|
Question 1133013: What is the greatest possible sum that can result from
BAD+MAD+DAM if different letters represent different digits chosen from 1, 3, 8 and 9?
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the sum will be BAD+ MAD + DAM.
the most significant digits have to be B, D, or M.
the next most significant digit has to be A.
the least significant digit will be D or M.
B, D, or M will have to be some combination of 9, 8, and 3.
in order for B, D, or M to be the most significant digit, than A has to be equal to 1.
the sum of the most significant digits will always be 2000 regardless of the value of B, D, or M.
the sum of the next most significant digits will always be 30 because the middle digit is always 1.
that leaves the sum of the least significant digits to be the deciding factor.
the least significant digit sum will be 2 * D + M.
if D = 9 and M = 8, this sum will be 26.
if D = 8 and M = 9, this sum will be 25.
if D = 3 and M = 9, this sum will be 15.
if D = 3 and M = 8, this sum will be 14.
the highest sum of the least significant digit will be when D = 9 and M = 8.
since A is always equal to 1, that means that B must be equal to 3.
your numbers have to be B = 3, A = 1 M = 8, D = 9
the sum is equal to BAD + MAD + DAM which is equal to 319 + 819 + 918 = 2056.
this was confirmed by an analysis using excel which calculated all the possible combinations and cam up with the maximum value of 2056 which occurred when B = 3, A = 1, M = 8, and D = 9.
i will confess that the analysis using excel was performed first.
that told me that the maximum value for the sum had to be 2056.
the logic was developed after the analysis using excel.
here's the results of the analysis using excel.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
(1) To get the maximum sum, the leading digits B, M, and D must be, in some order, the three largest given digits, 3, 8, and 9.
(2) That means the other letter A has to be 1.
So the three 3-digit numbers are 91?, 81?, and 31?.
Again, to get the maximum sum, the units digits have to be as large as possible. Since in each of the three numbers the digits are all different, the units digit of 91? can't be 9, so it can be at most 8; for the other two 3-digit numbers, the units digit can be 9.
And that fits with the given three 3-digit numbers, because two of them have the same units digit.
So D is 9 and M is 8; and that makes B 3.
So the three numbers are 319, 819, and 918; their sum is 2056.
|
|
|
| |
