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Who is afraid of 7 (or 23) ? (Posted on 2010-11-08) Difficulty: 3 of 5
No matter how we permute the 3 digits 1,2 4 , none of the 6 three- digit numbers (124,142,214,241,412,421) is a multiple of 7.

Clearly, out of 84 possible triplets created by choosing distinct non-zero digits
there must be several like this, i.e. generating a 6-pack of three-digit numbers,
non divisible by 7.

1. List those triplets.
2. Similar question re:4 digits and divisibility by 23.

No Solution Yet Submitted by Ady TZIDON    
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Comments: ( Back to comment list | You must be logged in to post comments.)
close to Charlie's list | Comment 4 of 5 |

I posted my reply for part 1 earlier, and it agrees with Charlie's for that part (30 triples).  I posted my general reply for part 2 recently, but after Charlie had posted, but before I examined it.

Checking my list (for part 2), I find I had the first 40 of Charlie's list for part 2, but did not get his last four (4567, 4569, 4578, 4589).  I'll have to give those another look.  Possibly I erred in coding so as to exclude a set with 4 as the lowest seed.     

(Edit) (note to Charlie)

I haven't found my error: forcing your last four (instead of the loops logic) did show them.  I was using only three decimal places when I divided each of the 24 variants by 23 to check for exact divisibility -- I don't have that handy MOD function -- but that does not seem to have been the problem.

 

Edited on November 8, 2010, 7:50 pm
  Posted by ed bottemiller on 2010-11-08 17:58:55

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