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

Question – Use Euclid division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8

Solution: Let x be any positive integer, which gives q as the quotient and r is the remainder when it is divided by 9.
Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,
a = 9q + r, 0 ≤ r < 9 we get,
So possible values of r are 0, 1, 2, 3, 4, 5, 6, 7, 8.

Now, when r = 0, x = 9q
Therefore, x³ = (9q)³ = 9 x 81q³ = 9m, where m = 81q³
When r = 1, x = 9q + 1
x³ = (9q + 1)³ = (9q)³ + 3.(9q)².1 + 3.9q.1² + 1³
9(81q³ + 3 x 9q² + 3q) + 1
= 9m + 1, where m = 81q³ + 3 x 9q² + 3q

When r = 2, x = 9q + 2
Hence x³ = (9q + 2)³
= (9q)³ + 3.(9q)².2 + 3 x 9q.2² + 2³
= 9(81q³ + 3.9.2q² + 3q.2²) + 8
= 9m + 8, where m = 81q³ + 3.9.2q² + 3q.2²

When r = 3, x = 9q + 3
Hence x³ = (9q + 3)³
= (9q)³ + 3.(9q)².3 + 3.9q.3² + 3³
= 9(81q³ + 3.9.q².3 + 3q.3²) + 27
= 9(81q³ + 3.9.q².3 + 3q.3² + 3)
= 9m, where m = 81q³ + 3.9.q².3 + 3q.3² + 3

When r = 4, x = 9q + 4
Hence x³ = (9q + 4)³
= (9q)³ + 3.(9q)².4 + 3.9q.4² + 4³
= 9(81q³ + 3.9.q².4 + 3q.4²) + 64
= 9(81q³ + 3.9.q².4 + 3q.4² + 7) + 1
= 9m + 1, where m = 81q³ + 3.9.q².4 + 3q.4² + 7

Continuing this process for r = 5, 6, 7 and 8, we get that x³ is of form 9m or 9m + 1 or 9m + 8.
Therefore, the cube of any positive integer is either of form 9m or 9m + 1 or 9m + 8, for some integer m.

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