**Real Numbers**

**(NCERT Ex 1.1)**

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

**(i) 135 and 225**

**(ii) 196 and 38220**

**(iii) 867 and 255**

**Answer**:

**(i) 135 and 225**

*Step 1*: Since 225 > 135, apply Euclid’s division lemma to 225 and 135 to get

225 = 135 × 1 + 90

*Step 2*: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to get

135 = 90 × 1 + 45

*Step 3*: Now the remainder r = 45 ≠ 0 so we apply Euclid’s division lemma to 90 and 45 to have

90 = 2 × 45 + 0

*Step 4*: The remainder has now become zero. Since the divisor at this stage is 45,

∴ HCF(135,225) =

**45**

**(ii) 196 and 38220**

*Step 1:*

**Since 38220 > 196, apply the division lemma to 38220 and 196 to get**

38220 = 196 × 195 + 0

Step 2: Since remainder is zero, the process stops. Since the divisor at this stage is 196,

∴ HCF(38220, 196) =

**196**

**(iii) 867 and 255**

*Step 1:*Since 867 > 255, we apply the division lemma to 867 and 255 to get

867 = 255 × 3 + 102

*Step 2*: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to have

255 = 102 × 2 + 51

*Step 3*: Now the new divisor is 102 and new remainder is 51. Apply the division lemma to get

102 = 51 × 2 + 0

*Step 4*: Since the remainder is zero, the process stops. Since the divisor at this stage is 51,

∴ HCF(867, 255) = 51

**Q2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.**

Answer: Let a be any odd positive integer and b = 6.

By applying Euclid’s algorithm, we have

a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

∴ a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Since a is an odd integer, it cannot take 6q, 6q + 2 and 6q + 4 as values.

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k

_{1}is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k

_{2}+ 1, where k

_{2}is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k

_{3}+ 1, where k

_{3}is an integer

It is clear from above three statements that, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers.

∴ any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5

**Q3: 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?**

Answer: HCF (616, 32) will give the maximum number of columns in which the Army contingent and the band can march.

Applying Euclid’s algorithm to find HCF (616, 32),

Step 1: 616 = 32 × 19 + 8

Step2: 32 = 8 × 4 + 0

Step3: Since the remainder is zero, the divisor at this stage is 8. ∴ HCF (616, 32) = 8

Thus, they can march in

**8 columns**each.

**Q4: 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 x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]**

*Hint*Answer: Let a be any positive integer and b = 3.

Applying Euclid's division lemma to a and b=3, we get

a = 3q + r for some integer q ≥ 0 and r = 0, 1, 2

⇒ a = 3q or 3q + 1 or 3q + 2

Squaring 'a' i.e.

a

^{2}= (3q)

^{2}or (3q +1 )

^{2}or (3q + 2)

^{2}

⇒ a

^{2}= 9q

^{2}or (9q

^{2}+ 6q + 1 ) or (9q

^{2}+ 12q + 4)

^{}

⇒ a

^{2}= 3(3q

^{2}) or 3(3q

^{2}+ 2q ) + 1 or 3(3q

^{2}+ 4q + 1) + 1

⇒ a

^{2}= 3 k

_{1}or 3k

_{2}+ 1 or 3k

_{3}+ 1

where k

_{1}= 3q

^{2}, k

_{2}= 3q

^{2}+ 2q and k

_{3}= 3q

^{2}+ 4q + 1

⇒ Since q = 1,2,3... are integers. Their products and sums will also be integers. ∴ k

_{1}, k

_{2}, and k

_{2}are also integers.

Thus we can say that the square of any positive integer is either of the form 3m or 3m + 1 for any integer m.

**Q5: Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.**

Answer: Let a be any positive integer and b = 3. Using Euclid’s division lemma we have

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ a = 3q or 3q +1 or 3q + 2

Therefore, every number can be represented as these three forms. Therefore three cases arise:

*Case 1: When a = 3q,*

a

^{3}= (3q)

^{3}= 27q

^{3}= 9(3q

^{3}) = 9m

Where m is an integer such that m = 3q

^{3}

*Case 2: When a = 3q + 1,*

a

^{3}= (3q + 1)

^{3}

a

^{3}= (27q

^{3}+ 27q

^{2}+ 9q + 1) = 9(3q

^{3}+ 3q

^{2}+ q) + 1

a

^{3}= 9m + 1

where m is an integer such that m = 3q

^{3}+ 3q

^{2}+ q

*Case 3: When a = 3q + 2*,

a

^{3}= (3q + 2)

^{3}

a

^{3}= (27q

^{3}+ 53q

^{2}+ 36q + 8) = 9(3q

^{3}+ 6q

^{2}+ 4q) + 8

a

^{3}= 9m +8

Where m is an integer such that m = 3q

^{3}+ 6q

^{2}+ 4q

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

__Similar Questions asked in Examinations__**Q6: Prove that 6 divides n(n + 1)(2n + 1) for any n ∈ N**

Answer: If n = 1 or 2, ⇒ values of expression n(n + 1)(2n + 1) are 6 or 30 which are divisible by 6.

∴ n ≥ 3.

Also n and n+1 are consecutive numbers and either of them is an even number.

∴ n or n+1 is divisible by 2

or 2 divides n(n + 1)(2n + 1)

Also n = 3k or 3k +1 or 3k + 2 for some k ∈ N

Case 1: when n = 3k,

n(n + 1)(2n + 1) = 3k(3k+1)(6k+1) which is divisible by 3

Case 2: when n = 3k + 1,

n(n + 1)(2n + 1)= (3k +1)(3k + 2)(6k + 3) = 3(3k +1)(3k + 2)(2k + 1) is divisible by 3.

Case 3: when n = 3k + 2,

n(n + 1)(2n + 1) = (3k + 2)(3k + 3)(6k +5) = 3(3k + 2)(k + 1)(6k +5) is divisible by 3.

∴ In all cases, 2 and 3 divides n(n +1)(2n + 1)

Since 2 and 3 have no common factor, thus we an say n(n + 1)(2n + 1)is divisible by 6.

not bad

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ReplyDeleteThe flow chart to finding HCF is very good.

ReplyDeleteYou can also watch these solutions in Video.

https://www.youtube.com/playlist?list=PLItJjtWTBUw6o6ZWYEfBNGnSlUZWD8mPY

Thank you so much!

Deletenot bad but i want more question for proving

ReplyDeletenot bad but i want more question for proving

ReplyDeleteHi

ReplyDeleteIn question 4 and 5,why are we taking 3q,3q+1,3q+2.....why not 2q,2q+1...or 5q,5q+1,5q+2 etc..??

ReplyDeleteIn question 4 and 5,why are we taking 3q,3q+1,3q+2.....why not 2q,2q+1...or 5q,5q+1,5q+2 etc..??

ReplyDelete