NCERT Solution: Real Numbers
(i) 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135 × 1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90 × 1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2 × 45+ 0
As there are no remainder so divisor 45 is our HCF.
(ii) 38220 > 196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 × 195 + 0
As there is no remainder so divisor 196 is our HCF.
(iii) 867 > 255 we always divide greater number with smaller one.
Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as
867 = 255 × 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as
255 = 102 × 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 × 2 + 0
As there is no remainder so divisor 51 is our HCF.
(i) 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135 × 1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90 × 1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2 × 45+ 0
As there are no remainder so divisor 45 is our HCF.
(ii) 38220 > 196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 × 195 + 0
As there is no remainder so divisor 196 is our HCF.
(iii) 867 > 255 we always divide greater number with smaller one.
Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as
867 = 255 × 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as
255 = 102 × 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 × 2 + 0
As there is no remainder so divisor 51 is our HCF.
Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6
6q + 0
6q + 1
6q + 2
6q +3
6q + 4
So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5
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.
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.
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,
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2
a2 = (9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
= 3 × (3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
Where k1, k2, and k3 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
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ 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,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ 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,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.