I am in the process of solving a warm-up problem (not graded) for a course I'm hoping to self-study this semester. I believe I have a solution sketch, and several older posts like here and here were helpful. But there is a crucial step to the argument that I do not understand. The problem is as follows.
At his death, a millionaire left his 10 children a million dollars in cash, all in $100, $10, $1 bills, 10-cent and 1-cent coins. Show that there is a way for them to split the fortune into ten stacks of equal value.
The starting point for me was to write let $a_{100}$ be the number of 100 dollar bills, $a_{10}$ the number of 10 dollar bills, $a_1$ the number of 1-dollar bills, $a_{0.10}$ the number of 10-cent coins, and $a_{.01}$ the number of 1 cent-coins. I'm not given the exact denominations that the inheritance is currently in, so I think these components are fixed at the start. So my starting point is: $$ 10^6 = a_{100} (100) + a_{10} (10) + a_{1} (1) + a_{0.10} (0.10) + a_{0.01} (0.01). $$ At this moment, my unit is dollars, but I can multiply through by $100$ to express this entire equation in cents which makes the coefficients easier to work with. $$ 10^8 = a_{100} (10^4) + a_{10} (10^3) + a_{1} (10^2) + a_{0.10} (10) + a_{0.01} (1) $$ The left-hand side is divisible by $10$, and the first four terms involve powers of $10$, so they are also divisible by $10$. If I subtract two quantities equivalent to $0$ mod $10$, I get another equivalent to $0$ mod $10$. So this means that $a_{0.01} \equiv 0 \bmod {10}$.
The number of 1-cent coins is a multiple of 10, so I can confidential assert that $\frac{a_{0.01}}{10}$ is an integer. What I want to say is, "without loss of generality, exchange ten 1-cent coins for a single 10-cent coin." By doing that, I will have $10\left(a_{10} + \frac{a_{0.01}}{10}\right)$ coins. Call this quantity $\lambda$. If I divide the above equation by $10$ I get: $$ 10^7 = a_{100} (10^3) + a_{10} (10^2) + a_{1} (10) + \lambda. $$ The left-hand side is divisible by 10, and the first three terms on the right-hand side are divisible by 10, so $\lambda$ is also divisible by 10 by the same argument. So I could continue by exchanging 10-cent coins for a commensurate number of dollars, showing that the number of dollar bills is divisible by 10, and the same for 10 dollar bills and 100 dollar bills.
What I do not fully understand is why this works. Why can I say this "without loss of generality"? My understanding is that I can invoke this in a case where I can reduce a complicated problem to a simpler one, and by solving a simpler one, I've also solved the more complicated one. My instinct was, after determining that I had a multiple of 10 of the 1-cent coins, to subtract this quantity from both sides because my left-hand side would still be divisible by 10. I could then repeat the same analysis, but with messier numbers. Is this the reason? Or is the argument that I can "work backwards" and recreate the original quantities of each denomination? I'm not able to see exactly how this would work.
