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**Soroban** Hello, Coach!

Sorry, your answer is incorrect . . .

There is an Amortization Formula: .$\displaystyle A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}$

. . where: .$\displaystyle \begin{Bmatrix}A & = &\text{periodic payment} \\ P &=& \text{principal} \\ i &=& \text{periodic interest rate} \\ n&=&\text{number of periods} \end{Bmatrix} $

We have: .$\displaystyle P \:=\:2,000,000,\;i \:=\:\frac{6\%}{12} \:=\:0.005,\;n \:=\:120$

Hence: .$\displaystyle A \;=\;2,000,000\cdot\frac{0.005\cdot1.005^{120}}{(1 .005)^{120}-1} \;=\;22,204.10039$

Therefore, his monthly payment will be: .$\displaystyle \$22,204.10$

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The derivation of the Formula is lengthy but quite elementary.

I think I can recreate the steps . . . if anyone is interested.