Prove that for each positive integer n, there exists an n-digit number having only odd digits and divisible by 5^n
(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)
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Posted by Charlie
on 2013-05-13 08:55:27 |
re: Trials
