Adding up numbers in Portuguese is strange
$begingroup$
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
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
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
asked Mar 20 at 4:09
ChaoticChaotic
621316
$endgroup$
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
alphametic
asked Mar 20 at 4:09
ChaoticChaotic
621316
asked Mar 20 at 4:09
ChaoticChaotic
621316
asked Mar 20 at 4:09
ChaoticChaotic
621316
asked Mar 20 at 4:09
ChaoticChaotic
621316
asked Mar 20 at 4:09
ChaoticChaotic
621316
621316
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
edited 2 days ago
Omega Krypton
4,8702444
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
$endgroup$
$begingroup$
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
2 days ago
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 $D=1$. $T=2I+1=9$.
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
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
edited 2 days ago
JonMark Perry
20.4k64099
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
edited 2 days ago
Omega Krypton
4,8702444
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
$endgroup$
$begingroup$
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
2 days ago
add a comment |
$begingroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
edited 2 days ago
Omega Krypton
4,8702444
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
$endgroup$
$begingroup$
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
2 days ago
add a comment |
$begingroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
edited 2 days ago
Omega Krypton
4,8702444
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
$endgroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
edited 2 days ago
Omega Krypton
4,8702444
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
edited 2 days ago
Omega Krypton
4,8702444
edited 2 days ago
Omega Krypton
4,8702444
edited 2 days ago
Omega Krypton
4,8702444
4,8702444
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
20.8k24791
$begingroup$
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
2 days ago
add a comment |
$begingroup$
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
2 days ago
$begingroup$
Perfect! Easy one?
$endgroup$
– Chaotic
Mar 20 at 4:50
$begingroup$
Perfect! Easy one?
$endgroup$
– Chaotic
Mar 20 at 4:50
1
1
$begingroup$
Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
$endgroup$
– El-Guest
2 days ago
$begingroup$
Not super easy, but there were a limited number of combinations to try. Good puzzle @Chaotic!
$endgroup$
– El-Guest
2 days ago
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 $D=1$. $T=2I+1=9$.
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
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 $D=1$. $T=2I+1=9$.
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
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 $D=1$. $T=2I+1=9$.
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
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 $D=1$. $T=2I+1=9$.
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
edited 2 days ago
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
answered 2 days ago
JonMark PerryJonMark Perry
20.4k64099
20.4k64099
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
add a comment |
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
2 days ago
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
edited 2 days ago
JonMark Perry
20.4k64099
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 2 days ago
JonMark Perry
20.4k64099
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 2 days ago
JonMark Perry
20.4k64099
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 2 days ago
JonMark Perry
20.4k64099
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
JonMark Perry
20.4k64099
edited 2 days ago
JonMark Perry
20.4k64099
edited 2 days ago
JonMark Perry
20.4k64099
20.4k64099
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
answered 2 days ago
NaveenGopal NolluNaveenGopal Nollu
1
1
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
NaveenGopal Nollu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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