Inspired by Nihar and Sandeep

When a non-negative integer xx is divided by 5, we get a remainder yy such that the number xx can be represented as x=q×5+yx=q×5+y. For an explicit example if x=41x=41, then we have x=8×5+1x=8×5+1. And note that the remainder, here, is 1.

If I give you three chances to enter the integer yy, what is the probability that you will get the qu​estion right on the third attempt?

Clarification:

  • You will have to come up with a valid option (a possible remainder which we get after division by 5) each time you enter an integer.

  • The remainder is defined being bounded in the interval [0,5)[0, 5)

×

Problem Loading...

Note Loading...

Set Loading...