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Positive Integer, Odd Digits, Divisibility (Posted on 2013-05-12) Difficulty: 4 of 5
Prove that for each positive integer n, there exists an n-digit number having only odd digits and divisible by 5^n

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.6667 (3 votes)

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Hints/Tips re: Trials | Comment 3 of 5 |
(In reply to Trials by Salil)

The following extended table shows the odd-digited number growing from right to left, maintaining the same trailing digits as the preceding minimum such number.

 n       5^n               n-digit number          multiple of 5^n
 1       5                 5                       1     
 2       25                75                      3     
 3       125               375                     3     
 4       625               9375                    15     
 5       3125              59375                   19     
 6       15625             359375                  23     
 7       78125             3359375                 43     
 8       390625            93359375                239     
 9       1953125           193359375               99     
 10      9765625           3193359375              327    
 11      48828125          73193359375             1499     
 12      244140625         773193359375            3167     
 13      1220703125        3773193359375           3091     
 14      6103515625        73773193359375          12087     
 15      30517578125       773773193359375         25355     
 16      152587890625      5773773193359375        37839     
 17      762939453125      15773773193359375       20675     
 18      3814697265625     515773773193359375      135207     
 19      19073486328125    7515773773193359375     394043     
 20      95367431640625    97515773773193359375    1022527     
 21      476837158203125   397515773773193359375   833651    
 22      2384185791015625  7397515773773193359375  3102743             
 23      11920928955078125 37397515773773193359375 3137131             

produced by the following:
 
   10   for N=1 to 23
   20     B=5^N:Num=B:Mini=10^(N-1):Maxi=10^N-1:Found=0
   30     repeat:Good=0
   40      if Num>Mini then
   50        :S=cutspc(str(Num))
   60        :Good=1
   70        :for I=1 to len(S)
   80          :if val(mid(S,I,1))@2=0 then Good=0:endif
   90        :next
  100        :if Good then print N,B,Num,Num//B:endif
  110      Num=Num+B
  120     until Good=1 or Num>Maxi
  130   next N

(with columns aligned manually)


  Posted by Charlie on 2013-05-13 08:55:27
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