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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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Solution computer solution | Comment 2 of 5 |

Part 1:

DEFDBL A-Z
CLS
FOR a = 1 TO 7
FOR b = a + 1 TO 8
FOR c = b + 1 TO 9
   n1 = a * 100 + b * 10 + c
   n2 = a * 100 + c * 10 + b
   n3 = b * 100 + a * 10 + c
   n4 = b * 100 + c * 10 + a
   n5 = c * 100 + b * 10 + a
   n6 = c * 100 + a * 10 + b
   IF n1 MOD 7 > 0 THEN
   IF n2 MOD 7 > 0 THEN
   IF n3 MOD 7 > 0 THEN
   IF n4 MOD 7 > 0 THEN
   IF n5 MOD 7 > 0 THEN
   IF n6 MOD 7 > 0 THEN
     PRINT a; b; c
   END IF
   END IF
   END IF
   END IF
   END IF
   END IF
NEXT
NEXT
NEXT

finds

 1  2  4
 1  2  5
 1  2  9
 1  3  6
 1  3  8
 1  4  6
 1  4  8
 1  4  9
 1  5  9
 1  6  7
 1  7  8
 2  3  4
 2  4  8
 2  5  6
 2  5  7
 2  5  8
 2  6  9
 2  7  9
 2  8  9
 3  4  5
 3  4  7
 3  4  9
 3  5  6
 3  6  8
 4  6  8
 4  8  9
 5  6  9
 5  7  9
 5  8  9
 6  7  8

Part 2:

DECLARE SUB permute (a$)
DEFDBL A-Z
CLS
FOR a = 1 TO 6
FOR b = a + 1 TO 7
FOR c = b + 1 TO 8
FOR d = c + 1 TO 9

  n$ = LTRIM$(STR$(a)) + LTRIM$(STR$(b)) + LTRIM$(STR$(c)) + LTRIM$(STR$(d))
  h$ = n$
  good = 1
  DO
   n = VAL(n$)
   IF n MOD 23 = 0 THEN good = 0: EXIT DO
   permute n$
  LOOP UNTIL n$ = h$
  IF good THEN PRINT a; b; c; d

NEXT
NEXT
NEXT
NEXT


(Listing for permute shown elsewhere on Perplexus.)

finding:

 1  2  3  4
 1  2  3  5
 1  2  3  6
 1  2  3  7
 1  2  4  9
 1  2  6  7
 1  2  6  8
 1  2  7  8
 1  3  4  9
 1  3  5  6
 1  3  5  8
 1  3  7  8
 1  3  8  9
 1  4  5  8
 1  4  8  9
 1  5  6  7
 1  5  6  9
 1  5  7  9
 1  6  8  9
 2  3  4  9
 2  3  5  7
 2  3  5  9
 2  3  7  9
 2  4  5  9
 2  4  6  7
 2  4  6  8
 2  5  6  9
 2  5  8  9
 2  6  8  9
 2  7  8  9
 3  4  5  6
 3  4  5  7
 3  4  6  7
 3  5  6  8
 3  5  7  8
 3  7  8  9
 4  5  6  7
 4  5  6  9
 4  5  7  8
 4  5  8  9
 
 If the second part had stuck with 7, instead of 23, we'd have the more exclusive:
 
 1  2  3  8
 1  3  8  9
 2  4  6  9


  Posted by Charlie on 2010-11-08 16:22:21
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