Question: What is the remainder when T=(1!)!
Question: What is the remainder when T=(1!)!^3 + (2!)^3 + (3!)^3 + … + (432!)^3 is divided by 144?
Answer/Solution
81
Steps/Work
144 =12 *12
(1!)^3=1
2!=2, therefore (2!)^3=2*2*2=8
3!=6, therefore (3!)^3=6*6*6=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 T=(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3 is divided by 144 is same as, when 1+8+216 =225 is divided by 144
now remainder when 225 is divided by 144 is 81. hence answer should be D
(1!)^3=1
2!=2, therefore (2!)^3=2*2*2=8
3!=6, therefore (3!)^3=6*6*6=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 T=(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3 is divided by 144 is same as, when 1+8+216 =225 is divided by 144
now remainder when 225 is divided by 144 is 81. hence answer should be D