A teller pays 43.34 on a a check for 34.43, and so be short 8.91 at the end of the day. Prove that such error always leads to an outage,in pennies, which is divisible by 9.

  • We are currently studying modular arithmetic and congruence and I came across this proof on my study set that I'm really not sure how to approach. We have gone over equivalence relations in mod arithmetic, equivalence classes (complete set of residues), and have gone through several mod arithmetic examples w/o variables. From our notes I know that a property for divisibility by 9 is that the sum of the digits is divisible by 9, although I am still not sure how to formulate a proof around this, any help is appreciated.
  • $\begingroup$ What is an outage in pennies? $\endgroup$ – siegehalver Feb 20 '16 at 20:42
  • $\begingroup$ Divisibility by 9? sum of the digits divisible by 11? No, my friend! $\endgroup$ – Max Feb 20 '16 at 20:43
  • $\begingroup$ A number is divisible by 9 if the sum of the digits is divisible by 9. $\endgroup$ – cpiegore Feb 20 '16 at 20:47
  • $\begingroup$ It is a bit silly to pose this question without saying what you mean by "such error". As it is, the question is unanswerable. $\endgroup$ – TonyK Feb 20 '16 at 21:22

I think you are trying to show that the difference between numbers abba and baab where a and b are digits is divisible by 9.

The first number is 1000a + 100b + 10b + a, The second is 1000b + 100a +10a +b

The difference is 1001a + 110b - 1001b - 110a = 891 (a-b). Since 891 is divisible by 9, the whole product is also divisible by 9.


Note that $\$43.34$ and $\$34.43$ are re-arrangements of each other; a permutation of digits. Note also that they each correspond to $4334$ and $3443$ pennies respectively.

This is actually an example of a "magic trick" I like to perform for classes on a random day throughout the semester (I cannot take credit for its creation! I do not know the original source...).

"Magic" trick:

  1. Pick a four digit number with at least two distinct digits (four is completely arbitrary).

  2. Rearrange this number any way you like, swapping at least one pair of digits (it can be any non-identity permutation).

  3. Subtract the smaller of these two numbers from the larger, the original and mixed-up. For example, if you pick $3572$ as your number and mix it up to form $7235$, you'll subtract $7325 - 3572 = 3753$. (It actually doesn't need to be $\text{larger} - \text{smaller}$, but negatives confuse the class, even though they don't matter mathematically)

  4. Now take the result of the subtraction, and keep one of its nonzero digits a secret. Tell me the other digits in any order you like. (I tell the class that picking $0$ is "too boring!" After reading and seeing how this trick works, think about why I won't let them pick $0$.)

I will know your secret digit by the time you've told me your last digit. How can I possibly do that?

"Magic" Revealed:

Say your four digit number is $d_3d_2d_1d_0$, which is of course the decimal representation for $d_3 \cdot 10^3 + d_2 \cdot 10^2 + d_1 \cdot 10^1 + d_0$. The permuted version has simply shuffled the digits around; say to $d_i \cdot 10^3 + d_j \cdot 10^2 + d_k \cdot 10^1 + d_\ell$.

Now, the subtraction looks like $(d_3 - d_i) \cdot 10^3 + (d_2 - d_j) \cdot 10^2 + (d_1 - d_k) \cdot 10^1 + (d_0 - d_\ell),$ even though this might not be a base-$10$ representation (if any of the subtractions are negative).

Since each power of $10$ is congruent to $1$ modulo $9$ (make sure you know why $10^k \equiv 1 \pmod {9}$), this just becomes $(d_3 - d_i) + (d_2 - d_j) + (d_1 - d_k) + (d_0 - d_\ell) \bmod 9$. But the various $d$'s have just been permuted, so of course $(d_3 + d_2 + d_1 + d_0) - (d_i + d_j + d_k + d_\ell) = 0$. So our sum is congruent to $0$ modulo $9$, and hence must be a multiple of $9$. In other words, the result of the subtraction must be a multiple of $9$, and so its digits must, if we repeatedly sum, add up to $9$, which lets me "magically" know the digit that was kept secret.

Can you see how this kind of reasoning applies to your situation, so that "such an error" always winds up accumulating errors that are a multiple of $9$ pennies?


You can think of 43.34 as 4334 pennies and 34.43 as 3443 pennies. The difference 891 is divisible by 9 since 8+9+1 = 18 which is divisible by 9.

  • $\begingroup$ By the way 891 IS also divisible by 11 but this is irrelevant because your problem is about divisibility by 9 $\endgroup$ – cpiegore Feb 20 '16 at 21:03

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