An eleven digit duodecimal positive integer N is constituted by each of the nonzero digits from 1 to B exactly once, such that N satisfies all the following conditions.
- The sum of the digits 1 and 2 and all the digits between them is equal to the duodecimal number 12.
- The sum of the digits 2 and 3 and all the digits between them is equal to the duodecimal number 23.
- The sum of the digits 3 and 4 and all the digits between them is equal to the duodecimal number 34.
- The sum of the digits 4 and 5 and all the digits between them is equal to the duodecimal number 45.
- The sum of the digits 5 and 6 and all the digits between them is equal to the duodecimal number 56.
If the first digit of N is more than the last digit, determine all possible value(s) that N can assume.
Note: Think of this puzzle as an extension of
An ID Number Problem.
Very similar to the original problem, and the solution is as well:
the base 12 numbers equate to base 10 numbers as following:
12 = 14
23 = 27
34 = 40
45 = 53
56 = 66
Since all 11 digits in total add up to 66 (and the first digit is higher), we know it looks like:
6 _ _ _ _ _ _ _ _ _ 5
since from 4 to 5, we need the remaining numbers to add up to 53 (and 6 is not included), we need to remove a 7 as well... therefore:
6 7 4 _ _ _ _ _ _ _ 5
since from 3 to 4, we need the remaining numbers to add up to 40 (and 5, 6, and 7 are not included), we need to remove an 8... therefore:
6 7 4 _ _ _ _ _ 3 8 5
since from 2 to 3, we need the remaining numbers to add up to 27 (and 4, 5, 6, 7, and 8 are not included), we need to remove a 9... therefore:
6 7 4 9 2 _ _ _ 3 8 5
since form 1 to 2, we need the remaining numbers to add up to 14 (and 3, 4, 5, 6, 7, 8, 9) are not included), we need to remove the A... therefore:
6 7 4 9 2 _ 1 A 3 8 5
leaving only B to place, so the only value for N could be:
67492B1A385
solution
