Given:
Two positive integers, $a = 306$ and $b = 657$.
The Highest Common Factor (HCF) of the two integers is given as: $HCF(306, 657) = 9$.

To Find:
The Least Common Multiple (LCM) of the two integers, $LCM(306, 657)$.

Step 1: Stating the Fundamental Theorem/Relationship
For any two positive integers $a$ and $b$, the relationship between their HCF and LCM is defined by the following mathematical identity:
$HCF(a, b) \times LCM(a, b) = a \times b$
[This property holds true for any two positive integers.]

Step 2: Substituting the Known Values
Substitute the given values into the identity:
$a = 306$
$b = 657$
$HCF(306, 657) = 9$
$9 \times LCM(306, 657) = 306 \times 657$

Step 3: Isolating the LCM
To find the $LCM(306, 657)$, divide both sides of the equation by $9$:
$LCM(306, 657) = \frac{306 \times 657}{9}$

Step 4: Performing the Arithmetic Calculation
First, simplify the fraction by dividing $306$ by $9$:
$306 \div 9 = 34$
[Since $9 \times 30 = 270$ and $306 - 270 = 36$, and $9 \times 4 = 36$, therefore $30 + 4 = 34$]
Now, multiply the result by $657$:
$LCM(306, 657) = 34 \times 657$

Step 5: Final Multiplication
$34 \times 657 = 34 \times (600 + 50 + 7)$
$= (34 \times 600) + (34 \times 50) + (34 \times 7)$
$= 20400 + 1700 + 238$
$= 22100 + 238$
$= 22338$

Final Answer: The LCM(306, 657) is 22338.

Given: The number $3825$.

To Find: The prime factorization of $3825$.

Step 1: Testing for divisibility by the smallest prime numbers.

We begin by testing the divisibility of $3825$ using the rules of divisibility for prime numbers in ascending order ($2, 3, 5, 7, 11, \dots$).

Check for $2$: The last digit of $3825$ is $5$, which is odd. Therefore, $3825$ is not divisible by $2$.

Check for $3$: The sum of the digits of $3825$ is $3 + 8 + 2 + 5 = 18$. Since $18$ is divisible by $3$, the number $3825$ is divisible by $3$.

$3825 \div 3 = 1275$

Step 2: Continuing the factorization of the quotient.

Now, we factorize $1275$.

Check for $3$: The sum of the digits of $1275$ is $1 + 2 + 7 + 5 = 15$. Since $15$ is divisible by $3$, $1275$ is divisible by $3$.

$1275 \div 3 = 425$

Step 3: Continuing the factorization of the new quotient.

Now, we factorize $425$.

Check for $3$: The sum of the digits of $425$ is $4 + 2 + 5 = 11$. Since $11$ is not divisible by $3$, $425$ is not divisible by $3$.

Check for $5$: The last digit of $425$ is $5$. According to the divisibility rule for $5$, any number ending in $0$ or $5$ is divisible by $5$.

$425 \div 5 = 85$

Step 4: Continuing the factorization of the new quotient.

Now, we factorize $85$.

Check for $5$: The last digit of $85$ is $5$.

$85 \div 5 = 17$

Step 5: Identifying the final prime factor.

The number $17$ is a prime number, meaning its only factors are $1$ and itself.

$17 \div 17 = 1$

Step 6: Expressing the number as a product of prime factors.

Combining all the prime factors obtained from the divisions above:

$3825 = 3 \times 3 \times 5 \times 5 \times 17$

Using exponential notation to group identical factors:

$3825 = 3^2 \times 5^2 \times 17^1$

Final Answer: The prime factorization of $3825$ is $3^2 \times 5^2 \times 17$.

Given: A composite number $156$.

To Find: The prime factorization of $156$, expressed as a product of its prime factors.

Step 1: Understanding Prime Factorization
Prime factorization is the process of expressing a composite number as a product of prime numbers. We will use the method of successive division by the smallest prime numbers ($2, 3, 5, 7, 11, \dots$) until the quotient becomes $1$.

Step 2: Sequential Division

We start by dividing $156$ by the smallest prime number, which is $2$:

$156 \div 2 = 78$

[Since $156$ is an even number, it is divisible by $2$ according to the divisibility rule for $2$.]

Next, we divide the quotient $78$ by $2$:

$78 \div 2 = 39$

[Since $78$ is an even number, it is divisible by $2$.]

Next, we check for divisibility of $39$ by the next prime number, $3$:

$39 \div 3 = 13$

[Since the sum of the digits of $39$ is $3 + 9 = 12$, which is divisible by $3$, $39$ is divisible by $3$ according to the divisibility rule for $3$.]

Finally, we check for divisibility of $13$ by the next prime number, $5, 7, 11, \dots$:

$13 \div 13 = 1$

[Since $13$ is a prime number, it is only divisible by $1$ and itself.]

Step 3: Compiling the Factors
Based on the divisions performed above, we can write $156$ as:

$156 = 2 \times 2 \times 3 \times 13$

Step 4: Expressing in Exponential Form
We group the repeated prime factors using exponents:

$156 = 2^2 \times 3^1 \times 13^1$

Final Answer: The prime factorization of $156$ is $2^2 \times 3 \times 13$.

