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All Odds (Posted on 2014-02-24) Difficulty: 2 of 5
Let n be an integer greater than 1. If all digits of 9997*n are odd, find the smallest possible value of n.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution computer solution | Comment 2 of 3 |

DEFDBL A-Z
CLS
FOR n = 3 TO 9999 STEP 2
  pr = n * 9997
  p$ = LTRIM$(STR$(pr))
  good = 1
  FOR i = 1 TO LEN(p$)
    IF INSTR("02468", MID$(p$, i, 1)) > 0 THEN good = 0
  NEXT
  IF good THEN PRINT n, pr
NEXT

finds

  n            9997*n
3335          33339995
3341          33399977
3355          33539935
3361          33599917
3401          33999797
3535          35339395
3541          35399377
3555          35539335
3561          35599317
3601          35999197
4001          39997997
5335          53333995
5341          53393977
5355          53533935
5361          53593917
5401          53993797
5535          55333395
5541          55393377
5555          55533335
5561          55593317
5601          55993197
6001          59991997

  Posted by Charlie on 2014-02-24 12:34:01
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