2
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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

multiplication

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.

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  • 3
    $\begingroup$ Can I use a computer to type my answer? :O $\endgroup$ – Ian MacDonald Jun 14 '15 at 3:19
3
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$S \times A$ 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 $51011975$.

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

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