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# Debt Race

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

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Age 16 to 18

Challenge Level

- Problem
- Getting Started
- Student Solutions

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