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 6? Options.
A) 2 / 3	B) 3 / 8	C) 1 / 2	D) 5 / 8	E) 3 / 4
<gadget id="calculator">96 / 2</gadget>
<output>48</output>

<gadget id="calculator">96 / 6</gadget>
<output>16</output>

<gadget id="calculator">48 + 16</gadget>
<output>64</output>

<gadget id="calculator">64 / 96</gadget>
<output>2/3 = around 0.666667</output>

<result>A</result>
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 6?
divide(add(divide(96, 2), divide(96, 6)), 96)
i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 6 = integer , therefore we have 48 / 96 numbers divisible by 6 o e o / 6 = not integer we can not forget multiples of 6 from 1 to 96 we have 16 numbers that are multiple of 8 therefore , 48 / 96 + 16 / 96 = 64 / 96 = 2 / 3 answer : a