Question: Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
| Class | 10th |
| Subject | Maths |
| Chapter | Real Numbers |
| Exercise | 1.1 |
| Previous Question | Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. |
Solution:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, putting the value of r, we get,
x = 3q
or
x = 3q + 1
or
x = 3q + 2
Now, by taking the cube of all the three above expressions, we get,
Case (i): When r = 0, then,
3 = (3)
3 = 27
3 = 9(3
3
) = 9m; where m = 3
3
Case (ii): When r = 1, then,
3 = (3 + 1)
3= (3)
3 + 1
3 + 3 × 3q × 1 (3q + 1) = 27
3 + 1 + 27
2 + 9q
Taking 9 as common factor, we get,
3 = 9(3
3 + 3
2 + ) + 1
Putting (3
3 + 3
2 + ) = m, we get,
3 = 9 + 1
Case (iii): When r = 2, then,
3 = (3 + 2)
3= (3)
3 + 2
3 + 3 × 3q × 2 (3q + 2) = 27
3 + 54
2 + 36 + 8
Taking 9 as common factor, we get,
3 = 9(3
3 + 6
2 + 4) + 8
Putting (3
3 + 6
2 + 4) = , we get,
3 = 9 + 8
Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is
of the form 9m, 9m + 1 or 9m + 8.