Given: The integer $140$.

To find: The prime factorization of $140$, expressed as a product of its prime factors.

Step 1: Understanding Prime Factorization
Prime factorization is the process of expressing a composite number as a product of prime numbers. A prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself. We will use the method of successive division by the smallest prime numbers ($2, 3, 5, 7, 11, \dots$).

Step 2: Sequential Division

We begin by dividing $140$ by the smallest prime number, which is $2$:

Next, we divide the quotient $70$ by the smallest prime number, which is $2$:

Next, we check if $35$ is divisible by $2$ or $3$. Since $35$ is odd and the sum of its digits ($3+5=8$) is not divisible by $3$, we move to the next prime number, $5$:

Finally, we divide the quotient $7$ by the prime number $7$:

Step 3: Compiling the Factors
The prime factors obtained are $2, 2, 5,$ and $7$.

We can write the product as:
$140 = 2 \times 2 \times 5 \times 7$

Step 4: Expressing in Exponential Form
Using the law of exponents, where $a \times a = a^2$:

Final Answer: The prime factorization of $140$ is $2^2 \times 5 \times 7$.

Given: Two integers, $a = 510$ and $b = 92$.

To Find: The Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the given integers, and to verify the relationship: $LCM \times HCF = \text{Product of the two numbers}$.

Step 1: Prime Factorization of the given numbers

We express each number as a product of its prime factors.

For $510$:
$510 = 2 \times 255$
$255 = 3 \times 85$
$85 = 5 \times 17$
$17 = 17 \times 1$
So, $510 = 2^1 \times 3^1 \times 5^1 \times 17^1$

For $92$:
$92 = 2 \times 46$
$46 = 2 \times 23$
$23 = 23 \times 1$
So, $92 = 2^2 \times 23^1$

Step 2: Determining HCF and LCM

HCF (Highest Common Factor): The HCF is the product of the smallest power of each common prime factor in the numbers.
Common prime factor is $2$. The smallest power of $2$ is $2^1$.
$HCF(510, 92) = 2^1 = 2$.

LCM (Least Common Multiple): The LCM is the product of the greatest power of each prime factor involved in the numbers.
Prime factors involved are $2, 3, 5, 17, 23$.
Greatest powers are $2^2, 3^1, 5^1, 17^1, 23^1$.
$LCM(510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1$
$LCM(510, 92) = 4 \times 3 \times 5 \times 17 \times 23$
$LCM(510, 92) = 12 \times 5 \times 17 \times 23$
$LCM(510, 92) = 60 \times 17 \times 23$
$LCM(510, 92) = 1020 \times 23$
$LCM(510, 92) = 23460$.

Step 3: Verification of the relationship

We need to verify if $LCM \times HCF = a \times b$.

Left Hand Side (LHS):
$LCM \times HCF = 23460 \times 2$
$LCM \times HCF = 46920$.

Right Hand Side (RHS):
Product of numbers $= 510 \times 92$
$510 \times 92 = 510 \times (90 + 2)$
$= 45900 + 1020$
$= 46920$.

Since $LHS = RHS$, the relationship is verified.

Final Answer: The HCF is $2$, the LCM is $23460$, and the relationship $LCM \times HCF = \text{Product of the two numbers}$ is verified as $46920 = 46920$.

Given: The set of integers $\{8, 9, 25\}$.

To Find: The Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the given integers using the prime factorisation method.

Step 1: Prime Factorisation of the given integers

We express each number as a product of its prime factors:

For $8$:
$8 = 2 \times 4$
$8 = 2 \times 2 \times 2 = 2^3$

For $9$:
$9 = 3 \times 3 = 3^2$

For $25$:
$25 = 5 \times 5 = 5^2$

Step 2: Determining the HCF

The Highest Common Factor (HCF) is defined as the product of the smallest power of each common prime factor involved in the numbers.

Prime factors of $8$: $2^3$
Prime factors of $9$: $3^2$
Prime factors of $25$: $5^2$

Since there are no common prime factors among $8, 9,$ and $25$ other than $1$, we can write:

$HCF(8, 9, 25) = 1$

Step 3: Determining the LCM

The Least Common Multiple (LCM) is defined as the product of the greatest power of each prime factor involved in the numbers.

The prime factors present are $2, 3,$ and $5$. The greatest powers are:

Greatest power of $2 = 2^3 = 8$
Greatest power of $3 = 3^2 = 9$
Greatest power of $5 = 5^2 = 25$

Therefore, $LCM(8, 9, 25) = 2^3 \times 3^2 \times 5^2$

$LCM = 8 \times 9 \times 25$

Calculating the product sequentially:

$8 \times 9 = 72$
$72 \times 25 = 72 \times \frac{100}{4} = 18 \times 100 = 1800$

Step 4: Verification of Results

The prime factorisation method yields:

HCF = $1$
LCM = $1800$

Final Answer: The HCF of 8, 9, and 25 is 1, and the LCM is 1800.

Given: The integer $7429$.

To Find: The prime factorization of $7429$.

