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 About two and seven (Posted on 2014-03-19)

1. What is the greatest number, consisting of distinct even digits only, divisible by 72?
2.What is the smallest 7-digit number (abcdefg) divisible by 72?
The digits a,b,c ... etc are distinct and none of them is 2 or 7.

 No Solution Yet Submitted by Ady TZIDON No Rating

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1. To be divisible by 72, it must be divisible by 9.  Thus the sum of its digits must be divisible by 9.  The only combination of distinct even digits that sums to 9 is (0,4,6,8).

To be divisible by 72, it must also be divisible by 8.  It is easy to observe that the largest possible number created from these digits, 8640, is divisible by 8, so this must be the solution.

2. Using similar reasoning as above, the sum of the digits of the number must be divisible by 9.  The sum of the 8 remaining digits (after removing 2 and 7) is divisible by 9, so the one digit we need to exclude must be either 0 or 9.  Since we're attempting to find the smallest possible solution, we'll exclude the 9.  Thus the set of digits in the solution is (0,1,3,4,5,6,8).

The number must also be divisible by 8, which is true if the number formed by the rightmost three digits is divisible by 8.  To the greatest extent possible, we want the digits of abcdefg to be increasing from left to right, and it turns out that 568 is divisible by 8, so the solution must be 1034568 (assuming we can't place 0 in the leading spot).

 Posted by tomarken on 2014-03-19 09:36:41

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