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Adding up numbers in Portuguese is strange
How does one go about picking words when creating Verbal Arithmetic puzzles?SPEND LESS MONEY alphameticHow do I solve cryptarithmetic puzzles?Letters = numbersNumbers = letters 2.0Does AAA+BBB+CCC+DDD=ABCD have a solution for distinct digits A,B,C,D?Arrange numbers to 3 different math operationsSum of three six digit numbersWhat are these numbers?
$begingroup$
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 Mar 20 at 12:40
Omega Krypton
4,9452544
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
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 $D=1$. $T=2I+1=9$.
answered Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
$begingroup$
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
edited Mar 20 at 8:15
JonMark Perry
20.4k64099
answered Mar 20 at 6:12
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 |
Your Answer
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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 Mar 20 at 12:40
Omega Krypton
4,9452544
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
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
edited Mar 20 at 12:40
Omega Krypton
4,9452544
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
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
edited Mar 20 at 12:40
Omega Krypton
4,9452544
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 Mar 20 at 12:40
Omega Krypton
4,9452544
answered Mar 20 at 4:42
El-GuestEl-Guest
20.8k24791
edited Mar 20 at 12:40
Omega Krypton
4,9452544
edited Mar 20 at 12:40
Omega Krypton
4,9452544
edited Mar 20 at 12:40
Omega Krypton
4,9452544
4,9452544
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
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 $D=1$. $T=2I+1=9$.
answered Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
$begingroup$
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 $D=1$. $T=2I+1=9$.
answered Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
$endgroup$
$begingroup$
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 $D=1$. $T=2I+1=9$.
answered Mar 20 at 5:16
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 Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
edited Mar 20 at 9:03
answered Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
answered Mar 20 at 5:16
JonMark PerryJonMark Perry
20.4k64099
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
edited Mar 20 at 8:15
JonMark Perry
20.4k64099
answered Mar 20 at 6:12
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 Mar 20 at 8:15
JonMark Perry
20.4k64099
answered Mar 20 at 6:12
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 Mar 20 at 8:15
JonMark Perry
20.4k64099
answered Mar 20 at 6:12
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 Mar 20 at 8:15
JonMark Perry
20.4k64099
answered Mar 20 at 6:12
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 Mar 20 at 8:15
JonMark Perry
20.4k64099
edited Mar 20 at 8:15
JonMark Perry
20.4k64099
edited Mar 20 at 8:15
JonMark Perry
20.4k64099
20.4k64099
answered Mar 20 at 6:12
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 Mar 20 at 6:12
NaveenGopal NolluNaveenGopal Nollu
1
answered Mar 20 at 6:12
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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