How many positive integers \(n\) satisfy the inequality \( \left\lceil \frac{n}{101} \right\rceil + 1 > \frac{n}{100} \, ?\)Recall that \(\lceil a \rceil\) is the least integer that is greater than or equal to \(a\).

appears to be 15049

I get 19999.

19999/101 = 198.00990099... ; ceil(19999/101) = 199; ceil(19999/101) + 1 = 200; 19999/100 = 199.99

ceil(20000/101)+1 = 200; 20000/100 = 200

.

I checked it over first 10 million integers.

It appears to transition from 15048 at 19998 to 15049 at 19999

and remain there.

You are right Rom. I didn't check them all carefully enough!