$A\tfrac{B}{C}=100$ is the equality where A,B,C are numbers generated by combining all digit values $0,1,2,3,4,5,6,7,8,9$ only once for each digit in total. For example;

$123\tfrac{4567}{890}$ is an example for A,B,C where $A=123$, $B=4567$, $C=890$.

is an example to assign unique digits only A,B,C values.

So how many possible different A,B,C values are there to make this equality true?

Example: If this question was asked without using $0$ digit, one of the answers would be;


Note: $01234$ and other numbers starting with $0$ cannot be accepted as a number since it is not valid 5 digit number.

Clarification: $A\tfrac{B}{C}$ is a mixed number meaning it is also equal to $A+\frac{B}{C}$.

  • $\begingroup$ Are you looking for all possible answers? If you are then that will probably require a script so the computer-puzzle tag should be added $\endgroup$ Commented May 23, 2017 at 19:41
  • $\begingroup$ @BeastlyGerbil I am not sure only script is a way to find an answer... and I know that there are not many ones. $\endgroup$
    – Oray
    Commented May 23, 2017 at 20:20
  • $\begingroup$ Is that supposed to be A*B/C or A+B/C? $\endgroup$
    – Kruga
    Commented May 24, 2017 at 7:27
  • $\begingroup$ @Kruga it is called "mixed number". That means A+B/C. $\endgroup$
    – Oray
    Commented May 24, 2017 at 11:21

1 Answer 1


Computer-assisted answer

How many digits does A have?

Clearly no more than 2, since then A+B/C has to be strictly bigger than 100. Clearly no fewer than 1, since there are no zero-digit numbers. Can A have just one digit? If so, then B/C (made from nine digits) has a value somewhere between 90 and 100. Clearly then B must be longer than C; it must be longer by an odd number of digits. If it's 3 or more digits longer than B/C is at least 102345/987>100, contradiction. But B could be just one digit longer; e.g., 4+98765/1023 is slightly >100. OK, so A can have just one digit; in that case it's clear that B/C is 9xxxx/1xxx because otherwise B/C<90 and is too small to reach 100 on adding a single digit.

On the other hand:

Can A have two digits? If so, then B/C (made from eight digits) has a value somewhere between 1 and 90. This is certainly possible if B,C have four digits each yielding a ratio of at most 9876/1023~=9.6. In this case A must actually begin with 9 because B/C is <10, so actually B/C is at most 8765/1023~=8.6. Alternatively, B could be two digits longer than C (note that the length difference must be an even number of digits) in which case it's at least 10234/987 which is a little bigger than 10. Nothing problematic or even difficult there. But clearly B can't be four or more digits longer than C.


the possible patterns are x + 9xxxx/1xxx, 9x+xxxx/xxxx, and xx+xxxxx/xxx.


in each case the fraction has to end up equalling an integer so e.g. in the first case we need 1xxx to divide 9xxxx. (Of course in that case the ratio has to be one of 92, ..., 99.)

This is all very well but doesn't seem like it constrains things enough for non-computerized searching to be bearable.

So, now that the computer-puzzle tag was added, I did the computer searching. I believe the complete list of solutions is


which raises the question of

whether there's a nice accessible proof that only this "shape" of answer can work.

  • $\begingroup$ You may want to add something for $0$ since some digits cannot be 0, for example you cannot have x0+xxxxx/xxx since xxxxx/xxx or any kind combination cannot be y0. $\endgroup$
    – Oray
    Commented May 24, 2017 at 11:57
  • $\begingroup$ 9x +xxxx/xxxx is really a case? $\endgroup$
    – Oray
    Commented May 24, 2017 at 12:17
  • $\begingroup$ Prima facie, yes. E.g., 98+1023/7654 < 100 < 98+7654/1023 so the sort of simple inequality-based reasoning used here can't rule out answers of that shape. $\endgroup$
    – Gareth McCaughan
    Commented May 24, 2017 at 12:19
  • $\begingroup$ ("here" = in my answer as it currently stands.) $\endgroup$
    – Gareth McCaughan
    Commented May 24, 2017 at 12:19
  • $\begingroup$ I believe you need to reconsider that again since, 90 is impossible as I said, for 91 and 92, you need to find 9 which is also impossible without 9 on the right side. (8034/1000=8.034), 99 impossible since you cannot have two same number B and C to make 99+1. $\endgroup$
    – Oray
    Commented May 24, 2017 at 12:25

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