how do we prove (n)whole cube - (n)whole square is divisible by 6? Can you explain the procedure as well?

Asked by  | 12th Sep, 2012, 09:26: PM

Expert Answer:

Please cross-check, the question should be :
"prove that n3 - n is divisible by 6, for any positive integer n".
 
The solution is as follows:
 

Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n.

When n = 6m,

n3 - n = (6m)3 - 6m = 216 m3 - 6m

= 6m(36m2 - 1)

= 6q, where q = m(36m2 -1)

n3 - n is divisible by 6

 

When n = 6m + 1,

n3 - n = n(n2 - 1) = n (n - 1) (n + 1)

= (6m + 1) (6m) (6m + 2)

= 6m(6m + 1) (6m + 2)

= 6q, where q = m(6m + 1) (6m + 2)

n3 - n is divisible by 6

 

When n = 6m + 2,

n- n = n (n - 1) (n + 1)

= (6m + 2) (6m + 1) (6m + 3)

= (6m + 1) (36 m2 + 30m + 6)

= 6m (36 m2 + 30m + 6) + 1 (36m2 + 30m + 6)

= 6[m (36m2 + 30m + 6)] + 6 (6m2 + 5m + 1)

= 6p + 6q,

where p = m (36m2 + 30m + 6)

q = 6m2 + 5m + 1

n3 - n is divisible by 6

 

When n = 6m + 3

n3 - n = (6m + 3)3 - (6m + 3)

= (6m + 3) [(6m + 3)2 - 1]

= 6m [6m + 3)2 - 1] + 3 [(6m + 3)2 - 1]

= 6 [m [(6m + 3)2 - 1] + 3 [36m2 + 36m + 8]

= 6 [m [(6m + 3)2 - 1] + 6 [18m2 + 18m + 4]

= 6p + 3q,

where p = m[(6m + 3)2 - 1]

q = 18m2 + 18m + 4

n3 - n is divisible by 6

 

When n = 6m + 4

n3 - n = (6m + 4)3 - (6m + 4)

= (6m + 4) [(6m + 4)2 - 1]

= 6m [(6m + 4)2 - 1] + 4 [(6m + 4)2 - 1]

= 6m [(6m + 4)2 - 1] + 4 [36m2 + 48m + 16 - 1]

= 6m [(6m + 4)2 - 1] + 12 [12m2 + 16m + 5]

= 6p + 6q,

where p = m [(6m + 4)2 - 1]

q = 2 (12 m2 + 16m + 5)

n3 - n is divisible by 6

When n = 6m + 5

n3 - n = (6m + 5) [(6m + 5)2 - 1]

= 6m [(6m + 5)2 - 1] + 5 [(6m + 5)2 - 1]

= 6m [(6m + 5)2 - 1] + 5 [36m2 + 60m + 24]

= 6p + 30q

= 6 (p + 5q),

where p = m [(6m + 5)2 - 1]

q = 6m2 + 10m + 4

n3 - n is divisible by 6

Hence, n3 - n is divisible by 6, for any +ve integer n.


Answered by  | 12th Sep, 2012, 10:41: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.