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Too early in the morning to have SODA?

Flock of Geese alphameticMultiplicative alphametic: This is too hardAn almost Shakespearian alphameticSquare dance alphameticAlphametic between Kennedy and NixonIt is as easy as A B C, Figure out U V C from the given relationshipLady Luck Powers Up Every Member to Sum Upto Non-Prime Number. Who am I?UVC wants to give you a Helping Hand in Solving these Unique Set of Pan digital Fractional-Decimal RelationsPlease figure out this Pan digital PrinceA Lollipop with Roots

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

9

$begingroup$

Each letter shown represent distinct digit...can vary from zero to nine.

$COCA$, $COLA$, $SODA$ are three concatenated numbers.

Figure these out from the following relation:

$COCA + COLA = SODA$

mathematics logical-deduction calculation-puzzle no-computers alphametic

share|improve this question

edited Jun 17 at 9:01

Omega Krypton

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asked Jun 17 at 8:31

UvcUvc

2,39155 silver badges2727 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  deleted image. feel free to rollback if needed :)
  $endgroup$
  – Omega Krypton
  Jun 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Am I the only one who thought "Too early in the morning" was a hint?
  $endgroup$
  – Mr Lister
  Jun 18 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t know..when I got up little early yesterday, title popped in my head
  $endgroup$
  – Uvc
  Jun 18 at 12:13
  

  
  
  
  

add a comment | 

9

$begingroup$

Each letter shown represent distinct digit...can vary from zero to nine.

$COCA$, $COLA$, $SODA$ are three concatenated numbers.

Figure these out from the following relation:

$COCA + COLA = SODA$

mathematics logical-deduction calculation-puzzle no-computers alphametic

share|improve this question

edited Jun 17 at 9:01

Omega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

asked Jun 17 at 8:31

UvcUvc

2,39155 silver badges2727 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  deleted image. feel free to rollback if needed :)
  $endgroup$
  – Omega Krypton
  Jun 17 at 8:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Am I the only one who thought "Too early in the morning" was a hint?
  $endgroup$
  – Mr Lister
  Jun 18 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t know..when I got up little early yesterday, title popped in my head
  $endgroup$
  – Uvc
  Jun 18 at 12:13
  

  
  
  
  

add a comment | 

$begingroup$

Each letter shown represent distinct digit...can vary from zero to nine.

$COCA$, $COLA$, $SODA$ are three concatenated numbers.

Figure these out from the following relation:

$COCA + COLA = SODA$

mathematics logical-deduction calculation-puzzle no-computers alphametic

share|improve this question

edited Jun 17 at 9:01

Omega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

asked Jun 17 at 8:31

UvcUvc

2,39155 silver badges2727 bronze badges

$endgroup$

share|improve this question

edited Jun 17 at 9:01

Omega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

asked Jun 17 at 8:31

UvcUvc

2,39155 silver badges2727 bronze badges

share|improve this question

edited Jun 17 at 9:01

Omega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

asked Jun 17 at 8:31

UvcUvc

2,39155 silver badges2727 bronze badges

edited Jun 17 at 9:01

Omega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

edited Jun 17 at 9:01

Omega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

asked Jun 17 at 8:31

UvcUvc

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asked Jun 17 at 8:31

UvcUvc

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3 Answers
3

active

oldest

votes

13

$begingroup$

Based on Omega Krypton's answer,

$2C+1=S,C+L=D+10$, $A=0,O=9$. (Note that $O=9$ so $C+L$ carries.)

We also need that these digits $C,L,D,S$ are distinct between $1sim 8$. ($0$ and $9$ are taken.)

If $C=1$ or $C=2$ then, since $Dge 1$ we have $Lge 9$ which is incorrect.

So $C=3$ and $S=7$. We have $L=8$ and $D=1$.

That is $3930+3980=7910$.

share|improve this answer

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  hi, nice try! +1
  $endgroup$
  – Omega Krypton
  Jun 17 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it!!....deceptively unique
  $endgroup$
  – Uvc
  Jun 17 at 8:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah deceptively. :)
  $endgroup$
  – r_64
  Jun 17 at 8:57
  

  
  
  
  

add a comment | 

17

$begingroup$

We have the following

 COCA
+COLA
-----
 SODA

First, from the ones column, we have $A+A implies A$ which is only possible if $A=0$.

Next, notice something similar in the

hundreds place; $O+O implies O$. Since $0$ is already taken and the only possibility without a carry over, we must have a carry over from the 10s, and $O=9$ is the only possibility. We will also have a carry over into the thousands.

Since we have a 4 digit number as the result, we know that

$0 lt C le 4$.

But:

-But $C=4 implies S=9$ which is already taken by $O$.

-And $C=1 implies L=9$ to achieve a carryover, which is taken by $O$.

