Question: What is the remainder when Q=(1!)!^3 + (2!)^3 + (3!)^3 + … + (432!)^3 is divided by 144?  Answers:  A) 0  B) 25  C) 49  D) 81  E) 100
144 =12 *12   (1!)^3=1   2!=2, therefore (2!)^3=2*2*2= 
<gadget id="calculator">2*2*2</gadget>
<output>8</output>
8   3!=6, therefore (3!)^3=6*6*6= 
<gadget id="calculator">6*6*6</gadget>
<output>216</output>
216   4!=24 therefore (4!)^3= 24*24*24 which is completely divisible by 144   thus from 4! onwards each term will be divisible by 144 and will leave the remainder of 0   now the remainder when Q=(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3 is divided by 144 is same as, when 1+8+216= 
<gadget id="calculator">1+8+216</gadget>
<output>225</output>
225 is divided by 144   now remainder when 225 is divided by 144 is 81. hence answer should be 81
<result>D</result>
Question: What is the remainder when Q=(1!)!^3 + (2!)^3 + (3!)^3 + … + (432!)^3 is divided by 144?