Each of the ID numbers issued to Mr.Cooper and Mr. Duncan is of the form ABCDEFGHIJ, with each of the letters representing a different digit from 0 to 9 inclusively, such that:
(i) BCD is divisible by 2.
(ii) CDE is divisible by 3.
(iii) DEF is divisible by 5.
(iv) EFG is divisible by 7.
(v) FGH is divisible by 11.
(vi) GHI is divisible by 13.
(vii) HIJ is divisible by 17.
Determine the ID numbers issued to each of the gentlemen, given that the ID number of Mr. Cooper is greater than that of Mr. Duncan.
Note: A is not 0, and C is greater than D.
*** While a solution may be trivial with the aid of a computer program, show how to derive it without one.
(In reply to Hmmmmm.... (spoilers)
I came to the same conclusion, Charlie, albeit by brute force.
I started by realizing that F must = 5 in both numbers. It cannot equal 0 because there is no number in the form of 0GH that contains unique digits and is also divisible by 11. From there, I sought out three-digit multiples of 7 that have a 5 in the tens place, then three-digit multiples of 11 that have a 5 in the hundreds place, etc.
By the time I got to the end, I had two numbers:
ABCD357289 (remaining digits: 0, 1, 4, 6)
ABCD952867 (remaining digits: 0, 1, 3, 4)
In both cases, for CDE to be divisible by three, C + D must also be divisible by three. This possible only with 0 and 6 in the first number, 0 and 3 in the second number. Since C is greater than D, and since D must be an even number, we now have:
In both cases, 1 and 4 are the leftover digits, but we don't have enough information to determine in which order they should be placed. I see four possible solutions, if Cooper's number is greater than Duncan's:
Posted by Jyqm
on 2008-08-19 20:30:07