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'Debt Race' printed from https://nrich.maths.org/
To solve this problem we must first set up a linear difference
equation and then solve the equation to find n when the debt
remaining = 0.
$A_n$ =amount owed in the nth year of repayment
I = Interest rate
L= Lump sum payed annually
In general $A_n= (1+I)A_{n-1}- L$
$A_n - (1+I)A_{n-1} = - L$
Set the right hand side = 0 and solve for the complementary
function.
$z - (1+ I) = 0$
$z = 1+I$
$A_{CF} = C(1+I)^{n}$
Now solve for the particular integral:
Let $A_n = K$ then $A_{n-1}$ also = K
Substituting into the general form we find:
$K -(1+I)K =-L \to K = \frac{L}{I}$
General solution:
$A_n = A_{CF} + A_{PI}= C(1+I)^n + \frac{L}{I}$
Now use the boundary condition that at n = 0 , $A_n =
100,000$
We find that the constant $c = (10^5 - \frac{L}{I})$
Therefore $A_n = (10^5 - \frac{L}{I}) (I +1)^n +\frac{L}{I}$
If we now set $A_n$ = 0 and solve for n we find:
$n =\frac{log(\frac{-LI^{-1}}{10^5 - LI^{-1}}) }{log(1+I)}$
If we now substitute the values of I and L for each person given in
the intitial question we find:
Person A: n=9.73years
Person B: n= 9.60 years
Person C: n= 9.59 years
Person D: n= 9.69 years
Person E: n= 9.92 years