# Reconstruct this multiplication to find the combination for the safe

You've solved it! No, wait a minute - you've solved the first step. You don't know the combination for the safe yet. All you know is that someone has helped you by using a pseudonym. Their name isn't really Artairs A Sebeenas. But if you write it like this you get a multiplication, and the product represented by "SEBEENAS" is the 8-digit number you need.

No two letters represent the same digit. What's the combination? Please show all work, and use of a computer is not allowed.

• Can I use a computer to type my answer? :O – Ian MacDonald Jun 14 '15 at 3:19

## 1 Answer

$$S \times A$$ ends with $$S$$

Therefore, if $$S = 1,3,7$$ or $$9$$, $$A = 1$$
If $$S = 5$$, $$A = 1,3,5,7,9$$
If $$S = 2,4,6$$ or $$8$$, $$A = 1$$ or $$6$$
If $$S = 0$$, $$A$$ is anything, but note that $$S$$ cannot be equal zero since it is the start of the answer.

$$A \times A + \text{carryover} = 10S + E$$

If $$S = 1,3,7,9,2,4,6$$ or $$8$$, $$A = 1$$ and $$S = 0$$, which is a contradiction.
If $$S = 2,4,6,8$$, $$A = 6$$ and $$S = 3$$ or $$4$$, which means that $$S = 4$$, $$A = 6$$
If $$S = 5$$, $$A = 1,3,5,7,9$$, and only $$A = 7$$ can achieve $$S=5$$.
Therefore, $$(S,A) = (4,6)$$ or $$(5,7)$$
Let these be Case 1 and Case 2 respectively.

Case 1

$$S = 4, A = 6$$
$$R \times A + \text{carryover}$$ ends with $$A$$

$$\text{carryover} = 2$$ (The $$2$$ in $$4 \times 6=24$$)
$$A = 6$$
$$6R + 2$$ ends with $$6$$
$$6R$$ ends with $$4$$
$$R = 4$$ (reject since $$S = 4$$) or $$9$$

$$S = 4$$, $$A = 6$$, $$R = 9$$
Now we have:
$$\text{69T6I94}$$
$$6$$
$$\text{4EBEEN64}$$
$$69 \times 6 + \text{carryover} = \text{4EB}$$

$$0 \leq \text{carryover} < 5$$
$$414 + \text{carryover} = \text{4EB}$$
$$E = 1$$

$$S = 4, A = 6, R = 9, E = 1$$
Now it is:
$$\text{69T6I94}$$
$$6$$
$$\text{41B11N64}$$
$$6 \times 6 + \text{carryover}$$ (of $$6 \times I + \text{carryover}$$) ends with $$1$$

(both) $$\text{carryover} = 5$$
$$6 \times I + 5$$ starts with $$5$$
$$I = 8$$ or $$9$$ (reject since $$R = 9$$)

$$S = 4, A = 6, R = 9, E = 1, I = 8$$
$$\text{69T6894}$$
$$6$$
$$\text{41B11N64}$$
$$N = 3$$
$$\text{69T6894}$$
$$6$$
$$\text{41B11364}$$

However, $$T$$ cannot exist, since $$6 \times T + 4$$ ends with $$1$$, so $$T$$ is not an integer.

Therefore Case 1 has no solutions.

Apply the same method to Case 2

$$R = 2$$
$$E = 0$$ or $$1$$
If $$E=0$$, $$I=2$$ (reject since $$R=2$$), so no sol if $$E=0$$
If $$E=1$$, $$I = 3$$ or $$4$$
If $$I = 3$$, $$N = 2$$ (reject)
If $$I = 4$$, $$N = 9$$, $$T = 8$$

Therefore:
$$7287425$$
$$7$$
$$51011975$$

Thus, your answer is $$51011975$$.

• Yes - well done! :-) – h34 Jun 14 '15 at 10:16