Seven Seas

Although the Pacific Ocean, Baltic Sea, and Arctic Sea represent three bodies of water, when added together, their sum yields seven Cs. Using the nine digits {0, 2, 3, 4, 5, 6, 7, 8, 9} substitute a different digit for each letter to make a true sum for PACIFIC + BALTIC + ARCTIC = CCCCCCC. Notice that 1 is not a possible digit for this problem. Your task is to provide the value of the word PACIFIC.
Source: mathcontest.olemiss.edu 10/1/2007

SOLUTION
This problem is similar to the problem titled “Peace” dated 3/24/2008.
Step 1 C+C+C=5+5+5
There are two possible choices: 0+0+0=0 \textup{ or }5+5+5=15. We drop the first choice C=0 because the sum of non-zero whole numbers is not equal to zero. Thus, C=5 and the carry digit is 1. Remember 1 is allowed as a carry digit.

Step 2 \textup{Carry }1+I+I+I
There is only one choice: 1+8+8+8=25. Thus, I=8 and the carry digit is 2.

Step 3 \textup{Carry }2+F+T+T
In this step we cannot use {1, 5, 8} because 1 is not allowed and {5, 8} are already taken.
3.1 F=2
2+2+0+0=4
2+2+3+3=10
2+2+4+4=12
2+2+6+6=16
\cdots
All the sums are even, so we move on to the next value for F.
3.2 F=3
\underline{2+3+0+0=5}
2+3+2+2=9
2+3+4+4=13
2+3+6+6=17
2+3+7+7=19
2+3+9+9=23
Thus, there is one possible choice, 2+F+T+T=2+3+0+0=5.
3.3 F=4
2+4+0+0=6
2+4+2+2=10
2+4+3+3=12
\cdots
All the sums are even, so we move on.
3.4 F=6
2+6+0+0=8
2+6+2+2=12
2+6+3+3=14
\cdots
All the sums are even, so we move on.
3.5 F=7
Remember we cannot use {1, 5, 8}.
2+7+0+0=9
2+7+2+2=13
\underline{2+7+3+3=15}
2+7+4+4=17
2+7+6+6=21
2+7+7+7=23
2+7+9+9=27
Thus, there is one possible choice, 2+F+T+T=2+7+3+3=15.
3.6 F=9
Remember we cannot use {1, 5, 8}.
2+9+0+0=11
\underline{2+9+2+2=15}
2+9+3+3=17
2+9+4+4=19
2+9+6+6=23
\underline{2+9+7+7=25}
Thus, there are two possible choices, 2+9+2+2=15 \textup { or }2+9+7+7=25
3.7 Summary
In summary, there are 4 possible choices for 2+F+T+T:
1.\;2+3+0+0=5
2.\;2+7+3+3=15
3.\;2+9+2+2=15
4.\;2+9+7+7=25
3.7.1
2+F+T+T=2+3+0+0

Carry

1

  

  

0

  

  

2

  

  

1

  

  

  

  

             
 

5

C

 

8

I

 

3

F

 

8

I

 

5

C

             
 

?

A

 

2

L

 

0

T

 

8

I

 

5

C

             
  

?

R

  

5

C

  

0

T

  

8

I

  

5

C

             
       

5

   

5

   

5

   

5

               

If we choose 2+F+T+T=2+3+0+0, then L=2.

Now, we must find what is Carry 1+5+A+R? Remember now we cannot use {0, 1, 2, 3, 5, 8}. If you work through all the possible values of A=\left \{4,6,7,9\right \}, then you will come to the conclusion that this choice has no solution.
3.7.2 2+F+T+T=2+7+3+3

Carry

1

  

  

1

  

  

2

  

  

1

  

  

  

  

             
 

5

C

 

8

I

 

7

F

 

8

I

 

5

C

             
 

?

A

 

1

L

 

3

T

 

8

I

 

5

C

             
  

?

R

  

5

C

  

3

T

  

8

I

  

5

C

             
       

5

   

5

   

5

   

5

               

This choice will force L=1 which is not allowed.
3.7.3 2+F+T+T=2+9+2+2

Carry

1

  

  

1

  

  

2

  

  

1

  

  

  

  

             
 

5

C

 

8

I

 

9

F

 

8

I

 

5

C

             
 

?

