Question 983638: Z N P M
B P C R
Y R P M
B Y C R
--------------
X L X X
The above is a normal sum operation, but each number from 0 – 9 were each replaced by a specific letter from the alphabet.
1. Can you identify what number is X? X ≡
2. Can to identify any other number/s?
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! I think the easiest thing to see on this puzzle can be found in the first column.
Since we are to assume that each number is a 4 digit number, by definition the first column cannot contain any 0s.
We also know that the result of Z + Y + B + B is a single digit number. So let's assume that B=1.
If B = 1, then Z+Y can be at minimum 2+3 = 5, so Z + Y + B + B >= 7.
Notice if B =2, then Z+Y is at minimum of 1+3, so Z+Y+B+B >= 8.
And if B = 3, then Z+Y is at minimum of 1+2, so Z + Y +B +B >= 9
Notice that in all of these examples, the 1st column is using the numbers 1,2, and 3, so while we don't know where these numbers go, we can consider them used.
If we look at M+R+M+R, this is the same as 2(M+R) which is an even value.
So X must be even.
The only value that satisfies X given these two things is 8. Hence X = 8.
We update the picture to look like this:
Z N P M
B P C R
Y R P M
B Y C R
--------------
8 L 8 8
And the numbers we have not used are [0,4,5,6,7,9].
Now we get a bit more complicated.
In order to get a units digit of 8, M+R+M+R =8 or M+R+M+R = 18 or M+R+M+R = 28 [38 is not possible, as our maximum is 34]. So to simplify this, essentially M+R = 4, 9 ,14
The sums that would satisfy these are
4: 0+4
9: 0+9, 4+5
14 9+5
Notice that in each of these cases, the numbers 6 and 7 are not used. So we know it is in one of the other two columns.
Since P+C+P+C+1 is not even, we can eliminate the possibility that M+R+M+R = 9.
So the sums are updated to just:
4: 0+4
14:9+5
Looking in the second column, we can see that for N + P + R + Y
the minimum value is : 0 + 4 + 5 + 1 = 10
and the maximum value is 9 + 5 + 6 + 7 = 27
So the carry digit for the 1st column is either a 1 or 2 (important that the carry digit cannot be a 0).
Remember that we have said that 1,2,3 are all being used in the 1st column, we just don't know where yet.
So either 1 + Z + Y + 2B = 8
or 2 + Z + Y + 2B = 8.
But notice that if 2 + Z + Y + 2B = 8, then Z+Y+2B = 6.
If B = 1, then Z+Y = 4, but we know that if B = 1, Z+Y = 5 [since 2+3=5] Contradiction.
If B = 2, then Z+Y = 2, but Z+Y = 4 [since 1+3 = 4]. Contradiction.
So the carry digit cannot be 2.
So since 1 + Z + Y + 2B = 8
If B = 1, then 1+ Z + Y + 2 = 8.
Z+Y = 5. Success!
Let's make sure that B=2 isn't viable.
If B = 2, then 2 + Z + Y + 4 = 8, so Z+Y = 2. But we know Z+Y should be 4 in this case. Contradiction.
Hence B = 1.
We update our chart.
1
Z N P M
1 P C R
Y R P M
1 Y C R
--------------
8 L 8 8
Let's remind ourselves of what we have.
Z = 2 or 3
Y = 2 or 3
M = 0, 4 or 9
R = 0, 4, or 9
P = 0, 4, 5, 6, 7, 9
C = 0, 4, 5, 6, 7, 9
N = 0, 4, 5, 6, 7, 9
B = 1
X = 8
We know that N+P+R+Y + [carry] has to be between 10-19
What is the minimum value that we get if 9 is in the second column?
Assuming that the carry is 0 and y = 2
We get 9 + 4 + 5 + 2 = 20. So it is impossible to get a carry of 1 if 9 is in the second column.
So we update our list:
Z = 2 or 3
Y = 2 or 3
M = 0, 4 or 9
R = 0, 4
P = 0, 4, 5, 6, 7
C = 0, 4, 5, 6, 7, 9
N = 0, 4, 5, 6, 7
B = 1
X = 8
Since R must be 0 or 4, then we know that M+R = 4, so M and R can only be 0 or 4. We know that the carry digit for the third column must be 0.
Z = 2 or 3
Y = 2 or 3
M = 0, 4
R = 0, 4
P = 5, 6, 7
C = 5, 6, 7, 9
N = 5, 6, 7
B = 1
X = 8
L = 5,6,7,9
1 ? 0
Z N P M
1 P C R
Y R P M
1 Y C R
--------------
8 L 8 8
Notice that P+C = 4,9,14.
But P+C=4 contradicts that M and R are only 0 or 4. And if P+C = 9 then P =0,9 or C = 0,9, but P or C cannot be 0.
Also for similar reasons P+C = 9 could be 4 and 5, but 4 is exclusive to the last column. So P+C cannot be 9. Hence P+C = 14 and since P cannot be 9, then C must be 9.
Hence P = 5.
Z = 2,3
Y = 2,3
M = 0, 4
R = 0, 4
P = 5
C = 9
N = 6, 7
B = 1
X = 8
L = 6,7
1 2 0
Z N 5 M
1 5 9 R
Y R 5 M
1 Y 9 R
--------------
8 L 8 8
If L = 6, then
2 + 7 + 5 + R + Y = 16
so 7 + R + Y = 9
(1) R + Y = 2
If L = 7, then
2 + 6 + 5 + R + Y = 17
(2) R + Y = 4
Dealing with R+Y = 2, Y cannot be 3. Y would be 2 and R = 0. Is this a unique solution?
Dealing with R+Y = 4, Y could be 2, but then R = 2, so no. If Y = 3, then R = 1 (which it can't be).
So Y = 2 and R = 0. Since R =0, M =4.
We found Y=2 and R=0 in the L = 6 case. Since L=6, N=7. This completes our solution.
So the final solution is as follows:
It is:
3754 B=1 C=9 L=6 M=4 N=7 P=5 R=0 X=8 Y=2 Z=3
1590
2054
+1290
-----
8688
Not sure how well this formatted, but I hope it helped! Honestly, if you have ever done a sudoku puzzle, this isn't far off the logic.
Please email me at swincher4391@yahoo.com if you have any trouble following.
Thanks!
Devin
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