# Cryptarithmetic: 1+1=0?

Transcription:

CRYPTARITHMETIC
1+1=0?

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

ONE
+ ONE
-----
=ZERO

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

There can be leading zeroes.

SHIVANSH SHARMA


For any clarification, comment down below.

• 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 :)
– Stiv
Commented Sep 7, 2020 at 8:00
• @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. Commented Sep 7, 2020 at 9:03

## 2 Answers

$$653+653=1306$$
$$673+673=1346$$
$$693+693=1386$$

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