There are $60$ different 3-letter combinations of A,B,C,D,E. Using each letter to represent a different integer from $1$ to $9$, what is the solution for ABCDE to equal the sum of all 3-letter combinations (ABC + ABD + ABE + BAC + BAD + . . .)?
As well as the correct solution, the best answer should include simplified expression for this problem.
Answer
First let's look at what all the 60 combinations look like. Consider the first digit - obviously, we have 5 choices for which letter to place there. Since there are a total of 60 combinations, each letter must appear there a total of 12 times. Therefore, we can add up everything contributing to the sum from just that place alone as 1200*(A+B+C+D+E). This argument can be applied to each digit separately, so we can evaluate the entire sum as 1332*(A+B+C+D+E) since 12+120+1200=1332.
Now, the constraint that each letter must represent a distinct number means that the sum A+B+C+D+E must fall somewhere in the range of 15 (1+2+3+4+5) to 35 (9+8+7+6+5), inclusive. That's actually a small enough range to check by hand, in which case you get
35964
as the unique answer for this problem.
No comments:
Post a Comment