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I am trying to solve this below problem:

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. (Note that this would not be true if there were $\$3$ bills).

So the best I can do is set this up as solving an equation in five variables. I want to find $a,b,c,d,e$ so that $$ 10^6 = a \cdot 100 + b \cdot 10 + c \cdot 1 + d \cdot 0.1 + e \cdot .01. $$ I can try to take congruences mod $10$. The LHS is clearly $0$, and any multiple of $10$ is certainly $0$, but $\frac{1}{10}$ is not even defined in $\mathbb{Z}/10\mathbb{Z}$, since that would amount to division by zero, so this strategy does not work.

Can anyone give me a hint on how to get started?

Mathematical Rookie
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    If you don't have integers for all your coefficients, multiply by powers of ten until they are. – Nij Aug 22 '21 at 00:53
  • Note that this will also be possible if some of the money is in $1,000 and in $10,000 and in $100,000 bills. – Hans Engler Aug 22 '21 at 01:03

1 Answers1

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First, prove that the number of 1-cent coins is divisible by 10. This follows from the fact that "cents place" of \$1 million is zero. Then, WLOG, exchange the 1-cent coins for an equivalent number of 10-cent coins.

Then, prove that the number of 10-cent coins is divisible by 10. This follows from the fact that there are no 1-cent coins, and the "10-cent place" of \$1 million is zero. Then, exchange the 10-cent coins for an equivalent number of \$1 coins.

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Then, prove there must be a multiple of 10 \$100 bills. This is because all the currency now is in \$100 bills.

Mark Saving
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  • I don't know how I would go about proving any of these, though. Would you mind walking me through one of them? – Mathematical Rookie Aug 22 '21 at 00:59
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    @HansEngler You're right, the trick is that you have to trade in the pennies for dimes after proving there's a multiple of ten of them, then trade in the dimes for $ bills after proving there's a multiple of 10 of them, etc. I forgot to add that part. – Mark Saving Aug 22 '21 at 01:04
  • So this proof sketch shows that a division into stacks of equal value would also be possible if there were 10,000 heirs :) – Hans Engler Aug 22 '21 at 01:07
  • @HansEngler Correct, because we could always amalgamate the fortune into piles of exactly $100 – Mark Saving Aug 22 '21 at 01:07
  • @MarkSaving I'm still not fully sure that I understand. I suppose the idea is that because $10^6$ ends in a $0$, if I had a number of pennies not a multiple of $10$, I'd have some "change" modulo $10$, and adding any multiple of either dimes, dollars, and so forth, also divisible by $10$ in cents, wouldn't recover this. The intuition is ok, but is there a way to formalize that? I feel that my understanding is a bit shaky. – Mathematical Rookie Aug 22 '21 at 01:35
  • @AlgebraMaster To make things more precise, we will show that for all n from 0 to 5, we can arrange the money into a collection of $n$-piles, where an $n$-pile is a collection of money work $10^m$ cents for some $m$ such that $n <= m <= 5$. We do this by induction. The money is already arranged into $0$-piles initially, so the base case is covered. For the inductive step, suppose we have arranged the money into $n$-piles. The total amount of money is 0 mod $10^{n+1}$ cents, so there must be a multiple of 10 piles with exactly $10^n$ cents in value. We can group these into groups of 10. – Mark Saving Aug 22 '21 at 04:12