Each of the ID numbers issued to Mr. Abbado and Mr. Brenner consists of the digits 1 to 9 with each digit occurring exactly once such that:
(i) The sum of the digits 1 and 2 and all the digits between them is 12.
(ii) The sum of the digits 2 and 3 and all the digits between them is 23.
(iii) The sum of the digits 3 and 4 and all the digits between them is 34.
(iv) The sum of the digits 4 and 5 and all the digits between them is 45.
Given that the ID number of Mr. Abbado is greater than that of Mr. Brenner, determine the ID numbers issued to each of the two gentlemen.
The sum of the digits 1  9 is 45, so clearly the digits 4 and 5 are in either the first or last place on each ID. Let's begin with Mr. Abbado's ID, which begins with 5 since it is great than Mr. Brenner's.
5 _ _ _ _ _ _ _ 4
If the sum all the digits between and including 3 and 4 is 34, then the digits outside this string total 11. 5 is already excluded, so we need digits totalling 6. The combinations 1+2+3 and 2+4 are impossible, as 3 and 4 are needed for the string totalling 34. 5+1 is also excluded, since 5 is already accounted for. So the only other number that appears outside the 3  4 string is 6 itself:
5 6 3 _ _ _ _ _ 4
If the sum of all the digits between and including 2 and 3 is 23, then the digits outside this string total 22. 5, 6 and 4  totalling 15  are already excluded, so we need didigts totalling 7. The combinations 1+2+4, 1+6, 2+5 and 3+4 are all impossible, since 4, 5 and 6 are already accounted for. So once again the only number we need is 7 itself:
5 6 3 _ _ _ 2 7 4
The only digits left are 1, 8 and 9. If the sum of all the digits between and including 1 and 2 is 12, then 9 must be the digit between them:
5 6 3 8 1 9 2 7 4
And we've got Mr. Abbado's number! To get Mr. Brenner's, of course, we just need to reverse the string:
4 7 2 9 1 8 3 6 5
Nice puzzle!

Posted by Jyqm
on 20070804 15:17:46 