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Divisible (Posted on 2011-09-30) Difficulty: 3 of 5
Five single-digit positive integers appear in a sequence.

The sum of the first two is divisible by the second, but not by the first.
The sum of the first three is divisible by the third, but not by the first or the second.
The sum of the first four is divisible by the fourth, but not by the first, second or third.
The sum of the first five is divisible by the fifth, but not by the first, second, third or fourth.

What are the five integers, in order?

See The Solution Submitted by Charlie    
Rating: 4.0000 (3 votes)

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Solution Eureka! (spoiler) | Comment 1 of 6
This is fairly easy to do using pencil and paper:

Note: 
1) The 5 digits must be different
2) The 1st, 2nd, 3rd and 4th digits cannot be 1, because 1 divides the sum of the 5 digits.
3) The first digit must be a multiple of the 2nd digit

The only possibility for the first two digits are therefore
  42
  62
  82
  93
  84

4) The third digit must divide the sum of the first two, but it cannot be a multiple of the 2nd digit.
  4 + 2 = 6, which makes 423 a possibility for the 1st 3 digits
  6 + 2 = 8, so nothing works as a 3rd digit
  8 + 2 = 10, which makes 825 a possibility
  9 + 3 = 12, which makes both 932 and 934 possibilities
  8 + 4 = 12, which makes 843 a possibility

 recapping, the first 3 digits must be
  423
  825
  932
  934 or
  843

5) The 4th digit must divide the sum of the first three. but it cannot be a multiple of any of the first 3
  4+2+3 = 9, so nothing works as a 4th digit
  8+2+5 = 15, but 3 cannot be the 4th digit, because 2 divides 18
  9+3+2 = 14, so 9327 are a possibility for the 1st 4 digits
  9+3+4 = 16, but 2 does not work because 3 divides 18
  8+4+3 = 15, but 4 does not work because 4 divides 20

recapping, the first 4 digits can only be 9327

6) The 5th digit must divide the sum of the first 4
    
  9+3+2+7 = 21, so the first 5 digits can only be 93271

Final answer
 


  Posted by Steve Herman on 2011-09-30 20:41:54
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