Xjyuk

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Too early in the morning to have SODA?

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

5

$begingroup$

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 19 hours ago

Omega Krypton

8,33821164

asked 20 hours ago

UvcUvc

2,041326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  deleted image. feel free to rollback if needed :)
  $endgroup$
  – Omega Krypton
  19 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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 19 hours ago

Omega Krypton

8,33821164

asked 20 hours ago

UvcUvc

2,041326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  deleted image. feel free to rollback if needed :)
  $endgroup$
  – Omega Krypton
  19 hours ago
  

  
  
  
  

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 19 hours ago

Omega Krypton

8,33821164

asked 20 hours ago

UvcUvc

2,041326

$endgroup$

share|improve this question

edited 19 hours ago

Omega Krypton

8,33821164

asked 20 hours ago

UvcUvc

2,041326

share|improve this question

edited 19 hours ago

Omega Krypton

8,33821164

asked 20 hours ago

UvcUvc

2,041326

edited 19 hours ago

Omega Krypton

8,33821164

edited 19 hours ago

Omega Krypton

8,33821164

asked 20 hours ago

UvcUvc

2,041326

asked 20 hours ago

UvcUvc

2,041326

3 Answers
3

active

oldest

votes

8

$begingroup$

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 19 hours ago

r_64r_64

3364

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  hi, nice try! +1
  $endgroup$
  – Omega Krypton
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it!!....deceptively unique
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah deceptively. :)
  $endgroup$
  – r_64
  19 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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 13 hours ago

TreninTrenin

8,1701748

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Easiest deductions for me =)
  $endgroup$
  – Montolide
  9 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Since

$A+A equiv A pmod 10$

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 16 hours ago

JonMark Perry

22.4k643103

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On the right track..I think it is unique
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Keep going..eventually you will get there
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  

add a comment | 

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8

$begingroup$

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 19 hours ago

r_64r_64

3364

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  hi, nice try! +1
  $endgroup$
  – Omega Krypton
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it!!....deceptively unique
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah deceptively. :)
  $endgroup$
  – r_64
  19 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

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 19 hours ago

r_64r_64

3364

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  hi, nice try! +1
  $endgroup$
  – Omega Krypton
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Got it!!....deceptively unique
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah deceptively. :)
  $endgroup$
  – r_64
  19 hours ago
  

  
  
  
  

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 19 hours ago

r_64r_64

3364

$endgroup$

share|improve this answer

answered 19 hours ago

r_64r_64

3364

answered 19 hours ago

r_64r_64

3364

answered 19 hours ago

r_64r_64

3364

7

$begingroup$

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 13 hours ago

TreninTrenin

8,1701748

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Easiest deductions for me =)
  $endgroup$
  – Montolide
  9 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

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 13 hours ago

TreninTrenin

8,1701748

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Easiest deductions for me =)
  $endgroup$
  – Montolide
  9 hours ago
  

  
  
  
  

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 13 hours ago

TreninTrenin

8,1701748

$endgroup$

 COCA
+COLA
-----
 SODA

share|improve this answer

answered 13 hours ago

TreninTrenin

8,1701748

answered 13 hours ago

TreninTrenin

8,1701748

answered 13 hours ago

TreninTrenin

8,1701748

5

$begingroup$

Since

$A+A equiv A pmod 10$

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 16 hours ago

JonMark Perry

22.4k643103

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On the right track..I think it is unique
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Keep going..eventually you will get there
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Since

$A+A equiv A pmod 10$

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 16 hours ago

JonMark Perry

22.4k643103

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On the right track..I think it is unique
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Keep going..eventually you will get there
  $endgroup$
  – Uvc
  19 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Since

$A+A equiv A pmod 10$

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 16 hours ago

JonMark Perry

22.4k643103

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164

$endgroup$

share|improve this answer

edited 16 hours ago

JonMark Perry

22.4k643103

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164

edited 16 hours ago

JonMark Perry

22.4k643103

edited 16 hours ago

JonMark Perry

22.4k643103

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164

answered 19 hours ago

Omega KryptonOmega Krypton

8,33821164