## Solution to: Word Sums

It is given that the digits 1 and 6 are the most frequently used, so we first look how often each letter occurs in the puzzle:

• M, V, and P occur once;
• T occurs twice;
• A, R, and E occur three times;
• S occurs four times;
• N and U occur five times.

So, either N=1 and U=6, or N=6 and U=1. Since even 9999 + 99999 + 999999 + 999999 is less than 6000000, N cannot be 6. So N=1 and U=6.

We now have:

```     mars
ve16s
6ra16s
sat6r1
------- +
1ept61e
```

R+6+6+R (in the 'tens' column) is even, but since its sum ends on a 1, 1 must have been carried from the 'units' column. We can also see that R=4 or R=9. The fact that 1 is carried from the 'units' to the 'tens' column gives the following possible values for S:

• S=3, so S+S+S+1=10, and therefore E=0;
• S=4, so S+S+S+1=13, and therefore E=3;
• S=5, so S+S+S+1=16, and therefore E=6: this is not possible because U=6.

Depending on the values of S and E, we now have:

```S=3 and E=0:    S=4 and E=3:
mar3            mar4
v0163           v3164
6ra163          6ra164
3at6r1          4at6r1
------- +       ------- +
10pt610         13pt613
```

From the previous step, we know that R=4 or R=9. We first look at the case S=3. If R would be 4, a value of 2 would be carried from the 'tens' to the 'hundreds' column. So, the sum of 2+A+1+1+6 must end on a 6, which gives A=6. This is not possible because U=6. So, if S=3, then R must be 9 and a value of 3 is carried from the 'tens' to the 'hundreds' column. So, the sum of 3+A+1+1+6 must end on a 6, which gives A=5.

We now look at the case S=4. In this case, R can only be 9. Then a value of 3 is carried from the 'tens' to the 'hundreds' column, which gives A=5.

Therefore, regardless of the values of S and E, we know that R=9 and A=5. Depending on the values of S and E, we now have:

```S=3 and E=0:    S=4 and E=3:
m593            m594
v0163           v3164
695163          695164
35t691          45t691
------- +       ------- +
10pt610         13pt613
```

In either case, a value of 1 is carried from the 'hundreds' to the 'thousands' column. In the case that S=3 and E=0, the sum of 1+M+0+5+T must end on T, which gives M=4. In the case that S=4 and E=3, the sum of 1+M+3+5+T must end on T, which gives M=1. This, however, is not possible because U=1. From this, it follows that S cannot be 4, so we now know that S=3, E=0, and M=4.

We now have:

```     4593
v0163
695163
35t691
------- +
10pt610
```

The only remaining digits are now 2, 7, and 8. Since a value of 1 is carried from the 'thousands' to the 'ten thousands' column, it follows that V=2 and P=7. For T, only the value 8 remains.

The complete addition is as follows:

```     4593
20163
695163
358691
------- +
1078610
```

Conclusion: neptune equals 1078610. Back to the puzzle
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