Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Let a and b be two positive integergs such that a is greater than b; then :
a = bq + r;
where q and r are positive integers 0 ≤ r < b.
Taking b = 3, we get
a = 3q + r ; where 0 ≤ r < 3.
⇒ Different values of integer a are 3q, 3q + 1 or 3q + 2.
Cube of 3q = (3q)3
= 27q3Â = 9(3q3) = 9m ;
where m is some integer.
Cube of 3q + 1 = (3q + 1)3
= (3q + 3(3q)2 × 1 + 3(3q) × 12+ l3
[∵ (a + b)3 = a3 + 3a2b + 3ab2 + 1]
= 27q3Â + 27q2Â + 9q + 1
= 9(3q3Â + 3q2Â + q) + 1
= 9m + 1; where m is some integer.
Cube of 3q + 2 = (3q + 2)3
= (3q)3 + 3(3q)2 × 2 + 3 × 3q × 22 + 23
= 27q3Â + 54q2Â + 36q + 8
= 9(3q3Â + 6q2Â + 4q) + 8 = 9m + 8; where m is some integer.
∴ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Given integers are 32 and 616.
Clearly 616 > 32. Therefore, applying Euclid’s division lemma to 616 and 32, we get
Since, the remainder 8 ≠0, we apply the division lemma, to get
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8.
Therefore, the maximum number of columns in which both 616 members (army contingent) and 32 members (army band) can march is 8.
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Use Euclid’s division algorithm to find the HCF of:
(i)135 and 225 (ii) 196 and 38220.
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.
Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b
Taking b = 3, we get
a = 3q + r; where 0 ≤ r < 3
When, r = 0 = ⇒ a = 3q
When, r = 1 = ⇒ a = 3q + 1
When, r = 2 = ⇒ a = 3q + 2
Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m.
⇒ Squares of 3q = (3q)2
= 9q2Â = 3(3q)2Â = 3 m where m is some integer.
Square of 3q + 1 = (3q + 1)2
= 9q2Â + 6q + 1 = 3(3q2Â + 2 q) + 1
= 3m +1, where m is some integer
Square of 3q + 2 = (3q + 2)2
= (3q + 2)2
= 9q2Â + 12q + 4
= 9q2Â + 12q + 3 + 1
= 3(3q2Â + 4q + 1)+ 1
= 3m + 1 for some integer m.
∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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