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 srebmuN desreveR (Posted on 2010-11-18)
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

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 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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