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Find All Numbers (Posted on 2006-07-25) Difficulty: 3 of 5
Find all four–digit numbers in which the product of the digits is equal to the sum of the digits, and the number itself is divisible by the sum of its digits.

No Solution Yet Submitted by Ravi Raja    
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Some Thoughts re: Confirmed solution; follow-on problem | Comment 6 of 7 |
(In reply to Confirmed solution; follow-on problem by Mike B)

New problems should be submitted through the regular process, but I shall address your challenge here anyway.

For 3-digit numbers, the only digits where the sum and the product of the digits are equal are (0,0,0) and (1,2,3).  The only 3-digit numbers comprised of the digits that are divisible by the sum of the digits are 312 and 132.

For 5-digit numbers, there are four sets of numbers where the sum and product are equal: (0,0,0,0,0), (1,1,1,3,3), (1,1,1,2,5) and (1,1,2,2,2).

  • For (0,0,0,0,0), with a sum and product of 0, there is no solution as one can not divide by 0.
  • For (1,1,2,2,2), with a sum and product of 8, there is only one of the 5-digit numbers evenly divisible by 8: 22112.
  • For (1,1,1,3,3), with a sum and product of 9, all combinations are evenly divisible by 9: 11133, 11313, 11331, 13113, 13131, 13311, 31113, 31131, 31311, 33111. Two of these are palindromic: 13131 and 31113.
  • For (1,1,1,2,5), with a sum and product of 10, there are no 5-digit numbers divisible by 10. 

-- For all the above solutions (except the palindromes), the negative of each number is also a solution.

Edited on July 26, 2006, 1:42 am
  Posted by Dej Mar on 2006-07-26 00:50:23

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