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
The reversals of 1 to 9 are 1 to 9 and add to 9*10/2 = 45
The reversals of 10 to 99 are 1 to 99 except the multiples of 10.
These add to 99*100/2  10*9*10/2 = 4500.
The reversals of 100 to 999 are 1 to 999 except multiples of 10.
These add to 999*1000/2  10*99*100/2 = 450000.
The reversals of 1000 to 9999 are 1 to 9999 except multiples of 10.
These add to 9999*10000/2  10*999*1000/2 = 45000000.
The reversal of 10000 itself is 1.
For a total of 45454546.

Posted by Jer
on 20101119 16:59:04 