Some positive integers n have the property that the sum
[ n + reverse(n) ] consists entirely of odd (decimal) digits.
For instance, 36 + 63 = 99 and 409 + 904 = 1313.
We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not allowed in either n or reverse(n).
There are 120 reversible numbers below one-thousand.
a. Evaluate how many reversible numbers are there
below 10k, k=2,3... up to 6 or 7 .
b. Analyze the results, aiming to find the relation (i.e. approximate function) between N(k) and k.
Source: Project Euler, modified.
(In reply to
Part b researched solution by Charlie)
Note that even in the verbal description of the algorithm, the mathblogger makes a mistake:
3 and 7 have the same kind of argumentation, and I have carried it out on 11 as well just to check. So for a number n = 4k + 3 for some integer k we have that the middle digit and the outer pair gives us 5 and 20 options.
Then we have sets of internal pairs which gives us 20 and 25 solutions. So that means we can generalize it to 20*5*(20*25)^(k-1) = 100*500^(k-1)
But for n=3, k=0. The mathblogger's formula gives 100*500^(-1) = .2, obviously not the number of satisfactory 3-digit n's, which he himself lists as 100 earlier on his page. The formula is correct for n = 4k - 1, the same set of cases but with k assigned differently. With this assignment, 100*500^(k-1) comes out to 100*500^0 = 100, the correct value. My modified program takes this into consideration to get answers that match reality.
Edited on August 8, 2015, 8:21 am
|
|
Posted by Charlie
on 2015-08-08 08:20:18 |
re: Part b researched solution
