Chapter 1: Real Numbers

Q&A -Ask Doubts and Get Answers

Q
Real Numbers Solutions

Question:

Use Euclidโ€™s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

Proof: Suppose a is any positive integer and b = 3. By using Euclid’s algorithm we get a = 3q + r , for some integer q \geq 0, and  r = 0,1,2 because of 0 \leq  r  < 3.

Now substituting the value of r, we get,

If r = 0, then a = 3q

Similarly, for  r = 1 and 2, the value of a is 3q + 1 and 3q + 2 respectively.

Now, we can find the cube of values  3q , 3q + 1 and 3q + 2.

So, we have the following cases.

Case I: When a = 3q

(3q)3 = 27q3 = 9 (3q3) = 9m, if we divide by 9, then the remainder is 0, where m = 3q3.

Case II: When a = 3q + 1

(3q + 1)3

= 27q3 + 27q2 + 9q + 1     [ \because (x + y)=  x+ 3x2y + 3xy+ y]

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, if we divide by 9, then the remainder is 1 and the value of m is q(3q2 + 3q + 1).

Case III: When a = 3q + 2

(3q + 2)3

= 27 q3 + 54q2 + 36q + 8     [ \because (x + y)=  x3  + 3x2y + 3xy+ y]

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, if we divide by 9, then the remainder is 8 and the value of m is q (3q2 + 6q + 4).

Hence, the cube of any positive integer is either in form of 9 m or 9 m + 1 or, 9 m + 8.

VIDEO EXPLANATION


Related Questions for Study

What our students and parents say about us!

Choose EduSakshamยฎ
Embrace Better Learning

edusaksham-mobile-app