1) Find the HCF and LCM of smallest prime number and smallest composite number.

(Composite number: A composite number is a positive integer that can be formed by multiplying
 two smaller positive integers. Equivalently, it is a positive integer that has at least one 
 divisor other than 1 and itself. ... For example, the integer 14 is a composite number because
 it is the product of the two smaller integers 2 × 7.)

Solution
Smallest prime number = 2
Smallest composite number = 4 = $2^2$.

therefore HCF = 2 
and LCM = 4
 
 2)Find the HCF and LCM of first two consecutive Odd Composite Numbers.
 
 
 Solution

First two consecutive Odd Composite Numbers are 9 and 15.
Hence  9= 32
and 15 = 3× 5

Therefore HCF= 3
and LCM = $3^2×5$ = 45


3) Find the HCF and LCM of 404 and 96 and verify that HCF×LCM = product of two given numbers.


Solution 
404 = $2^2×101$
96= $2^5×3$ 

HCF = $2^2$ = 4
LCM = $2^5×3×101 $ = 9696

Therefore HCF×LCM = 4×9696 = 38784.
and Product of two given numbers = 404 × 96 = 38784.

Hence proved.


4) Find the HCF of 1260 and 7344 using Euclid algorithm .
 


5)Show that every positive odd integer is of the form $(4q+1)$ or $(4q+3)$
   where $q$ is some integer.

Solution 

Let $a$ be positive odd interger and $b$ = 4.
Using Euclid division algorithm $a=4q+r$
Since $0≤ r<4$, so possible remainder is 0,1,2 and 3

Therefore $a$ can be $4q$ , $4q+1$ ,$4q+2$, $4q+3$, where $q$ is the quotient.
Since $a$ is odd, so $a$ cannot be $4q$ or $4q+2$

Therefore any odd positive integer is of the form $(4q+1)$ or $(4q+3)$ where $q$ is some integer.


6) The product of HCF and LCM is 8064. If one of the number is 96, find the HCF and LCM of two numbers.

Solution 
Product of HCM × LCM = 8064.
One of the number =96.
Let other number be $x$
Since HCF×LCM = product of two given number.
8064 = 96×$x$
$x$ = $8064/96$ = 84

Hence the two given numbers are 96 and 84.

By factorisation method,
96 = $2^5$×3
84= $2^2$×3×7

Therefore HCF = $2^2×3$=12
LCM = $2^5×3×7$=672


7) If the HCF of two numbers is 3 and the product of two numbers is 819. 
Find the LCM.

Solution 
Since HCF×LCM = product of two given numbers.
Therefore 3 × LCM = 819.
LCM = $819/3$ = 273.


8) Find the HCF and LCM of 21,39,78.

Solution 
Factors of the numbers are:
21= 3 × 7
39= 3 ×13
78=2×3×13

Hence HCF = 3 
LCM = 2×3×7×13 = 546

9) Given that √2 is irrational, prove that $(5+3√2)$ is irrational.

Solution 
Let us assume that $(5+3√2)$ is rational. 
Then there exist co-prime positive integers $a$ and $b$ such that 
$(5+3√2) = a/b$
$3√2 = a/b - 5$
$3√2 = {a-5b}/b$
$√2 = {a-5b}/{3b}$
=> Since we assume that √2 is rational, therefore ${a-5b}/{3b}$ is rational.
This contracts the fact √2 is irrational,so our assumption is incorrect.
Hence $(5+3√2)$ is an irrational number.

10) If HCF of 65 and 117 is expressed in the form $65n-117$, then find 
  the value of $n$.

Solution:
Using euclid division algorithm,
117=65×1+52
65=52×1+13 
52=13×4+0 
The divisor at the stage for which the reminder is 0, is the required HCF.
Hence HCF of 65 and 117 is 13.
Since HCF =>$65n-117=13$
$65n=130$
$n=2$