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 Digits 1-9 (Posted on 2002-06-05)
Can you arrange the digits 1-9 in any order so that:
• The number formed by the first two digits is divisible by 2.
• The number formed by the first three digits is divisible by 3.
• The number formed by the first four digits is divisible by 4
• and so on up to nine digits...
•  See The Solution Submitted by Rhonda Wendel Rating: 3.1429 (7 votes)

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 Problem Resolution | Comment 12 of 13 |
(In reply to puzzle solution by K Sengupta)

Let pqrstuvwx = N, where letters p through x represent digits 1 through 9, let pqrst represent a five digit number, and so on. Since 5 divides pqrst , we conclude t=5.

It may be readily observed that, q, s, u and w are all even, and
accordingly  p, r, t, v and x are odd.

Since pqrs and pqrstuvw are divisible by 4 and their ten's digits are odd, we conclude that s and w are 2 and 6 (in some order), and the ten's digits of  q and u are 4 and 8 (in some order). Also, 3 divides p+q+r, 3 divides p+q+r+s+t+u, and 3 divides p+q+r+s+t+u+v+w+x = 45. Accordingly,  3 divides s+t+u  and 3 divides  u+v+x.

This implies inter-alia that, stu is either 258 or 654....(i)

Since 8 divides uvw and u is even, it follows that  that 8 divides gh.....(ii)

Now, pqrstuv is divisible by 7, so that (tuv + p - qrs) is divisible
by  7......(iii)

(i), (ii) and (ii) are simultaneously satisfied, whenever
pqrstuvwx = 381654729.

Consequently, the required number is 381654729.

Edited on June 19, 2007, 2:52 pm
 Posted by K Sengupta on 2007-06-19 05:59:58

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