(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
(i)135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to
obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to
obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the
division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii)196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to
obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii)867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to
obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102
to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the
division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 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.
Let a be any positive integer and b = 6.
Then, by Euclid’s algorithm, \(a = 6q + r\) for some integer \(q ≥ 0\), and \(r = 0, 1, 2, 3, 4, 5\) because \(0 ≤ r < 6\).
Therefore, \(a = 6q \text{ or } 6q + 1 \text{ or } 6q + 2 \text{ or } 6q + 3 \text{ or } 6q + 4 \text{ or } 6q + 5\)
Also, \(6q + 1 = 2 × 3q + 1 = 2k^1 + 1\), 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
Clearly, \(6q + 1, 6q + 3, 6q + 5\) 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.
And therefore, any odd integer can be expressed in the form \(6q + 1, \text{ or } 6q + 3, \text{ or } 6q + 5\).
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?
Solution:
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
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.
Solution 4:
Let \({a}\) be any positive integer and \({b = 3}\).
Then \({a = 3q + r}\) for some integer q ≥ 0}\)
And \({r = 0, 1, 2}\) because \({0 ≤ r < 3}\)
Therefore, \({a = 3q}\) or \({3q + 1}\) or \({3q + 2}\)
Or,
\({a^2= (3q)^2}\) 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 \({(3q^2 + 2q) +1}\) or \({3 (3q^2+ 4q+ 1)+ 1}\)
\({ a^2=3k_1}\) or \({3k_2 + 1}\) or \({3k_3 + 1 }\)
Where \({k_1, k_2}\), and \({k_3}\) are some positive integers
Hence, it can be said that the square of any positive integer is either of the form \({3m}\) or \({3m + 1}\).
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.
Solution 5:
Let \({a}\) be any positive integer and \({b = 3}\)
\({a = 3q + r}\), where \({q ≥ 0}\) and \({0 ≤ r < 3}\)
\({\therefore a =3q}\) or \({3q +1}\) or \({3q + 2}\)
Therefore, every number can be represented as these three forms.
There are three cases.
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 \({a = 3q + 1}\),
\({a^3 = (3q +1)^3}\)
\({a^3 = 27q^3 + 27q^2 + 9q + 1}\)
\({a^3= 9(3q^3 + 3q^2 + q) + 1}\)
\({a^3 = 9m + 1}\)
Where m is an integer such that \({m =(3q^3 + 3q^2 + q)}\)
Case 3: When \({a = 3q + 2}\),
\({a^3 = (3q +2)^3}\)
\({a^3 = 27q^3 + 54q^2 + 36q + 8}\)
\({a^3= 9(3q^3 + 6q^2 + 4q) + 8}\)
\({a^3 = 9m + 8}\)
Where m is an integer such that \({m = (3q^3 + 6q^2 + 4q)}\)
Therefore, the cube of any positive integer is of the form \({9m, 9m + 1}\),or \({9m + 8}\).
