According to Gregorian Calender(which is in use now, in many countries),decide whether a given year is a leap year or not. Give 'true' if the given year is a leap year and 'false',if not.
Note: In the Gregorian calendar 3 criteria must be taken into account to identify leap years:
The year is evenly divisible by 4; If the year can be evenly divided by 100, it is NOT a leap year, unless; The year is also evenly divisible by 400. Then it is a leap year.
Solution Stats
Problem Comments
3 Comments
Solution Comments
Show comments
Loading...
Problem Recent Solvers194
Suggested Problems
-
114313 Solvers
-
Find the peak 3n+1 sequence value
2568 Solvers
-
431 Solvers
-
middleAsColumn: Return all but first and last element as a column vector
647 Solvers
-
658 Solvers
More from this Author1
Problem Tags
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Kumar, in problem #4 there is one to many right parenthesis. George...
Actually I just realized it is the "isequal" is missing in test #4.
Thanks George.Am extremely sorry for that.I just corrected it.