Distance rate time problem of two mice Mouse A and Mouse B are separated by a distance of 1.62 meters underground. They decide to meet by digging all the way through. Mouse A will double his speed every day, that is, he starts to dig 2 cm the first day, 4 cm the second day, and so on. Mouse B will dig at a constant speed of 6 cm/day. How many cm will Mouse A have dug when they finally meet?
I'm not sure how to approach this question in the easiest/fastest method. Should I write a distance rate time equation? Or is there another way?
 A: Hint:  You should write an equation for the total distance dug after $n$ days.  Do you know how to sum the geometric series of A's digging?  Add that to B's digging.  Then you just need to find an $n$ that totals 1.62 meters (watch the units).  It will be small enough you can search by hand.
A: Hint: $M_1 = 2 + 4 + 8 + 16 + \dots$ How can each term be expressed in terms of the day ($i$) , what is the terms for the $n^{th}$ day and what is the sum of $n$ terms?
A: For this problem, there are multiple factors that can help figure out the answer.
Firstly,

 Because the problem is small enough, it works to calculate the progress one day at a time. For example: on day $1$, Mouse $A$ travel $2cm$, and Mouse $B$ travels $6cm$, shrinking the gap between them to $1.54m$ or $154cm$. This can be continued until they meet.

There is also a way to calculate the sum of Mouse $A$'s total digging:

 As Mouse $A$ travels $2cm$ the first day, and then twice as far every day after, we can construct a geometric series and then a geometric sum to determine how far in total Mouse $A$ has traveled. Because Mouse $A$ goes $2cm$ on the first day, $f(1)=2$, and because the distance doubles daily, $f(n)=2f(n-1)$. This information allows us to make the geometric sum of $f(n)=2^{n+1}-2$. Adding Mouse $B$'s travel distance gives us $f(n)=2^{n+1}+6n-2$ We can now calculate the total distance traveled after any number of days $n$.

Using the information we have gathered so far, the rest of the calculation is quite simple:

 Plugging in 162 for $f(n)$ gives us $162=2^{n+1}+6n-2\ \rightarrow\ 164=2(2^n)+6n\ \rightarrow\ 82=2^n +3n\ \rightarrow\ n=6.$  For how to solve  $82=2^n+3n$: https://www.symbolab.com/solver/step-by-step/82%3D%5Cleft(2%5E%7Bn%7D%5Cright)%2B3n

This problem could also be solved one day at a time as follows:

$Day\ 1:\ Mouse\ A\ goes\ 2cm,\ Mouse\ B\ goes\ 6cm,\ the\ mice\ are\ 154cm\ apart.$
$Day\ 2:\ A=4,\ B=6,\ D=144$
$Day\ 3:\ A=8,\ B=6,\ D=130$
$Day\ 4:\ A=16,\ B=6,\ D=108$
$Day\ 5:\ A=32,\ B=6,\ D=70$
$Day\ 6:\ A=64,\ B=6,\ D=0$

