cycles will meet. This is equivalent to $\boxed{205}$ years from $2014$ .

I am not very good at number theory so I solved it using some logical reasoning. Also I was forced to solve it because I wanted that disgusting image disappear from my homepage. :D

Let $x_{1}$ and $x_{2}$ denote the years of emergence of two broods. Then we have,

$x_{1}=1998+17n_{1}$

$x_{2}=2011+13n_{2}$

where $n_{1}$ and $n_{2}$ are integers. We want smallest positive integer solutions for $n_{1}$ and $n_{2}$ such that

$x_{1}=x_{2}$

This gives,

$17n_{1}-13n_{2}=13$

Now by inspection we see that one solution in integers is $n_{1}=0$ and $n_{2}=-1$ .

Now to get positive solution, we should increase $n_{1}$ by certain amount. But this amount should be properly chosen so as to get integer solutions. Let $d$ be this increment. Then to keep the balance, we have to change $n_{2}$ also properly.

For example, let $t_{1}$ and $t_{2}$ be one particular solution of the equation $ax+by=c$ . Then if we increase $t_{1}$ by $d$ , then we must decrease $t_{2}$ by $\frac{ad}{b}$ . Thus to get integer solutions, $\frac{ad}{b}$ should be an integer. In our case, $a$ and $b$ are relatively prime so we must have smallest possible $d$ equal to $13$ which gives $n_{1}=13$ and $n_{2}=16$ . This gives $x_{1}=x_{2}=2219$ . This means it will take $2219-2014=\boxed{205}$ years more for both broods emerge together.

Keep on adding 17 to 1998 and 13 to 2011 and check when the two numbers are same and they are same at 2219 ,so subtract 2014 from 2219 , you get 205.

The last time the two broods emerged together was 1998 (2011-13=1998). So the next time they would meet would be the value of the lowest common multiple of 17 and 13 in years. As 17 and 13 are prime numbers the lowest common multiple is 17 x 13 which equals 221, so the next time the broods emerged together would be 221 years on from 1998. This means they emerge in 2219. 2219 is 205 years on from 2014 (2219 - 2014 = 205). So the answer is 205 .

For the premises of this problem, we will be counting centered at the year $2000$ . This means that the year $1998$ will be represented by $-2$ , and $2011$ will be represented by $11$ . We can represent the life cycles of each brood of cicadas with a relationship that relates the cycle number ( $x$ or $y$ ) to the year .

The first brood: $(-2, 15, 32, 49, 66 ...)$ -> $(17x - 19)$ , where $x$ is the cycle number.

The second brood: $(11, 24, 37, 50, 63 ...)$ -> $(13y - 2)$ , where $y$ is the cycle number.

We can now set the two expressions equal to each other and manipulate them to come up with a solution:

$17x - 19 = 13y - 2$

Rearranging, we get:

$x = \frac{13y}{17} + 1$

This yields that the smallest value for $y$ such that the two expressions are equal is $y = 17$ , resulting in $x = 14$ .

Substituting the value, $x = 14$ into the expression, $17x - 19$ , we get that the first year that the two broods will emerge together is the year $2219$ .

Therefore, the broods will emerge $\boxed{205}$ years from now.

first brood has his first cycle in 2015, and second started in 2011, so there is 4 years gap, we have to cover this gap to make their cycle in same year, Now LCM of 13,17 is 17*13=221, we have to make 4 years to cover the gap so by hit n trial if we subtract each integer from their LCM it will give us the desired result... i.e. 221-17=204, 221-13=208, 208-204=4, so this 4 year gap will be coverd after 204 years of 2015 means in 2219 i.e after 205 years of 2014...

Its awesomely easy. Just notice the dates. The 2nd brood last appeared in 2011. So when did it second last appear. Yups on 1998! The same as the last time the 1st brood appeared.So now we have a common starting point. Now just HCF of 17 and 13 = 221. But we have to find no. of years from now (2014). So no. of years= [221- (2014-1998)] =221-16=205. 205 is the answer. Hurray!

$\DeclareMathOperator{\lcm}{lcm}$ The year 2011 is 13 years after 1998. Call 1998 year 0. The cycles meet at year 0; our goal is to find the next year during which they meet. Mathematically, this is equivalent to $lcm(13,17)$ . 13 and 17 are both prime, so the smallest integer solution is $13 \times 17 = 221$ years after year 0. Finally, we calculate how many years after 2014 this year occurs. $2014 - 1998 = 16$ , and $221 - 16 = \boxed{205}.$

After reading the second paragraph again you will see that the first brood appeared in 1998 and appears every 17 years and the second brood appeared in 2011 and appears every 13 years. this means that the 2nd brood also appeared in 1998!!

2011-13=1998

See!! So actually they will appear 17x13 (which is 221) year later together BUT from 1998. And it is 2014 (16 years from 1998).

So the 2 broods will appear after 221-16=205 years.

Hence the answer is 205!!!!

Both broods emerged before in 1998........take lcm of 13 & 17...it will be 221 & now it is 2014,.........so 2014 subtracted by 1998 is 16 & 221 minus 16 is 205.... so answer is 205

1998+17n = 2011+13m 17n=13(m+1) m=16, n=13 Hence no of years from 2014 = 13*m - 3 = 205

For the broods to emerge simultaneously, it must be that $2011+13x = 1998+17y$ for integers $x, y$ from the given data. Rearranging this equation, we obtain $13(x+1) = 17y$ . This is a simple Diophantine equation with solutions of the form $x = 17k+16, y = 13k+13$ for some positive integer $k$ . To find when the broods first meet, set $k = 0$ . This gives $x = 16, y = 13$ . Going back to the original equation, we see that $2011+13 \cdot 16 = 1998 + 17 \cdot 13 = 2219$ . $2219-2014 = 205$ , so the answer is $\boxed{205}$ years.