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Topic: A duodecimal digit arrangement and integer puzzle (Read 431 times) |
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K Sengupta
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A duodecimal digit arrangement and integer puzzle
« on: Apr 24th, 2009, 11:48pm » |
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An eleven digit positive duodecimal integer T is constituted by each of the nonzero duodecimal digits from 1 to B exactly once, such that T satisfies all the following conditions:
(A) The sum of the digits 1 and 2 and all the digits between them is equal to the duodecimal number 12.
(B) The sum of the digits 2 and 3 and all the digits between them is equal to the duodecimal number 23.
(C) The sum of the digits 3 and 4 and all the digits between them is equal to the duodecimal number 34.
(D) The sum of the digits 4 and 5 and all the digits between them is equal to the duodecimal number 45.
(E) The sum of the digits 5 and 6 and all the digits between them is equal to the duodecimal number 56.
Given that the first digit of T is less than the last digit, determine all possible value(s) that T can assume.
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pex
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Re: A duodecimal digit arrangement and integer puz
« Reply #1 on: Apr 29th, 2009, 5:49am » |
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Nice one.
| hidden: | Note that 112 + 212 + ... + B12 = 5612. Thus, (E) is equivalent to "The digits 5 and 6 are located in the first and last positions". Given that the first digit of T is less than the last digit, T must be of the form
5 _ _ _ _ _ _ _ _ _ 6.
Now proceed to (D). It says that the sum of the digits not in the block between 4 and 5 (inclusive) is 5612 - 4512 = 6610 - 5310 = 1310. One of those digits is clearly a 6; this leaves us with the problem of expressing 13 - 6 = 7 as a sum of distinct positive integers, none of which is 4, 5, or 6. It is clear that the only possibility is that the only other "missing" digit is 7. Therefore, we update the form of T to
5 _ _ _ _ _ _ _ 476.
By similar reasoning, it follows from (C) that the digits positioned between 5 and 3 need to sum to 8 without using 3, 4, 5, 6, and 7; again, the only option is the single digit 8 and the form of T is further restricted to
583 _ _ _ _ _ 476.
Continuing, the use of (B) leads to
583 _ _ _ 29476,
and applying (A), we obtain the following unique solution: T equals
583A1B29476. |
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