The greatest number of cartons is the HCF of 144 and 90
\(144={ 2 }^{ 4 }\times { 3 }^{ 2 }\)
\(90=2\times { 3 }^{ 2 }\times 5\)
\(HCF=2\times { 3 }^{ 2 }=18\)
∴ The greatest number of cartons = 18

Let n be any positive integer.
By Eucild's division lemma, 11= 5q + r, \(0\le r<5\)
n = Sq, Sq + 1, Sq + 2,5q + 3 or 5q + 4, where \(q\epsilon w\) q is a whole number
now    n² = (5q)² = 25q² = 5(5q²) = 5m
n²= (5q + 1)²= 25q² + 10q + 1= 5m + 1
n² = (5q + 2)²= 25q² + 20q + 4 = 5m+ 4
Similarly  n²= (5q + 3)² = 5m + 4
and   n² =(5q + 4)² = 5m + 1
Thus square of any positive integer cannot be of the form 5m + 2 or 5m + 3.

LCM of 18, 24 and 36 is 72

ஃ Required number = 9,99,936.

Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer. On dividing n by 5, let q be the quotient and r be the remainder.
Then, n = 5q + r, where \(0\le r<5\)
[ by Euclid's division lemma]
\(\Rightarrow \)  n = 5q + r, where r = 0, 1, 2, 3, 4
\(\Rightarrow \)  n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4
If n = 5q, then n is only divisible by 5.
If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) which is divisible by 5. So, (n + 4) is only divisible by 5.
If n = 5q + 2, then n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) which is divisible by 5. So, (n + 8) is only divisible by 5.
If n = 5q + 3, then n + 12 = 5q + 3 + 12  = 5q + 15 = 5(q + 3) which is divisible by 5. So, (n + 12) is only divisible by 5.
If n = 5q + 4, then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4) which is divisible by 5. So, (n + 16) is only divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

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.

Let \(\sqrt { q } \) be a rational number and its simplest form is a/b. where q is a prime, a and b are coprime integers and \(b\neq 0\).
Now, \(\sqrt { q } =\frac { a }{ b } \)
On squaring both sides, we get
\(q=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \Rightarrow q{ b }^{ 2 }={ a }^{ 2 }\)              .... (i) (1)
Since, qb² is divisible by q, so a² is also divisible by q.
\(\Rightarrow \)  a is also divisible by q.    [by using theorem 1] ... (ii)
Now, a can be written as a = qc for some integer c.
On substituting a = qc in Eq.(i) , we get
qb² = (qc)² \(\Rightarrow \) qb² = q²c² \(\Rightarrow \)  qc²
\(\Rightarrow \) b² is divisible by q.
\(\Rightarrow \) b is divisible by q.     [ by using theorem 1] ... (iii) (1)
From Eqs.(ii) and (iii), q is a common factor of a and b. But this contradicts our assumption that a and b have no common factor. So, our assumption that \(\sqrt { q } \) is a rational number, is wrong. Hence, \(\sqrt { q } \) is an irrational number, if q is prime.

Here, \(1.2\overline { 3 } \) = 1.23333... is a non-terminating repeating decimal.
So, \(1.2\overline { 3 } \) is a rational number and 3/4 is in the form of \(\frac{p}{q}\), where \(q\neq 0\). Thus, \(\frac{3}{4}\) is a rational number.
We know that, the sum of two rational numbers is a rational number.
Therefore, \(\left( 1.2\overline { 3 } +\frac { 3 }{ 4 } \right) \) is a rational number.

Since, the numbers 3, 5, 15, 25 and 75 divide the numbers 525 and 3000, therefore we can write the given numbers as, 525 = 75 x 7 and 3000 = 75 x 40 = 75 x 5 x 2³ .
Since, highest common factor among these is 75, therefore HCF = 75

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.

Let  \(3\sqrt { 2 } \)  be a rational number. Then, it will be of the form  \(\frac { p }{ q } \) , where p, q are coprime integers and  \(q\neq 0\).
Now, \(\frac { p }{ q } \) = \(3\sqrt { 2 } \)   \(\Rightarrow \quad \frac { p }{ 3q } =\sqrt { 2 } \)
Since, p is an integer and 3q is also an integer \(\left( 3q\neq 0 \right) \).
So, \(\frac { p }{ 3q } \) is a rational number.
\(\Rightarrow \quad \sqrt { 2 } \) is a rational number.
But this contradicts the fact that \(\sqrt { 2 } \) is an irrational number.
Hence, \(3\sqrt { 2 } \) is an irrational number.

We know that, 
HCF x LCM = Product of two numbers
\(\Rightarrow \)  5 x 63 x a = 35 x 45  \(\therefore \quad a=\frac { 35\times 45 }{ 5\times 63 } =5\)

Let the two number are a and b.
Given, HCF(a, b) = 2
and product (a x b) = 120
We know that, 
HCF (a, b) x LCM (a, b) = Product of a and b
\(\therefore \)   2 x LCM (a, b) = 120
\(\Rightarrow \)  LCM (a, b) = \(\frac{120}{2}\) = 6
Hence, the required LCM is 60.

The prime factorisation of 120 = 2 x 2 x 2 x 3 x 5
= 2³ x 3 x 5
and prime factorisation of 144
= 2 x 2 x 2 x 2 x 3 x 3 = 2⁴ x 3²
By using fundamental theorem of arithmetic,
Now, LCM(120, 144) = 2⁴ x 3² x 5 = 720
and HCF(120, 144) = 2³ x 3 = 24

Smallest composite number = 4 = 2 x 2 = 2²
and smallest prime number = 2 = 2¹
\(\therefore \)  HCF (4, 2) = 2¹ = 2
[since, HCF is the product of the smallest power of each common prime factor involved in the 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.