-And $C=2 implies Lin8,9$. But $L=9$ is taken, and $L=8 implies D=0$ is also taken.

Thus,

$C=3$.

Also, we know

$S=7$ because the hundreds will carry over, and we also know that in order to carry over the 10s, we need $Lge 7$. But $L=7$ and $L=9$ are taken leaving only $L=8$, and thus, $D=1$.

Thus, the solution is;

COCA+COLA=SODA, 3930+3980=7910

share|improve this answer

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Easiest deductions for me =)
  $endgroup$
  – Montolide
  Jun 17 at 18:41
  

  
  
  
  

add a comment | 

8

$begingroup$

Since we know that

$A+A equiv A pmod 10$

Therefore $A$

$=0$

Hundreds value must carry since $O neq 0$

Therefore

$O+O+1 equiv O pmod 10$

Therefore $O$

$=9$

We now get

$2C+1=S$
$C+L=D$

And since $S<9$

$0<C<4$

Then there are many possibilities... any relations I missed out?

share|improve this answer

edited Jun 18 at 9:10

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On the right track..I think it is unique
  $endgroup$
  – Uvc
  Jun 17 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Keep going..eventually you will get there
  $endgroup$
  – Uvc
  Jun 17 at 8:55
  

  
  
  
  

add a comment | 

Your Answer

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13

$begingroup$

Based on Omega Krypton's answer,

$2C+1=S,C+L=D+10$, $A=0,O=9$. (Note that $O=9$ so $C+L$ carries.)

We also need that these digits $C,L,D,S$ are distinct between $1sim 8$. ($0$ and $9$ are taken.)

If $C=1$ or $C=2$ then, since $Dge 1$ we have $Lge 9$ which is incorrect.

So $C=3$ and $S=7$. We have $L=8$ and $D=1$.

That is $3930+3980=7910$.

share|improve this answer

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  hi, nice try! +1
  $endgroup$
  – Omega Krypton
  Jun 17 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it!!....deceptively unique
  $endgroup$
  – Uvc
  Jun 17 at 8:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah deceptively. :)
  $endgroup$
  – r_64
  Jun 17 at 8:57
  

  
  
  
  

add a comment | 

13

$begingroup$

Based on Omega Krypton's answer,

$2C+1=S,C+L=D+10$, $A=0,O=9$. (Note that $O=9$ so $C+L$ carries.)

We also need that these digits $C,L,D,S$ are distinct between $1sim 8$. ($0$ and $9$ are taken.)

If $C=1$ or $C=2$ then, since $Dge 1$ we have $Lge 9$ which is incorrect.

So $C=3$ and $S=7$. We have $L=8$ and $D=1$.

That is $3930+3980=7910$.

share|improve this answer

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  hi, nice try! +1
  $endgroup$
  – Omega Krypton
  Jun 17 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it!!....deceptively unique
  $endgroup$
  – Uvc
  Jun 17 at 8:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah deceptively. :)
  $endgroup$
  – r_64
  Jun 17 at 8:57
  

  
  
  
  

add a comment | 

$begingroup$

Based on Omega Krypton's answer,

$2C+1=S,C+L=D+10$, $A=0,O=9$. (Note that $O=9$ so $C+L$ carries.)

We also need that these digits $C,L,D,S$ are distinct between $1sim 8$. ($0$ and $9$ are taken.)

If $C=1$ or $C=2$ then, since $Dge 1$ we have $Lge 9$ which is incorrect.

So $C=3$ and $S=7$. We have $L=8$ and $D=1$.

That is $3930+3980=7910$.

share|improve this answer

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

$endgroup$

share|improve this answer

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

answered Jun 17 at 8:54

r_64r_64

41655 bronze badges

17

$begingroup$

We have the following

 COCA
+COLA
-----
 SODA

First, from the ones column, we have $A+A implies A$ which is only possible if $A=0$.

Next, notice something similar in the

hundreds place; $O+O implies O$. Since $0$ is already taken and the only possibility without a carry over, we must have a carry over from the 10s, and $O=9$ is the only possibility. We will also have a carry over into the thousands.

Since we have a 4 digit number as the result, we know that

$0 lt C le 4$.

But:

-But $C=4 implies S=9$ which is already taken by $O$.

-And $C=1 implies L=9$ to achieve a carryover, which is taken by $O$.

-And $C=2 implies Lin8,9$. But $L=9$ is taken, and $L=8 implies D=0$ is also taken.

Thus,

$C=3$.

Also, we know

$S=7$ because the hundreds will carry over, and we also know that in order to carry over the 10s, we need $Lge 7$. But $L=7$ and $L=9$ are taken leaving only $L=8$, and thus, $D=1$.

