All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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.)
Solution computer solution | Comment 2 of 4 |
(In reply to Q-1 ? by ed bottemiller)

For part 1, the computer solution is really overkill, as the number is the very first whose square is above 100000.

My part 2 answer differs from ed bottemiller's:

list
   10   for K=1 to 10000
   20    Ks=cutspc(str(K))
   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   ?sum
run
 317     713
 45454546

this last number, 45,454,546, being the answer for part 2.


  Posted by Charlie on 2010-11-18 22:09:55
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information