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Karnataka State Syllabus Class 10 Maths Chapter 8 Real Numbers Ex 8.1
Question 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) Step 1: Consider a = 225 and , b = 135
By Euclid’s lemma.
a = (b × q) + r
225 = (135 × 1) + 90
Step 2: Consider a = 135 and b = 90
Here q = 1, r = 45
By Euclid’s lemm
a = (b × q) + r
135 = (90 × 1) + 45
Step 3: Consider a = 90 and b = 45
Here q = 2, r = 0
By Euclid’s lemma
a = (b × q) + r
90 = (45 × 2) + 0
Since the remainder is zero,
∴ Last divisor is HCF
∴ HCF of 135 and 225 is 45.
ii) 196 and 38220
Step 1: Consider a = 38220 and b = 196. Here q = 195, r = 0.
By Euclid’s lemma
a = (b × q) + r
38220 = (196 × 195) + 0
∴ HCF of 196 and 38220 is 196
iii) 867 and 255
Step 1: Consider a = 867 and b = 255
Here q = 3, r = 102
By Euclid’s lemma
a = (b × q) + r
867 = (255 × 3) + 102
Step 2: Consider a = 255 and b = 102
Here q = 2 and r = 51
By Euclid’s lemma
a = (b × q) + r 255 = (102 × 2) + 51
Step 3 : Consider a = 102 and b = 51
Here q = 2, r = 0
By Euclid’s lemma
a = (b × q) + r
102 = (51 × 2) + 0
∴ HCF of 867 and 225 is 51
Question 2.
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 positive odd integer and b = 6.
Then, by Euclid’s algorithm, a = bq + r, For some integer q ≥ 0 and 0 ≤ r < 6
∴ a = 6q + r [remainders are 0, 1, 2, 3, 4, 5].
Thus, a can be of the form 6q (or) 6q + 1 (or) 6q + 2 (or) 6q + 3 (or) 6q + 4 (or) 6q + 5 where q is some Quotient.
Since a is odd integer, so a cannot be of the form 6q, (or) 6q + 2 (or) 6q + 4. (since they are even)
Thus a is of form 6q + 1 (or) 6q + 3 (or) 6q + 5 where q is some integer.
Hence, any odd positive integer is of the form 6q + 1 (or) 6q + 3 (or) 6q + 5, where q is some integer.
Question 3.
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 of 616 and 32.
Step 1: Euclid’s division lemma
a = 616 and b = 32
616 = 32 × 19 + 8
616 = 616.
Step 2:
Euclid’s division lemma,
a = 32 & b = 8.
a = bq + r
32 = 8 × 4 + 0
32 = 32
Here, Remainder is zero. So, HCF of (616, 32) = 8.
Hence, maximum number of columns is 8.
Question 4.
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.
[Hint : 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.]
Answer:
i) When n = 3m,
n2 = 9m2
= 3 × 3m2
= 3q (∵ q = 3m2)
ii) If n = 3m + 1
n2 = (3m + 1)2
= 9m2 + 6m + 1
= 3 (3m2 + 2m) + 1
= 3q + 1 (∵ 3m2 + 2m = q)
∴ Square of any positive integer is either of the form 3m or 3m + 1 for some integer ‘m’.
Question 5.
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’ and ‘b’ are positive integers,
a = b × q + r
Let b = 3. ∴ a = 3q + r
i) If r = 0 then a = 3q
ii) If r = l then a = 3q + 1
iii) If r = 2 then a = 3q + 2
If we consider cubes of these,
i) If a = 3q, then
a3 = (3)3
= 27q3 = 9(3q)3 = 9m (∵ m = 3q2)
ii) If a = 3q + 1, then
(a)3 = (3q + 1)3 = 27q3 + 1 + 27q3 + 9q
= 9(3q3 + 3q2 + q) + 1
= 9m + 1
(∵ m = 3q3+ 3q2 + q)
iii) If a = 3q + 2, then
(a)3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8
= 9m + 8
∴ Cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.