Thus, the solution is;

COCA+COLA=SODA, 3930+3980=7910

share|improve this answer

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Easiest deductions for me =)
  $endgroup$
  – Montolide
  Jun 17 at 18:41
  

  
  
  
  

add a comment | 

17

$begingroup$

We have the following

 COCA
+COLA
-----
 SODA

First, from the ones column, we have $A+A implies A$ which is only possible if $A=0$.

Next, notice something similar in the

hundreds place; $O+O implies O$. Since $0$ is already taken and the only possibility without a carry over, we must have a carry over from the 10s, and $O=9$ is the only possibility. We will also have a carry over into the thousands.

Since we have a 4 digit number as the result, we know that

$0 lt C le 4$.

But:

-But $C=4 implies S=9$ which is already taken by $O$.

-And $C=1 implies L=9$ to achieve a carryover, which is taken by $O$.

-And $C=2 implies Lin8,9$. But $L=9$ is taken, and $L=8 implies D=0$ is also taken.

Thus,

$C=3$.

Also, we know

$S=7$ because the hundreds will carry over, and we also know that in order to carry over the 10s, we need $Lge 7$. But $L=7$ and $L=9$ are taken leaving only $L=8$, and thus, $D=1$.

Thus, the solution is;

COCA+COLA=SODA, 3930+3980=7910

share|improve this answer

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Easiest deductions for me =)
  $endgroup$
  – Montolide
  Jun 17 at 18:41
  

  
  
  
  

add a comment | 

$begingroup$

We have the following

 COCA
+COLA
-----
 SODA

First, from the ones column, we have $A+A implies A$ which is only possible if $A=0$.

Next, notice something similar in the

hundreds place; $O+O implies O$. Since $0$ is already taken and the only possibility without a carry over, we must have a carry over from the 10s, and $O=9$ is the only possibility. We will also have a carry over into the thousands.

Since we have a 4 digit number as the result, we know that

$0 lt C le 4$.

But:

-But $C=4 implies S=9$ which is already taken by $O$.

-And $C=1 implies L=9$ to achieve a carryover, which is taken by $O$.

-And $C=2 implies Lin8,9$. But $L=9$ is taken, and $L=8 implies D=0$ is also taken.

Thus,

$C=3$.

Also, we know

$S=7$ because the hundreds will carry over, and we also know that in order to carry over the 10s, we need $Lge 7$. But $L=7$ and $L=9$ are taken leaving only $L=8$, and thus, $D=1$.

Thus, the solution is;

COCA+COLA=SODA, 3930+3980=7910

share|improve this answer

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

$endgroup$

 COCA
+COLA
-----
 SODA

share|improve this answer

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

answered Jun 17 at 14:53

TreninTrenin

8,2801717 silver badges4949 bronze badges

8

$begingroup$

Since we know that

$A+A equiv A pmod 10$

Therefore $A$

$=0$

Hundreds value must carry since $O neq 0$

Therefore

$O+O+1 equiv O pmod 10$

Therefore $O$

$=9$

We now get

$2C+1=S$
$C+L=D$

And since $S<9$

$0<C<4$

Then there are many possibilities... any relations I missed out?

share|improve this answer

edited Jun 18 at 9:10

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On the right track..I think it is unique
  $endgroup$
  – Uvc
  Jun 17 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Keep going..eventually you will get there
  $endgroup$
  – Uvc
  Jun 17 at 8:55
  

  
  
  
  

add a comment | 

8

$begingroup$

Since we know that

$A+A equiv A pmod 10$

Therefore $A$

$=0$

Hundreds value must carry since $O neq 0$

Therefore

$O+O+1 equiv O pmod 10$

Therefore $O$

$=9$

We now get

$2C+1=S$
$C+L=D$

And since $S<9$

$0<C<4$

Then there are many possibilities... any relations I missed out?

share|improve this answer

edited Jun 18 at 9:10

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On the right track..I think it is unique
  $endgroup$
  – Uvc
  Jun 17 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Keep going..eventually you will get there
  $endgroup$
  – Uvc
  Jun 17 at 8:55
  

  
  
  
  

add a comment | 

$begingroup$

Since we know that

$A+A equiv A pmod 10$

Therefore $A$

$=0$

Hundreds value must carry since $O neq 0$

Therefore

$O+O+1 equiv O pmod 10$

Therefore $O$

$=9$

We now get

$2C+1=S$
$C+L=D$

And since $S<9$

$0<C<4$

Then there are many possibilities... any relations I missed out?

share|improve this answer

edited Jun 18 at 9:10

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

$endgroup$

share|improve this answer

edited Jun 18 at 9:10

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges

answered Jun 17 at 8:48

Omega KryptonOmega Krypton

8,70822 gold badges1111 silver badges6666 bronze badges