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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 \(a\).

 Apr 5, 2019
 #1
avatar+6045 
+3

appears to be 15049

 Apr 5, 2019
 #2
avatar+28224 
+5

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

 

.

 Apr 5, 2019
edited by Alan  Apr 5, 2019
 #3
avatar+6045 
+4

I checked it over first 10 million integers.

 

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

 

and remain there.

Rom  Apr 5, 2019
 #4
avatar+28224 
+5

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

Alan  Apr 5, 2019

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