Ex 1.1 Class 10 Maths 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.
Solution
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Step 1:
Since 225 is greater than 135, we can apply Euclid's division lemma to
a = 225 and b = 135 to find q and r such that
225 = 135q + r, 0 ≤ r < 135
So, dividing 225 by 135 we get 1 as the quotient and 90 as remainder.
i. e 225 = (135 × 1) + 90
Step 2:
Remainder r is 90 and is not equal to 0, we apply Euclid's division lemma to b = 135 and r = 90 to find whole numbers q and r such that
135 = 90 × q + r, 0 ≤ r < 90
So, dividing 135 by 90 we get 1 as the quotient and 45 as remainder.
135 = (90 × 1) + 45
Step 3:
Again, remainder r is 45 and is not equal to 0, so we apply Euclid's division lemma to b = 90 and r = 45 to find q and r such that
90 = 45 × q + r, 0 ≤ r < 45
So, dividing 90 by 45 we get 2 as the quotient and 0 as remainder.
90 = (2 × 45) + 0
Step 4:
Since the remainder is zero, the divisor at this stage will be HCF of
(135, 225)
Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is
45.
(ii) 196 and 38220
Given 38220 > 196 so applying Euclids Division Lemma,
we get 38220 = 196 × 195 + 0 …………… (i)
Remainder becomes zero. So, the divisor of this stage i.e 196 is HCF of 38220 and 196
HCF ( 196 , 38220) = 196
(iii) 867 and 255
Given integers are 867 and 255 and 867 > 255 So, applying Euclid’s division algorithm
We get 867 = 255 × 3 + 102
Here, remainder is 102. So, we again apply Euclid’s division algorithm on division 255 and remainder 102.
255 = 102 × 2 + 51
Here, remainder is 51. So, we again apply Euclid’s division algorithm on division 102 and remainder 51.
102 = 51 × 2 + 0
Here, remainder is 0. So, HCF(867, 255) = 51
Q.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 be any positive integer such that a = 6q + r
Then, by Euclid's algorithm a = 6q + r, for some integer q≥0 and where 0≤r<6 the possible remainders are 0, 1, 2, 3, 4, 5
i.e, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
If a = 6q or 6q + 2 or 6q + 4, then a is an even integer.
Also, an integer can be either even or odd.
Q.3 An army contingent of 616 members is to march behind an army band of 32 members in 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. To find the maximum number of columns, we have to find the HCF of 616 and 32.
