how many days per calendar year is the month a factor of the day
solution : 124 all 31 days in january , because 1 is a factor of all integers 14 of the 28 days in february ( because every other day is divisible by 2 ) 10 of the 31 days in march ( the multiples of 3 between 3 and 30 ) 7 days in april ( the multiples of 4 between 4 and 28 ) 6 days in may ( the multiples of 5 between 5 and 30 ) 5 days in june ( the multiples of 6 between 6 and 30 ) 4 days in july ( the multiples of 7 between 7 and 28 ) 3 days in august ( the multiples of 8 between 8 and 24 … so , so close if only the month were 32 days long ! ) 3 days in september ( 9 th , 18 th , 27 th ) 3 days in october ( 10 th , 20 th , 30 th ) 2 days in november ( 11 th , 22 nd ) 2 days in december ( 12 th , 24 th ) and there you have it ! as you can see , as you move through the months , your job starts to get easier as you pick up on the patterns and are able to reuse some of your previous insights … this is a very valuable skill to remember on test day ! e
add(floor(divide(add(multiply(const_3, const_10), const_1), multiply(const_3, const_4))), add(floor(divide(multiply(const_3, const_10), add(const_10, const_1))), add(floor(divide(add(multiply(const_3, const_10), const_1), const_10)), add(floor(divide(multiply(const_3, const_10), multiply(const_3, const_3))), add(floor(divide(add(multiply(const_3, const_10), const_1), multiply(const_2, const_4))), add(floor(divide(add(multiply(const_3, const_10), const_1), add(add(add(const_1, const_4), const_1), const_1))), add(floor(divide(add(multiply(const_3, const_10), const_1), add(const_1, const_4))), add(floor(divide(multiply(const_3, const_10), const_4)), add(divide(multiply(const_3, const_10), add(add(const_1, const_4), const_1)), add(floor(divide(add(multiply(const_3, const_10), const_1), const_3)), add(add(multiply(const_3, const_10), const_1), subtract(multiply(const_3, const_10), const_2))))))))))))