Let the following number denote P, where the digits are formed by concatenating all possible permutations of 10 distinct digits in increasing order (0123456789,0123456798, ...,9876543210).
P = 1234567890123456798...9876543210
Is P divisible by 11?
Even positions within the substrings are also even positions in the full concatenation, and the same is true of the odd positions. Counting the initial zero does not affect the result, and it's easier to consider it present in the concatenation.
There are 3,628,800 permutations of the ten digits and therefore 36,288,000 digits in the full concatenation. Most importantly, though, at each digit position in the groups of ten each of the ten digits will occur the same 362,880 times. Regardless of the number, all digit positions have the same total, so the sum in the odd positions is the same as the sum of the even positions, and are thus both congruent to zero mod 11 and so is the difference between the two.
Yes, P is divisible by 11.

Posted by Charlie
on 20190404 10:58:06 