Using Euclid's Division Lemma
a = bq + r,\(0\le r\)
963 = 657 x 1 + 306
657 = 306 x 2 + 45
306 = 45 x 6 + 36
45 = 36 x 1 + 9
36 = 9 x 4 + 0
HCF (657,963) = 9
Now, 9 = 657x + 963 x (- 15)
⇒  657x = 9 + 963 x 15
= 9 + 14445
⇒ 657x = 14454
⇒  x = 14454/657 = 22

By Euclid's division algorithm,
510 = 92 x 5 + 50
92 = 50 x l + 42
50 = 42 x 1 + 8
42 = 8 x 5 + 2
and  8=2 x 4 + 0
HCF (510, 92) = 2
92 = 2² x 2³
510 = 2 x 3 x 5 x 17
LCM (510, 92) = 2² x 2³ x 3 x 5 x 17
= 23460
HCF (510, 92) x LCM (510,92)
= 2 x 23460 = 46920
Product of two numbers = 510 x 92 = 46920
⇒ HCF x LCM = Product of two numbers

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 numbers are 969,2059 and remainders are 9,11 respectively. Then, new numbers after subtracting remainders are
969 - 9 = 960 and 2059 - 11 = 2048
Here, 2048 > 960
Now, by using Euclid's division lemma, we get
2048 = (960 x 2) + 128
Here, remainder = 128 \(\neq \) 0
So, on taking 960 as new dividend and 128 as new divisor and then apply Euclid's division lemma, we get
960 = (128 x 7) +64
Again, remainder = 64 \(\neq \) 0
So, on taking 128 as new dividend and 64 as new divisor and then apply Euclid's division lemma, we get
128 = (64 x 2) + 0
Since, the remainder has now become zero and the last divisor is 64.
\(\therefore \)  HCF of 2048 and 960 is 64.
Hence, the required largest number is 64.

Let us assume, to the contrary, that \(\sqrt 2\) is rational.
So, we can find integers r and s (≠ 0) such that \(\sqrt 2\) =\(\frac{r}{s}\) .
Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get \(\sqrt 2\) = \(\frac{a}{b}\) , where a and b are coprime.
So, b\(\sqrt 2\) = a.
Squaring on both sides and rearranging, we get 2b ² = a ² .Therefore, 2 divides a ² .
Now, by  it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c² , that is, b² = 2c² .
This means that 2 divides b² , and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that \(\sqrt 2\) is rational.
So, we conclude that \(\sqrt 2\) is irrational.

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.

Given, \(a+\sqrt { b } =c+\sqrt { d } \)
If a = c, then \(\sqrt { b } =\sqrt { d } \)
\(\Rightarrow \) b = d      [squaring both sides]
If \(a\neq c\), then there exists a positive rational number x such that a = c + x.
Now, \(a+\sqrt { b } =c+\sqrt { d } \)
\(\Rightarrow \quad c+x+\sqrt { b } =c+\sqrt { d } \quad \quad \left[ \because \quad a=c+x \right] \)
\(\Rightarrow \quad x+\sqrt { b } =\sqrt { d } \)
\(\Rightarrow \quad \left( x+\sqrt { b } \right) ^{ 2 }=\left( \sqrt { d } \right) ^{ 2 }\)    [squaring both sides]
\(\Rightarrow \quad { x }^{ 2 }+b+2x\sqrt { b } =d\quad \Rightarrow { x }^{ 2 }+2x\sqrt { b } +b-d=0\)
\(\Rightarrow \quad 2x\sqrt { b } =d-b-{ x }^{ 2 }\Rightarrow \sqrt { b } =\frac { d-b-{ x }^{ 2 } }{ 2x } \)
Since, d, x and b are rational numbers and x > 0.
So, \(\frac { d-b-{ x }^{ 2 } }{ 2x } \) is a rational number.
\(\Rightarrow \) \(\sqrt { b } \) is a ratonal number.
\(\Rightarrow \)  b is the square of a rational number.
from Eq.(i)
\(\sqrt { d } =x+\sqrt { b } \quad \Rightarrow \quad \sqrt { d } \) is a rational number.
\(\Rightarrow \) d is the square of a rational number.
Hence, either a = c and b = d or b and d are the squares of rationals.

Here, HCF of 616 and 32 gives the maximum number of columns, in which the two groups can march.
So, on applying Euclid's division lemma to 616 and 32, we get
616 = (32 x 19) + 8
[\(\because \)  dividend = divisor x quotient + remainder]
Here, remainder = 8 \(\neq \) 0, so new dividend is 32 and new divisor is 8.
So, again on applying Euclid's division lemma to 32 and 8, we get
32 = (8 x 4) + 0
Here, remainder is zero and the divisor is 8.
So, HCF of 616 and 32 is 8.
Hence, maximum number of columns is 8, in which they can march.

a is any positive integer

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.

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

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

\(3\times 12\times 101+4=4(3\times 3\times 101+1)\)
 = 4(909+1)
 = 4(910)
= a composite number
[∵  Product of more than two factors]

Given HCF (306, 1,314) =18
LCM (306, 1,314) = ?
Let  a = 306
b = 1,314
We know that
a x b = LCM (a, b) x HCF (a, b)
⇒ 306 x 1,314 = LCM (a, b) x 18
\(\Rightarrow \quad LCM(a,b)=\frac { 306\times 1,314 }{ 18 } \)
∴ LCM (306, 1,314) = 22,338

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

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

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.