Exercise 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
Sol:
(1) 135 and 225
a = 225, b = 135 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
225 = 135 ×1 + 90
135 = 90 ×1 + 45
90 = 45 × 2 + 0 {when we get r=0, our computing get stopped}
b = 45 {b is HCF}
Hence: HCF = 45
Sol:
(ii) 196 and 38220
a = 38220, b = 196 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
Sol:
(iii) 867 and 255
a = 867, b = 255 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Sol:
Let a is the positive odd integer
Where b = 6,
When we divide a by 6 we get reminder 0, 1, 2, 3, 4 and 5, {r < b}
Here a is odd number then reminder will be also odd one.
We get reminders 1, 3, 5
Using Euclid’s division algorithm
So we get
a = 6q + 1, 6q+3 and 6q+5
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?
Sol:
Maximum number of columns = HCF (616, 32)
a = 616, b = 32 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
616 = 32 ×19 + 8 {when we get r=0, our computing get stopped}
32 = 8 × 4 + 0
b = 8 {b is HCF}
HCF = 8
Hence: Maximum number of columns = 8
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.]
Solution :
To Show :
a2 = 3m or 3m + 1
a = bq + r
Let a be any positive integer, where b = 3 and r = 0, 1, 2 because 0 ≤ r < 3
Then a = 3q + r for some integer q ≥ 0
Therefore, a = 3q + 0 or 3q + 1 or 3q + 2
Now we have;
⇒ a2 = (3q + 0)2 or (3q + 1)2 or (3q +2)2
⇒ a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
⇒ a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 3 + 1
⇒ a2 = 3(3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
Let m = (3q2) or (3q2 + 2q) or (3q2 + 4q + 1)
Then we get;
a2 = 3m or 3m + 1 or 3m + 1
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.
Solution:
Let , a is any positive integer
By using Euclid’s division lemma;
a = bq + r where; 0 ≤ r < b
Putting b = 9
a = 9q + r where; 0 ≤ r < 9
when r = 0
a = 9q + 0 = 9q
a3 = (9q)3 = 9(81q3) or 9m where m = 81q3
when r = 1
a = 9q + 1
a3 = (9q + 1)3 = 9(81q3 + 27q2 + 3q) + 1
= 9m + 1 where m = 81q3 + 27q2 + 3q
when r = 2
a = 9q + 2
a3 = (9q + 2)3 = 9(81q3 + 54q2 + 12q) + 8
= 9m + 2 where m = 81q3 + 54q2 + 12q
⇒ The End
