Question

Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8. 

(OR) 

Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.

Answer 1

Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r; 

now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8 

So a = bq + r 

⇒ a = 9q + r (for b = 9)

now cube of a = a³ + (9q + r)³ 

= (9q)³ + 3.(9q) ³r + 3. 9q.r + r ³

= 9³q³ + 3.9² (q²r) + 3.9(q.r) + r ³

= 9[9².q³ + 3.9.q² r + 3.q.r] + r³

a³ = 9m + r³ (where ‘m’ = 9² q³ + 3.9.q2r + 3.q.r) 

if r = 0 ⇒ r³ = 0 then a³ = 9m + 0 = 9m 

and for r = 1 ⇒ r³ = l³ then a³ = 9m + 1 

and for r = 2 ⇒ r³ = 2³ then a³ = 9m + 8 

for r = 3 ⇒ r³ , = 3³ ⇒ a³ = 9m + 27 = 9(m) where m = (9m +3)

for r = 4 ⇒ r³ = 4³ ⇒ a³ = 9m + 64 = (9m + 63) + 1 = 9m + 1 

for r = 5 ⇒ r³ = 125 ⇒ a³ = 9m + 125 = (9m + 117) + 8 = 9m + 8 

for r = 6 ⇒ r³— 216 ⇒ a³ = 9m + 216 = 9m + 9(24) = 9m 

for r = 7 ⇒ r³ = 243 

⇒ a³ = 9m + 9(27) = 9m 

for r = 8 ⇒ r³ = 512 

⇒ a³ = 9m + 9(56) + 8 = 9m + 8

So from the above it is clear that a³ is either in the form of 9m or 9m + 1 or 9m + 8. 

Hence proved.