A

 

1

L

 

2

T

 

8

I

 

5

C

             
  

?

R

  

5

C

  

2

T

  

8

I

  

5

C

             
       

5

   

5

   

5

   

5

               

This choice will also force L=1 which is not allowed.
3.7.4 2+F+T+T=2+9+7+7

Carry

1

  

  

2

  

  

2

  

  

1

  

  

  

  

             
 

5

C

 

8

I

 

9

F

 

8

I

 

5

C

             
 

?

A

 

?

L

 

7

T

 

8

I

 

5

C

             
  

?

R

  

5

C

  

7

T

  

8

I

  

5

C

             
       

5

   

5

   

5

   

5

               

Find Carry 2+8+L+5. Remember now we cannot use {1, 5, 7, 8, 9}.
If you work through all the values of L=\left \{0,2,3,4,6\right \}, you will come to the conclusion that there is only one possible choice, L=0 .

Step 4 Carry 1+5+A+R
In this step we cannot use {0, 1, 5, 7, 8, 9}.

Carry

1

  

  

2

  

  

2

  

  

1

  

  

  

  

             
 

5

C

 

8

I

 

9

F

 

8

I

 

5

C

             
 

?

A

 

0

L

 

7

T

 

8

I

 

5

C

             
  

?

R

  

5

C

  

7

T

  

8

I

  

5

C

             
       

5

   

5

   

5

   

5

               

4.1 A=2
If you work through all the possible values of B=\left \{3,4,6\right \}, then you will come to the conclusion that this choice has no solution.
4.2 A=3
1+5+3+2=11
1+5+3+3=12
1+5+3+4=13
\underline{1+5+3+6=15}
There is one possible choice, 1+C+A+R=1+5+3+6=15.
4.3 A=4
If you work through all the possible values of B=\left \{2,3,6\right \}, then you will come to the conclusion that this choice has no solution.
4.4 A=6
1+5+6+2=14
\underline{1+5+6+3=15}
1+5+6+4=16
1+5+6+6=18
There is one possible choice, 1+C+A+R=1+5+6+3=15.
4.5 Summary
In summary, there are 2 possible choices for 1+C+A+R:
1.\;1+5+3+6=15
2.\;1+5+6+3=15
4.5.1 1+C+A+R=1+5+3+6

Carry

1

  

  

1

  

  

2

  

  

2

  

  

1

  

  

  

  

       
 

3

A

 

5

C

 

8

I

 

9

F

 

8

I

 

5

C

       
 

?

B

 

3

A

 

0

L

 

7

T

 

8

I

 

5

C

       
  

3

A

  

6

R

  

5

C

  

7

T

  

8

I

  

5

C

       
       

5

   

5

   

5

   

5

   

5

         

This choice will lead to 1+3+B+3=1+3+8+3=15. But, B=8 is not allowed because 8 is already taken.
4.5.2 1+C+A+R=1+5+6+3

Carry

1

  

  

1

  

  

2

  

  

2

  

  

1

  

  

  

  

       
 

6

A

 

5

C

 

8

I

 

9

F

 

8

I

 

5

C

       
 

?

B

 

6

A

 

0

L

 

7

T

 

8

I

 

5

C

       
  

6

A

  

3

R

  

5

C

  

7

T

  

8

I

  

5

C

       
       

5

   

5

   

5

   

5

   

5

         

This choice gives 1+6+B+6=1+6+2+6=15. Thus, B=2.

Step 5 P=4
The only value left is 4, so P=4.

Check

Carry

1

  

  

1

  

  

1

  

  

2

  

  

2

  

  

1

  

  

  

  

     
 

4

P

 

6

A

 

5

C

 

8

I

 

9

F

 

8

I

 

5

C

     
       

2

B

 

6

A

 

0

L

 

7

T

 

8

I

 

5

C

     
  

  

  

  

6

A

  

3

R

  

5

C

  

7

T

  

8

I

  

5

C

     
 

5

   

5

   

5

   

5

   

5

   

5

   

5

       

 Answer: PACIFIC=4658985.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , . Bookmark the permalink.

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