NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Ans : (i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 x 1 + 90 
Since remainder \( 90 \neq 0 \), we apply the division lemma to 135 and 90 to 
obtain 
135 = 90 x 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the 
division lemma to obtain 
90 = 2 x 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 x 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 x 3 + 102 
Since remainder \(102 \neq 0\) we apply the division lemma to 2S5 and 102 to obtain 
255 = 102 x 2+ 51 
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 
102 = 51 x 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. 

Ans : 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 \(\leq\) r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 x 3q +1 = \(2 k_{1}+1, \text { where } k_{1}\) is a positive integer
\(6 q+3=(6 q+2)+1=2(3 q+1)+1=2 k_{2}+1, \text { where } k_{2}\) is an integer
\(6 q+5=(6 q+4)+1=2(3 q+2)+1=2 k_{3}+1, \text { where } k_{3}\) 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,
or 6q + 3,
or 6q + 5 

Ans : 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 x 19 + 8 
32 = 8 x 19 +8 
The HCF (616, 32) is 8. 
Therefore, they can march in 8 columns each. 

Ans : Let a be any positive integer and b = 3. 
Then a = 3q + r for some integer \(q \geq 0\)
And r = 0, 1, 2 because \(0 \leq r<3\)
Therefore, a = 3q or 3q + 1 or 3q + 2 
Or,
\(\begin{array}{l}{a^{2}=(3 q)^{2} \text { or }(3 q+1)^{2} \text { or }(3 q+2)^{2}} \\ {a^{2}=\left(9 q^{2}\right) \text { or } 9 q^{2}+6 q+1 \text { or } 9 q^{2}+12 q+4} \\ {=3 \times\left(3 q^{2}\right) \text { or } 3\left(3 q^{2}+2 q\right)+1 \text { or } 3\left(3 q^{2}+4 q+1\right)+1} \\ {=3 k_{1} \text { or } 3 k_{2}+1 \text { or } 3 k_{3}+1}\end{array}\)
Where \(k_{1}, k_{2}, \text{and } k_{3} \) 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 

Ans : Let a be any positive integer and b = 3 
\(\begin{array}{l}{a=3 q+r, \text { where } q \geq 0 \text { and } 0 \leq r<3} \\ {\therefore a=3 q \text { or } 3 q+1 \text { or } 3 q+2}\end{array}\)
Therefore, every number can be represented as these three forms. 
There are three cases. 
Case 1: When a = 3q, 
\(a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m\),
Where m  \( \text { is an integer such that } m=3 q^{3}\)


\(\begin{array}{l}{\text {Case } 2 : \text { when } a=3 q+1} \\ {a^{3}=(3 q+1)^{3}} \\ {a^{3}=27 q^{3}+27 q^{2}+9 q+1} \\ {a^{3}=9\left(3 q^{3}+3 q^{2}+q\right)+1} \\ {a^{3}=9 m+1} \\ {\text { Where } m \text { is an integer such that } m=\left(3 q^{3}+3 q^{2}+q\right)}\end{array}\)
\(\begin{array}{l}{\text {Case } 3 : \text { when } a=3 q+2} \\ {a^{3}=(3 q+2)^{3}} \\ {a^{3}=27 q^{3}+54 q^{2}+36 q+8} \\ {a^{3}=9\left(3 q^{3}+6 q^{2}+4 q\right)+8} \\ {a^{3}=9 m+8} \\ {\text { Where } m \text { is an integer such that } m=\left(3 q^{3}+6
q^{2}+4 q\right)}\end{array}\)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.