5 bells begin to toll together at 12 : 00 noon and toll respectively at intervals of 3 , 4 , 5 , 6 , and 7 seconds . not including the toll at 12 : 00 noon , how many more times will all 5 bells toll together before 1 : 00 pm ( i . e . one hour later ) ?
the least common multiple is 2 * 2 * 3 * 5 * 7 = 420 . 3600 seconds / 420 = 8 + remainder . the answer is b .
floor(divide(const_3600, multiply(multiply(multiply(3, 4), 5), 7)))