if an integer e is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that e ( e + 1 ) ( e + 2 ) will be divisible by 8?  Select.
A) 1 / 4 B) 3 / 8 C) 1 / 2 D) 5 / 8 E) 3 / 4
<gadget id="calculator">96 / 8</gadget>
<output>12</output>

<gadget id="calculator">12 / 8</gadget>
<output>3/2 = around 1.5</output>

<gadget id="calculator">(3/2) * 2</gadget>
<output>3</output>

<gadget id="calculator">3 + 2</gadget>
<output>5</output>

<gadget id="calculator">5 / 8</gadget>
<output>5/8 = around 0.625</output>

<result>D</result>
if an integer e is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that e ( e + 1 ) ( e + 2 ) will be divisible by 8?
divide(add(multiply(divide(divide(96, 8), 8), 2), 2), 8)
for e total numbers 8 * 12 there are 12 numbers divisible by 8 - > 3 * 12 ( if 8 is an example - ( 6 , 78 ) , ( 7 , 89 ) , ( 8 , 910 ) ) and 12 numbers divisible by 4 but not divisible by 8 - > 2 * 12 ( if 4 is an example ( 2 , 34 ) and ( 4 , 56 ) ) the answer 5 / 8 - > d