Given capacities of two tankers are 850 L and 680 L.
Here, 850 > 680
Now, 850 = (680 x 1) + 170
[by Euclid's division lemma]
Here, remainder = 170 \(\neq \) 0. So, new dividend is 680 and divisor is 170.
Now, 680 = (170 x 4) + 0
[by Euclid's division lemma]
Here, remainder is zero and divisor is 170.
So, the HCF of 850 and 680 is 170.
Hence, the maximum capacity of the required container is 170 L.

We have: 6 = 2 \(\times\) 3, 72 = 2³ \(\times\) 3², 120 = 2³ \(\times\) 3 \(\times\) 5
Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2¹ \(\times\) 3¹ = 2 \(\times\) 3 = 6
2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So,    LCM (6, 72, 120) = 2³ \(\times\) 3² \(\times\) 5¹ = 360
Remark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM.

Proof : Let the prime factorisation of a be as follows :
a = p₁ p₂ . . . p_n , where p₁ ,p₂ , . . ., p_n are primes, not necessarily distinct. Therefore, a² = ( p₁ p₂ . . . p_n )( p₁ p₂ . . . p_n ) = p²₁ p²₂ . . . p ²_n .
Now, we are given that p divides a ² . Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a ² . However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁ ,p₂ , . . ., p_n . So p is one of p₁ , p₂ , . . ., p_n .
Now, since a = p₁ p₂ . . . p_n ,p divides a.
We are now ready to give a proof that \(\sqrt 2\) is irrational.
The proof is based on a technique called ‘proof by contradiction’. 

True, because the product of any two consecutive numbers, say n(n + 1) will always be even as one out of n or (n+1) must be even.

First, use Euclid's division algorithm for two larger numbers any of three and get the HCF of these two. After that take the third number and resulting HCF of two numbers and apply again Euclid's division algorithm and get the required HCF.
(i)  63  (ii) 15

\(a={ x }^{ 3 }{ y }^{ 2 }=x\times x\times x\times y\times y\) and \(b={ x }{ y }^{ 3 }=x\times y\times y\times y\)
HCF (a,b) = xy²

We have, (7 x 11 x 13) + 13 = 1001 + 13 = 1014
\(\Rightarrow \) 1014 = 2 x 3 x 13 x 13
We know that, a number is called composite number, if it has atleast one factor other than 1 and the number itself. Here, 1014 is the product of more than two prime numbers, i.e. 2, 3 and 13.
So, it is a composite number.
Now, (7 x 6 x 5 x 4 x 3 x 2 x 1) + 5 = 5045
\(\Rightarrow \) 5045 = 5 x 1009
Thus, it is the product of prime factors 5 and 1009.
Hence, it is also a composite number.

 \(\frac { 13 }{ 3125 } =\frac { 13 }{ 5\times 5\times 5\times 5\times 5 } \)
On multiplying by 2⁵  in numerator and denominator, we get
\(\frac { 13 }{ 3125 } =\frac { 13\times 2\times 2\times 2\times 2\times 2 }{ 5\times 2\times 5\times 2\times 5\times 2\times 5\times 2\times 5\times 2 } \)
\(=\frac { 13\times 32 }{ 10\times 10\times 10\times 10\times 10 } \)
\(=\frac { 416 }{ 100000 } \) = 0.00416

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.

Let a = n² - 1, where n = 1, 3, 5, ...
At n = 1, then a = (1)² - 1 = 1 - 1 = 0
At n = 3, then a = (3)² - 1 = 9 - 1 = 8
At n = 5, then a = (5)² - 1 = 25 - 1 = 24
which is divisible by 8.
Hence, n is an odd integer.

36

We, have

Product of prime factors of 140
= 2 x 2 x 5 x 7= 2² x 5 x 7

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

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

336 and 54
336 = 2 x 168
= 2 x 168
= 2 x 2 x 84
= 2 x 2 x 2 x 42
= 2 x 2 x 2 x 2 x 21
= 2 x 2 x 2 x 2 x 7 x 3 x 1
Therefore 336 = 2 x 2 x 2 x 2 x 7 x 3 .......(A)
54 = 2 x 27
= 2 x 3 x 9
= 2 x 3 x 3 x 3
Therefore 54 = 2 x 3 x 3 x 3 ......(B)
From (A) and (B) HCF of 336 and 54 = 2 x 3 = 6
LCM of 336 and 54 = 2 x 3 x 2 x 2 x 2 x 7 x 3 x 3 = 2⁴ x 3³ x 7 = 3024
Product of 336 and 54 = 18144
Product of LCM and HCF = 6 x 3024 = 18144
Therefore it is proved that LCM X HCF = Product of the two numbers.