if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 16 ?
"i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 16 = integer , therefore we have 48 / 96 numbers divisible by 8 o e o / 16 = not integer we can not forget multiples of 16 from 1 to 96 we have 6 numbers that are multiple of 16 therefore , 48 / 96 + 6 / 96 = 54 / 96 = 9 / 16 answer : d"
divide(add(divide(96, 2), divide(96, 16)), 96)