## 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: $PACIFIC=4658985$.