
Q1 Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 Ans: (i) Here, we have to find H.C.F of 135 and 225
First divide divide the larger integer smaller integer
Since, 225 > 135
Therefore, by Euclid’s Division algorithm
225 = 135 × 1 + 90 (i)
Here 90 ≠ 0, so proceed the same procedure further
Again by E.D.L, (E.D.L = Euclid’s division algorithm)
135 = 90 × 1 + 45 (ii)
As we know, 45 ≠ 0 therefore, again by E.D.L
90 = 45 × 2 + 0 (iii)
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 45.
From (i), (ii) and (iii)
H.C.F (225, 135) = H.C.F (135, 90) = H.C.F (90, 45) = 45.
(ii) Here, we have to find H.C.F of 38220 and 196
First divide the larger integer smaller integer
Since, 3822 > 196
Therefore by Euclid’s Division Algorithm
38220 = 196 × 195 + 0
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 196.
Hence, H.C.F (38220, 196) = 196.
(iii) Here, we have to find H.C.F of 867 and 255
First divide the larger integer smaller integer
Since, 867 > 255
Therefore, by Euclid’s Division algorithm
867 = 255 × 3 + 102 (i)
Remainder 102 ≠ 0, so proceed the same procedure further using E.D.L
255 = 102 × 2 + 51 (ii)
Here, 51 ≠ 0 again using E.D.L = 51 × 2
102 = 51 × 2 + 0 (iii)
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 51.
From (i), (ii) and (iii)
H.C.F (867, 255) = H.C.F (255, 102) = H.C.F (102, 51) = 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. Ans: Let a be any positive integer b = 6. Then by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0So, values of r we get, r = 0, 1, 2, 3, 4, 5
When, r = 0, then, a = 6q + 0,
similarly for r = 1, 2, 3, 4, 5 the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5 respectively.
If a = 6q, 6q+2, 6q+4, then a is even and divisible by 2.
Therefore, any positive odd integer is in the form of 6q+1, 6q+3, 6q+5
Where q is some integer.
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? Ans: Total no. of army contingent members = 616
No. of army band members = 32
To find max. numbers of the same columns in which the both groups march. We have to find it.
Their highest common factor. since, 616 > 32 then, by Euclid’s algorithm
616 = 32 × 19 + 8 (i)
Here 8 ≠ 0 then by again using Euclid algorithm
32 = 8 × 4 + 0 (ii)
Here, r =0 so we cannot proceed further. The divisor at this Stage is 8.
So the no. of columns is 8.
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. [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.] Ans: Let x be any positive integer and b = 3.
Then, by euclid’s algorithm
x = 3q + r, where q ≥ 0 and r = 0, 1, 2 [0 ≤ r ≤ b]
Case (i) : For r = 0, x = 3q, = x^{2} = 9q^{2}, taking 3 as common,
x^{2} = 9q^{2} = 3 (3q^{2}), which is of the form 3m, where m = 3q^{2}.
Case (ii) : For r = 1, x = 3q + 1
x^{2} = 9q^{2} + 1 + 6q, taking 3 as common,
= 3 (3q^{2} + 2q) + 1, which is of the form 3m + 1, where m = 3q^{2} + 2q
Case (iii) : For r = 2, 3q + 2
x^{2} = 9q^{2} + 4 + 12q = (9q^{2} + 12q + 3) + 1, taking 3 as common,
= 3 (3q^{2} + 4q + 1) + 1, which is of the form 3m +1, where m = 3q^{2} + 4q + 1
Hence, x^{2} is either of the form 3m, 3m + 1 for some 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. Ans: Let a be any positive integer and b = 3. Then, by euclid’s algorithm
a = 9q + r, where q ≥ 0 and r = 0, 1, 2 [ 0 ≤ r ≤ b ]
For r = 0, x = 3q, or
For r = 1, x = 3q +1
For r = 2, x = 3q + 2
Now by taking the cube of all the three above terms, we get,
Case (i) : when r = 0, then,
x^{3} = (3q)^{3} = 27q^{3} = 9 (3q^{3}) = 9m; where m = 3q
Case (ii) : when r = 1, then,
x^{3} = (3q +1)^{3} = (3q)^{3} + 1^{3} +3 × 3q × 1 (3q + 2 ) = 27q^{3} + 1 +27q^{2} + 9q
Taking 9 as common factor, we get,
x^{3} = 9 (3q^{3} +3q^{2} + q) +1
Putting (3q^{3} + 3q^{2} + q) = m, we get,
x^{3} = 9m + 1
Case (iii) : when r = 2, then,
x^{3} = (3q + 2)^{3} = (3q)^{3} + 2^{3} + 3 × 3q × 2 (3q + 2) = 27q^{3} + 54q^{2} + 36q + 8
Taking 9 as common factor , we get
x = 9 (3q + 6q + 4q) + 8
Putting (3q + 6q + 4q) = m, we get,
x = 9m + 8,
Therefore, it is proved that the cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.