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srebmuN desreveR (Posted on 2010-11-18) Difficulty: 3 of 5
Let us read a number k from right to left and call the result RVR(k); e.g. RVR(1)=1 , RVR(45)=54 , RVR(1200000000)=21.

1.Find the smallest positive integer k for which RVR(k) > (k^2 mod 100000).
2. Find the sum of the 10000 RVRs i.e. for all integers from 1 to 10000 inclusive.

Based upon problems from projectkhayyam.blogspot.com

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: computer solution | Comment 3 of 4 |
(In reply to computer solution by Charlie)

I can replicate ed's answer of  499950001 with this program:

list
   10   for K=1 to 10000
   20    Ks=cutspc(str(K)):Ks=right("00000000"+Ks,5)
   30    Kr=0
   40    for I=len(Ks) to 1 step -1
   50       Kr=10*Kr+val(mid(Ks,I,1))
   60    next I
   70    if Kr>(K^2)@100000 then if Flag=0 then print K,Kr:endif:Flag=1
   75    Sum=Sum+Kr
   80   next K
   90   print Sum
OK
run
 1       10000
 499950001
OK

but that requires that the reverse of, say, as shown, 1 is 10000; that is, using enough leading zeros to make a 5-digit number before reversing it, so the reverse of, say, 45, would be 54000, rather than the 54 shown in the example.


  Posted by Charlie on 2010-11-18 22:19:48
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