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  • Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Show that the cube of a positive integer of the form 6q + r, q is an integer and   r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Answers (1)

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is remainder when we divide A by 5, therefore:
0 \leq r < 6,    r = 0, 1, 2, 3,4,5
A = 6q + r
Here A =  6q + r
A3 = (6q + r)3 = 216q3 + r3 + 3.6q.r (6q + r)
  [\because (a + b)3 = a3 + b3 + 3ab (a + b)]
A3 = (216q3 + 108q2r + 18qr2) + r3                  …(1)
Put r = 0, 1, 2, 3, 4, 5

Case 1:
A = 6q
A2 = 216q3
A3 =6 (36q3)
=6 (m )              

Case 2:
A = 6q + 1
A3 = (216q3 + 108q2 + 18q) + 1
=6(36q3 + 18q2 + 3q) + 1
a3 = 6m + 1            
{m = 36q3 + 18q2 + 3q}

Case 3:
A = 6q + 2
A3 = (216q3 + 216q3 + 72q) + 8
= (216q3 + 216q2 + 72q + 6) + 2
= 6m + 2

Case 4:
A = 6q + 3
A3=216q3 + 324q2 + 162q + 24 + 3
=6(36q3 + 54q2 + 27q + 4) + 3
= 6m + 3  
{m = 36 q3 + 54q2 + 27q + 4}

Case 5:
A = 6q + 4
A3 = (216 q3 + 432 q2 + 288q) + 64
=6(36q3 + 72q2 + 48 q) + 60 + 4
=6(36q3 + 72q2 + 48q + 10) + 4
=6m + 4          
{m = 36q3 + 72q2 + 48q + 10}

Case 6:
A = 6q + 5
A3 = (216 q3 + 540 q2 + 450 q) + 125
= 216 q3 + 540 q2 + 450 q + 120 + 5
= 6 (36q3 + 90q2 + 75 q + 20) + 5
= 6m + 5                     
(where m = 36 q3 + 90q2 + 75q + 20)
Hence, cube of a positive integer of the form 6q + r, q is a integer can be express in the form 6m + r where r = 0, 1, 2, 3, 4 and 5.

 

 

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infoexpert27

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