Question 662277
DISCLAIMER: I am not into finance. I am just a math/science-loving old chemist.
 
I agree that to calculate the interest accrued over 1 day, you would multiply times {{{0.18/365}}},
and that to calculate the balance after 1 day, you would multiply times {{{1+0.18/365}}},
and that to calculate the balance after 1 month, you would multiply times {{{(1+0.18/365)^(365/12)}}}.
{{{(1+0.18/365)^(365/12)=1.01511}}} (conveniently rounded) as per my computer.
The monthly interest rate would be 0.01511 (as a decimal), or 1.511% per month.
 
On my computer, using Excel, the function
PMT(0,01511,60,8000)=$203.72
calculates the monthly payment needed to pay off
$8000, in 60 months, at an interest of 1.511%,
making payments at the end of each month.
 
Bank people probably have even easier ways to calculate it,
but I like to understand the algebra in it,
and here is what I understand:
One month after starting the loan,
with the interest accrued, the balance would be
${{{8000*1.01511}}}, but at that point Joe pays ${{{x}}},
and the new balance is ${{{8000*1.01511-x}}}
A month later, the same thing happens, and the balance becomes
${{{(8000*1.01511-x)*1.0511-x=8000*1.01511^2-1.01511*x-x}}}
Calling {{{1.01511=R}}} to save ink,
the balance is ${{{8000R^2-xR-x}}} after 2 months,
${{{8000R^3-xR^2-xR-x}}} after 3 months,
${{{8000R^4-xR^3-xR^2-xR-x}}} after 4 months, and so on.
After 5 years (60 months), the balance is
${{{8000R^60-xR^59-xR^58-xR^57}}}.........{{{-xR^2-xR-x}}}=${{{0}}}
So, {{{8000R^60=x}}}({{{R^59+R^58+R^57}}}.........{{{+R^2+R+1)}}})
It is not hard to prove that the sum in brackets can be calculated as
{{{R^59+R^58+R^57}}}.........{{{+R^2+R+1=(R^60-1)/(R-1)}}}
So we can write the equation above, using the simpler sum calculation, as
{{{8000R^60=x(R^60-1)/(R-1)}}}
And now we can solve for {{{x}}}
{{{8000R^60=x(R^60-1)/(R-1)}}} --> {{{x=8000R^60(R-1)/(R^60-1)}}}
The computer calculated for me
{{{R^60=1.01511^60=2.45906}}}  (conveniently rounded),
{{{R^60-1=1.01511^60-1=2.45906-1=1.45906}}}  (conveniently rounded),
and we know {{{R-1=1.01511=0.01511}}}
So the good enough approximation is {{{x=8000*2.45906*0.1511/1.45906=203.73}}}
(Using {{{R=0.015}}} yields {{{x=203.15}}}, which is less accurate).