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## Discussion of Problem 1209. 1, 10, 100, 1000...

1s are located at (i-th Tri-angular number + 1)th position
Posted by A. A. Noman Ansary 23 Dec 2020 01:12
Number of zeros in bracket:  0     1     2      3      4
Sequence                  : (1)   (10)  (100)  (1000)  (10000)
1 is found at             : T0+1   T1+1  T3+1   T4+1    T5+1
Value of Ti+1             : 0+1    1+1   3+1    6+1     10+1

The i-th triangular number is the sum of the i natural numbers from 1 to i.

Ti = i(i+1)/2

i-th '1' is located at (Ti+1)-th position.
Let, z = (Ti + 1) = (i(i+1)/2) + 1
-> z = (i^2 + i + 2)/2
-> 2z = i^2 + i + 2
-> (1)x(i^2) + (1)x(i^1) + (-2(z-1)x(i^0) = 0
So, now if we solve for i using quadratic formula,
i = (-1 +- squareRoot(8z-7))/2 [do calculation on your own]
z = {1,2,4,7,11,...}
plugin these values and you will find that squareRoot(8z-7) = an integersquare number for any z. So, a number belongs to z only and if only squareRoot(8z-7) can produce an integer. And here z is the position number where the digits are 1.

[code deleted]

Edited by author 24.12.2020 19:00

Edited by moderator 14.02.2021 18:16