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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)

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Solution Analytic Solution to part 2 Comment 4 of 4 |
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 2010-11-19 16:59:04
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