Real Numbers

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Question
CBSEENMA10006078

Use Euclid’s division algorithm to find the HCF of:

(i)135 and 225 (ii) 196 and 38220.

Solution

(i) Given integers are 135 and 225, clearly 225 > 135. Therefore, by applying Euclid’s division lemma to 225 and 135, we get

If. Since, the remaindei 90 ≠ 0, we apply division lemma to 135 and 90, to ge.

We consider the new divisor 90 and new remainder 45 and apply division lemma to get

The remainder of this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 45 is the HCF of 135 and 225.
(ii) Given integers are 196 and 38220. Therefore by applying Euclid’s division lemma to 196 and 38220, we get


The remainder at this step is zero. So. our procedures stops and divisor at this stage i.e. 196 is the HCF of 196 and 38220.

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Question
CBSEENMA10006079

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution

Let a be any odd positive integer and b = 6. Let q be a quotient and r be remainder.

Therefore, applying division lemma, we have 

Question
CBSEENMA10006081

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?

Solution

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.

 

Question
CBSEENMA10006082

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 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

= 9q2 = 3(3q)2 = 3 m where m is some integer.

Square of 3q + 1 = (3q + 1)2

= 9q2 + 6q + 1 = 3(3q2 + 2 q) + 1

= 3m +1, where m is some integer

Square of 3q + 2 = (3q + 2)2

= (3q + 2)2

= 9q2 + 12q + 4

= 9q2 + 12q + 3 + 1

= 3(3q2 + 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.