Let a be an arbitrary positive integer. Then, by Euclid's division lemma, corresponding to the positive integers a and 4, there exist non-negative integers q and r such that 
a = 4q + r, where \(0\le r<4\)
\(\Rightarrow \)  a³ = (4q + r)³ = 64q³ + r³ + 12qr² + 48q²r
[ \(\because \)  (A + B)³ = A³ + B³ + 3AB² + 3A²B]
\(\Rightarrow \)  a³ = 64q³ + 48q²r + 12qr² + r³
where,  \(0\le r<4\)         .... (i)
The possible values of r are 0, 1, 2, 3.
If r = 0, then from Eq.(i), we get
a³ = 64q³ = 4(16q³) \(\Rightarrow \)  a³ = 4m
where, m = 16q³ is an integer.
If r = 1, then from Eq.(i), we get
a³ = 64q³ + 48q² + 12q + 1
= 4(16q³ + 12q² + 3q) + 1 = 4m + 1
where, m = (16q³ + 12q² + 3q) is an integer.
If r = 2, then from Eq.(i), we get
a³ = (64q³ + 96q² + 48q) + 8
\(\Rightarrow \)  a³ = 4(16q³ + 12q² + 12q + 2)
\(\Rightarrow \)  a³ = 4m, where m = (16q³ + 12q² + 12q + 2)​​​​​​​ is an integer.
If r = 3, then from Eq.(i), we get
a³ = 64q³ + 144q² + 108q + 27
\(\Rightarrow \)  a³ = 64q³ + 144q² + 108q + 24 + 3
\(\Rightarrow \)  a³ = 4(16q³ + 36q² + 27q + 6) + 3 = 4m + 3
where, m = (16q³ + 36q² + 27q + 6)​​​​​​​ is an integer.
Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

Given, rational number is \(\frac{1}{13}\).
Here, prime factors of 13 are not of the form 2ⁿ 5^m.
So, it will not have a terminating decimal expansion.
[ by using theorems 3 and 4]
Now, \(\frac { 1 }{ 13 } =0.076923076923...=\overline { 0.076923 } \)
Thus, \(\frac{1}{13}\) has non-terminating repeating decimal expansion.

We have, (3 x 5 x 7) + 7 = 105 + 7 = 112
\(\therefore \)  Prime factors of 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7.
Hence, it is a composite number.

If 6" ends with 0, then it must have 5 as a factor. But we know that only prime factor of 6ⁿ are 2 and 3.
∴  6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
From the fundamental theorem of arithmetic, we know that the prime factorization of every composite numbers is unique.
∴  6ⁿ can never end with 0.

Fundamental theorem of arithmetic: Every composite number can be expressed as the product of powers of primes and this factorization in unique.
2520 = 2³ x 3² x 5 x 7
10530 = 2 x 3⁴ x 5 x 13
LCM = 2³ x 3⁴ x 5 x 7 x 13
= 294840

We have, 26 and 91

\(\therefore \)  Prime factors of 26 = 2 x 13
and prime factors of 91 = 7 x 13
Now, LCM of 26 and 91 = 2 x 7 x 13 = 182
and HCF of 26 and 91 = 13
For verification
LCM x HCF = 182 x 13 = 2366
and product of two numbers = 26 x 91 = 2366
Thus, LCM x HCF = Product of two numbers.

Let us assume, to the contrary, that 5 - \( \sqrt{3}\) is rational.
That is, we can find coprime a and b (b \(\neq\)0) such that 5 - \( \sqrt{3}\) = \(\frac{a}{b}\)
Therefore, 5 - \(\frac{a}{b}=\sqrt{3}\)
Rearranging this equation, we get \(\sqrt{3}\) = 5 - \(\frac{a}{b}\)=\(\frac{5b-a}{b}\)
Since a and b are integers, we get 5 - \(\frac{a}{b}\) is rational, and so \( \sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is irrational.
This contradiction has arisen because of our incorrect assumption that 5 - \( \sqrt{3}\) is rational.
So, we conclude that 5 - \( \sqrt{3}\) is irrational.

Let the number of columns be x.
x is the largest number, which should divide both 104 and 96
104 = 96 x 1 + 8 1
96 = 8 x 12 + 0
∴  HCF of 104 and 96 is 8
Hence, 8 columns are required.

The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8,9,10
∴  LCM = 2 x 2 x 3 x 2 x 3 x 5 x 7
 =2520

\(\frac { 6 }{ 1250 } =\frac { 6 }{ 2\times { 5 }^{ 4 } } \)
 \(=\frac { 6\times { 2 }^{ 3 } }{ 2\times { 2 }^{ 3 }\times { 5 }^{ 4 } } \)
 \(=\frac { 6\times { 2 }^{ 3 } }{ { 2 }^{ 4 }\times { 5 }^{ 4 } } \)
\(=\frac { 6\times { 2 }^{ 3 } }{ (10)^{ 4 } } =0.0288\)
\(\frac { 6 }{ 1250 } \) will terminate after 4 decimal places.

\(\frac { 3 }{ 8 } =\frac { 3 }{ { 2 }^{ 3 } } \)
 \(=\frac { { 3\times 5 }^{ 3 } }{ { 2 }^{ 3 }\times { 5 }^{ 3 } } \)
 \(=\frac { 375 }{ 10^{ 3 } } =\frac { 375 }{ 1000 } \)
 \(=0.375\)

A rational number is either terminating or nonterminating repeating.
An irrational number is non-terminating and non-repeating

Since ​​​​​​ ​\(\frac { 1 }{ 7 } +\frac { 7 }{ 100 } =\frac { 1 }{ 100 } =0.01\)
Thus smallest rational number is \(\\ \frac { 7 }{ 100 } \)

The decimal expansion of a rational number terminates, if the denominator of rational no p/q, when p and q are co-primes and q can be expressed as 2^m5ⁿ  where In and n are non-negative integers.
e.g. \(\frac { 3 }{ 10 } =\frac { 3 }{ 2^{ 1 }\times 5^{ 1 } } =0.3\)

Length = 8 m 50 cm = 850 cm
breadth = 6 m 25 cm = 625 cm
height = 4 m 75 cm = 475 cm
length of the longest rod is equal to

HCF (625, 850) = 25
∵ 25 divides 475
∴ HCF(62, 850, 475)=25

Hence HCF (1,656,4,025)= 23

We know that a x b = HCF (a, b) x LCM (a, b)
⇒ 1,800 = 12 x LCM (a, b)
⇒  LCM(a,b)=\(\frac { 1,800 }{ 12 } =150\)

HCF of 3³ x 5 and 3² x 5²
= 3² x 5
= 9 x 5
= 45

The smallest prime number is 2 and the smallest composite number is 2² Hence, required HCF (2²,2) = 2.

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then least prime factor of (a + b) is 2.

The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number

13; x = - 2, y =1

Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 10⁴ = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2⁴ x 5⁴ = (2 x 5)⁴
So, the prime factors of q are 2 and 5.