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 4?
Choose the correct choice from the following choices: A) 1 / 4 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 / 4</gadget>
<output>24</output>

<gadget id="calculator">48 + 24</gadget>
<output>72</output>

<gadget id="calculator">72 / 96</gadget>
<output>3/4 = around 0.75</output>

<result>E</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 4?
divide(add(divide(96, 2), divide(96, 4)), 96)
"i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 4 = integer , therefore we have 48 / 96 numbers divisible by 8 o e o / 4 = not integer we can not forget multiples of 8 from 1 to 96 we have 24 numbers that are multiple of 4 therefore , 48 / 96 + 24 / 96 = 72 / 96 = 3 / 4 answer : e"