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.
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