enter image description here



Figure out which letter represents which digit
and the number of solutions possible for the following equation


Each letter represents a single digit.
Same letters represent same digits.
Different letters represent different digits.

There can be leading zeroes.


For any clarification, comment down below.

  • 4
    $\begingroup$ Hi Shivansh, did not downvote but just to point out that as nice as it is to have a swish-looking image, text in images cannot be read by screen readers and other accessibility tools. In future, maybe consider at least adding a text version alongside the image to ensure it can be read by all. Thanks :) $\endgroup$
    – Stiv
    Commented Sep 7, 2020 at 8:00
  • $\begingroup$ @Stiv Thanks for the suggestion. Actually, I created this just for fun and thought that I could post it here too. But from next time, I'll also add a text version. $\endgroup$ Commented Sep 7, 2020 at 9:03

2 Answers 2




Just adding some reasoning to Lukas Rotter's solutions

Suppose first $Z=0$.
Since we have the sum of two three-digit numbers being equal to a three-digit number, it follows that $O < 5$ and since $ZERO$ is even, $O$ must either be $2$ or $4$. Given that $ERO > ONE$, it follows that $E > O$ so if $O=2$ then $E=6$ and if $O=4$ then $E=7$.
However, looking at the sum in the hundreds place, we must have $O+O \equiv E \mod 10$ or $O+O+1 \equiv E \mod 10$, neither of which can be true in either of the above cases. Hence $Z > 0$ and automatically $Z=1$ is the only possibility because doubling a three-digit number can give you at most $1998$.

Suppose now then that $Z=1$.
If we double a three-digit number and get a four-digit number then the three-digit number must be at least $500$. Hence $O \geq 5$. Since $O$ is the last digit of the 4-digit number it must be even. Hence, $O \in \{6,8\}$.

First suppose $O=8$.
Then $E$ must be $4$ or $9$ but notice again from the $E$ in $ZERO$ that we will have $O+O \equiv E \mod 10$ or $O+O+1 \equiv E \mod 10$ neither of which can be true here. Hence, this leads to no solutions.

Now suppose $O=6$.
This means that $E$ must be $3$ or $8$. Again, looking at the the $E$ in $ZERO$, we have $O+O \equiv E \mod 10$ or $O+O+1 \equiv E \mod 10$ and this can only be true if $E=3$ and there is a $1$ carried over from the summation in the 'tens' place.
Given that $3+3=6$, we must have $N+N \equiv R \mod 10$ and $N \geq 5$ to ensure carrying over the $1$. Additionally, $N$ cannot be $6$ to clash with $O$ and $N$ cannot be $8$ because this will force $R$ to be $6$, again clashing with $O$. Hence there are three remaining possibilities which are $N \in \{5,7,9\}$.
Analysing each possibility in turn, we find that they all give rise to solutions.
$653 + 653 = 1306$
$673 + 673 = 1346$
$693 + 693 = 1386$
and this covers all possibilities


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