an integer n between 1 and 99 , inclusive , is to be chosen at random . what is the probability that n ( n + 1 ) will be divisible by 3 ?
"n ( n + 1 ) to be divisible by 3 either n or n + 1 must be a multiples of 3 . in each following group of numbers : { 1 , 2 , 3 } , { 4 , 5 , 6 } , { 7 , 8 , 9 } , . . . , { 97 , 98 , 99 } there are exactly 2 numbers out of 3 satisfying the above condition . for example in { 1 , 2 , 3 } n can be : 2 , or 3 . thus , the overall probability is 2 / 3 . answer : d ."
multiply(divide(divide(99, 3), 99), const_2.0)