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My friend from Portugal said that 2 + 2 is 8 in Portuguese. And he could prove it. I did not understand it at first. But then he showed me the following equation:

  D O I S
  D O I S
+ ________
  O I T O

Ah, now it is clear. Two is "dois" and eight is "oito" in Portuguese. And the equation above makes perfect sense. Could you solve the single one solution that explain this?

Rules: D, O, I, S and T are a number between 0 - 9. Each one is unique.

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From $S+S\equiv O\pmod{10}$, we know $O$ is even. From $O+O$, we know $O\le4$ because $2O$ doesn't carry (otherwise $O$ would be odd because $D+D$ is even). Therefore $I$ is doubly even ($I\equiv 2O\pmod{10}$ and $O\equiv 2S\pmod{10}$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=S\pm5$ (we have $D+D=S+S\equiv O \pmod{10}$), and $2D$ doesn't carry (otherwise there would be five letters in the answer), and so $S=6$ (because $O=2$, $S$ is either $1$ or $6$) and $D=1$. Finally, $T=2I+1=9$, so the sum is $1246+1246=2492$.

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  • $\begingroup$ You haven't given a value for $T$. $\endgroup$
    – ZanyG
    Mar 20 '19 at 5:54
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    $\begingroup$ »Therefore I is …,« why? how can one know that whithout looking ahead? $\endgroup$
    – aschipfl
    Mar 15 at 0:30
  • $\begingroup$ @aschipfl; $O\equiv2S\pmod{10}$ is even, and $I\equiv2O\pmod{10}$, therefore $I\equiv4S\pmod{10}$. $\endgroup$
    – JMP
    Mar 15 at 1:59
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    $\begingroup$ My point is, how can you exclude a carry to be involved in $I$ at that point already? $\endgroup$
    – aschipfl
    Mar 15 at 2:39
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    $\begingroup$ @aschipfl; $I$ is the next digit after $T$ in $I+I=T$, and the carry bit would have to roll over into it, which means that $I$ is now odd. $\endgroup$
    – JMP
    Mar 15 at 2:59
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Could it be

   1246 

 + 1246 

 ------ 

   2492

Then

D=1, O=2, I=4, S=6, T=9

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2
  • $\begingroup$ Perfect! Easy one? $\endgroup$
    – Chaotic
    Mar 20 '19 at 4:50
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    $\begingroup$ Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic! $\endgroup$
    – El-Guest
    Mar 20 '19 at 11:28
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Although this post is quite similar to JMP's answer, I still want to share it here:

4th column: $O = 2 S$, therefore $O$ is even.
1st column: $D < 5$, because $2 D < 10$.
2nd column: $2 O$ must not carry, hence $O < 5$, so either $O = 2$ or $O = 4$.
4th column: Therefore $D = O / 2$, so either $1$ or $2$.
1st column: $S = D + 5$, so $2 S$ will carry and will make $T$ odd.
3rd column: $I \neq 9$, because then $2 I + 1 = 19$, leading to duplicate $T = 9$ as well.
3rd column: $I \neq 8$, because then $2 I + 1 = 17$, and the carry would force $I$ odd.
Hence $O \neq 4$, so the only possibility left is $O = 2$.
Thus, $D = 1$, $S = D + 5 = 6$, $I = 2 O = 4$ and $T = 2 I + 1 = 9 $.

Therefore, the result is:

$1246$
$1246$
--------
$2492$

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Let's start as a basic Brute Force

Consider D as 1

 Since D and S gives a Unit Place as O

 So,S is D+5 ;S is 6

 So, Here O is 2

 Given I = O + O;     So,I=4

 Since T is I+I with a carry of 1 ;T is 9

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