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Adding up numbers in Portuguese is strange

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9

$begingroup$

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.

alphametic

share|improve this question

asked Mar 20 at 4:09

ChaoticChaotic

621316

$endgroup$

add a comment | 

9

$begingroup$

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.

alphametic

share|improve this question

asked Mar 20 at 4:09

ChaoticChaotic

621316

$endgroup$

add a comment | 

$begingroup$

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.

alphametic

share|improve this question

asked Mar 20 at 4:09

ChaoticChaotic

621316

$endgroup$

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

share|improve this question

asked Mar 20 at 4:09

ChaoticChaotic

621316

share|improve this question

asked Mar 20 at 4:09

ChaoticChaotic

621316

asked Mar 20 at 4:09

ChaoticChaotic

621316

asked Mar 20 at 4:09

ChaoticChaotic

621316

3 Answers
3

active

oldest

votes

10

$begingroup$

Could it be

 1246 

 + 1246 

 ------ 

 2492 

Then

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

share|improve this answer

edited Mar 20 at 12:40

Omega Krypton

5,5892849

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect! Easy one?
  $endgroup$
  – Chaotic
  Mar 20 at 4:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
  $endgroup$
  – El-Guest
  Mar 20 at 11:28
  

  
  
  
  

add a comment | 

13

$begingroup$

From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.

share|improve this answer

edited Mar 20 at 9:03

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You haven't given a value for $T$.
  $endgroup$
  – ZanyG
  Mar 20 at 5:54
  

  
  
  
  

add a comment | 

0

$begingroup$

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

share|improve this answer

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1

$endgroup$

add a comment | 

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10

$begingroup$

Could it be

 1246 

 + 1246 

 ------ 

 2492 

Then

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

share|improve this answer

edited Mar 20 at 12:40

Omega Krypton

5,5892849

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect! Easy one?
  $endgroup$
  – Chaotic
  Mar 20 at 4:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
  $endgroup$
  – El-Guest
  Mar 20 at 11:28
  

  
  
  
  

add a comment | 

10

$begingroup$

Could it be

 1246 

 + 1246 

 ------ 

 2492 

Then

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

share|improve this answer

edited Mar 20 at 12:40

Omega Krypton

5,5892849

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect! Easy one?
  $endgroup$
  – Chaotic
  Mar 20 at 4:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
  $endgroup$
  – El-Guest
  Mar 20 at 11:28
  

  
  
  
  

add a comment | 

$begingroup$

Could it be

 1246 

 + 1246 

 ------ 

 2492 

Then

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

share|improve this answer

edited Mar 20 at 12:40

Omega Krypton

5,5892849

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

$endgroup$

share|improve this answer

edited Mar 20 at 12:40

Omega Krypton

5,5892849

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

edited Mar 20 at 12:40

Omega Krypton

5,5892849

edited Mar 20 at 12:40

Omega Krypton

5,5892849

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

answered Mar 20 at 4:42

El-GuestEl-Guest

22.2k35193

13

$begingroup$

From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.

share|improve this answer

edited Mar 20 at 9:03

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You haven't given a value for $T$.
  $endgroup$
  – ZanyG
  Mar 20 at 5:54
  

  
  
  
  

add a comment | 

13

$begingroup$

From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.

share|improve this answer

edited Mar 20 at 9:03

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You haven't given a value for $T$.
  $endgroup$
  – ZanyG
  Mar 20 at 5:54
  

  
  
  
  

add a comment | 

$begingroup$

From $S+S=O$ we know $O$ is even. From the $O+O$ part, we know $Ole4$ because $2O$ doesn't carry (otherwise $O$ would also have to be odd). Therefore $I$ is doubly even ($I=2O$), and because $2I$ doesn't carry, $I=4$ and $O=2$. $D=Spm5$ and $2D$ doesn't carry, so $S=6$ and $D=1$. $T=2I+1=9$.

share|improve this answer

edited Mar 20 at 9:03

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

$endgroup$

share|improve this answer

edited Mar 20 at 9:03

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

answered Mar 20 at 5:16

JonMark PerryJonMark Perry

21.1k641100

0

$begingroup$

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

share|improve this answer

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1

$endgroup$

add a comment | 

0

$begingroup$

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

share|improve this answer

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1

$endgroup$

add a comment | 

$begingroup$

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

share|improve this answer

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1

$endgroup$

share|improve this answer

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

edited Mar 20 at 8:15

JonMark Perry

21.1k641100

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1

answered Mar 20 at 6:12

NaveenGopal NolluNaveenGopal Nollu

1