Adding up numbers in Portuguese is strange
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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.
alphametic
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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
$endgroup$
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
$endgroup$
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
621316
add a comment |
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3 Answers
3
active
oldest
votes
$begingroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
$endgroup$
$begingroup$
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 |
$begingroup$
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 $D=1$. $T=2I+1=9$.
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You haven't given a value for $T$.
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– ZanyG
Mar 20 at 5:54
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
New contributor
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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
$endgroup$
$begingroup$
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
$endgroup$
$begingroup$
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
$endgroup$
Could it be
1246
+ 1246
------
2492
Then
D=1, O=2, I=4, S=6, T=9
edited Mar 20 at 12:40
Omega Krypton
4,9452544
4,9452544
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
Mar 20 at 11:28
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
Mar 20 at 11:28
$begingroup$
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
Mar 20 at 11:28
$begingroup$
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 |
$begingroup$
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 $D=1$. $T=2I+1=9$.
$endgroup$
$begingroup$
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 $D=1$. $T=2I+1=9$.
$endgroup$
$begingroup$
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 $D=1$. $T=2I+1=9$.
$endgroup$
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 $D=1$. $T=2I+1=9$.
edited Mar 20 at 9:03
answered Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
20.4k64099
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
Mar 20 at 5:54
add a comment |
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
Mar 20 at 5:54
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
Mar 20 at 5:54
$begingroup$
You haven't given a value for $T$.
$endgroup$
– ZanyG
Mar 20 at 5:54
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
edited Mar 20 at 8:15
JonMark Perry
20.4k64099
20.4k64099
New contributor
answered Mar 20 at 6:12
NaveenGopal NolluNaveenGopal Nollu
1
1
New contributor
New contributor
add a comment |
add a comment |
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