Step 1: Testing for divisibility by small prime numbers.

We test the divisibility of $7429$ by the sequence of prime numbers: $2, 3, 5, 7, 11, 13, 17, \dots$

1. Divisibility by 2: The last digit is $9$ (odd), so it is not divisible by $2$.

2. Divisibility by 3: The sum of the digits is $7 + 4 + 2 + 9 = 22$. Since $22$ is not divisible by $3$, $7429$ is not divisible by $3$.

3. Divisibility by 5: The last digit is not $0$ or $5$, so it is not divisible by $5$.

4. Divisibility by 7: $7429 \div 7 = 1061$ with a remainder of $2$. Not divisible.

5. Divisibility by 11: The alternating sum of digits is $7 - 4 + 2 - 9 = -4$. Not divisible by $11$.

6. Divisibility by 13: $7429 \div 13 = 571$ with a remainder of $6$. Not divisible.

7. Divisibility by 17: $7429 \div 17 = 437$. Since the remainder is $0$, $17$ is a prime factor.

Step 2: Factoring the quotient $437$.

We continue testing for divisibility starting from $17$ (since prime factors are non-decreasing):

1. Divisibility by 17: $437 \div 17 = 25$ with a remainder of $12$. Not divisible.

2. Divisibility by 19: $437 \div 19 = 23$. Since the remainder is $0$, $19$ is a prime factor.

Step 3: Factoring the quotient $23$.

Since $23$ is a prime number, it is only divisible by $1$ and itself.

Step 4: Compiling the prime factors.

The prime factorization is obtained by multiplying the prime divisors found in the previous steps:

$7429 = 17 \times 437$

$7429 = 17 \times 19 \times 23$

[Justification: By the Fundamental Theorem of Arithmetic, every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.]

Final Answer: 17 \times 19 \times 23

Given: Three integers: $17$, $23$, and $29$.

To find: The Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the given integers using the prime factorisation method.

Step 1: Prime Factorisation of the given integers

We express each number as a product of its prime factors:

For $17$: Since $17$ is a prime number, its only factors are $1$ and itself.
$17 = 17^1 \times 1^1$

For $23$: Since $23$ is a prime number, its only factors are $1$ and itself.
$23 = 23^1 \times 1^1$

For $29$: Since $29$ is a prime number, its only factors are $1$ and itself.
$29 = 29^1 \times 1^1$

Step 2: Finding the HCF

The Highest Common Factor (HCF) is defined as the product of the smallest power of each common prime factor in the numbers.

Comparing the prime factors of $17$, $23$, and $29$:
The prime factors are $17$, $23$, and $29$. There are no common prime factors other than $1$.

Therefore, $HCF(17, 23, 29) = 1$.

Step 3: Finding the LCM

The Least Common Multiple (LCM) is defined as the product of the greatest power of each prime factor involved in the numbers.

The prime factors involved are $17$, $23$, and $29$. The greatest power of each is $1$.

$LCM(17, 23, 29) = 17^1 \times 23^1 \times 29^1$

Calculation:
First, multiply $17$ and $23$:
$17 \times 23 = 17 \times (20 + 3) = 340 + 51 = 391$

Next, multiply the result by $29$:
$391 \times 29 = 391 \times (30 - 1)$
$= (391 \times 30) - (391 \times 1)$
$= 11730 - 391$

Performing the subtraction:
$11730 - 391 = 11339$

Therefore, $LCM(17, 23, 29) = 11339$.

Final Answer: The HCF is 1 and the LCM is 11339.

Given: Two integers, $a = 26$ and $b = 91$.

To Find: The Highest Common Factor (HCF) and Least Common Multiple (LCM) of 26 and 91, and to verify the relationship: $LCM \times HCF = a \times b$.

Step 1: Prime Factorization of the given numbers

To find the HCF and LCM, we first express each number as a product of its prime factors.

For 26:
$26 = 2 \times 13$
[Since 26 is an even number, we divide by the smallest prime 2, leaving 13, which is prime.]

For 91:
$91 = 7 \times 13$
[Since 91 is not divisible by 2, 3, or 5, we test 7: $91 \div 7 = 13$. Both 7 and 13 are prime numbers.]

Step 2: Determining HCF and LCM

HCF (Highest Common Factor): The HCF is the product of the smallest power of each common prime factor in the numbers.
Common prime factor: $13^1$
$HCF(26, 91) = 13$

LCM (Least Common Multiple): The LCM is the product of the greatest power of each prime factor involved in the numbers.
Prime factors involved: $2^1, 7^1, 13^1$
$LCM(26, 91) = 2 \times 7 \times 13$
$LCM(26, 91) = 14 \times 13$
$LCM(26, 91) = 182$

Step 3: Verification of the relationship $LCM \times HCF = a \times b$

Left Hand Side (LHS):
$LCM \times HCF = 182 \times 13$
Calculation:
$182 \times 10 = 1820$
$182 \times 3 = 546$
$1820 + 546 = 2366$
$LHS = 2366$

Right Hand Side (RHS):
$Product of numbers = 26 \times 91$
Calculation:
$26 \times 90 = 2340$
$26 \times 1 = 26$
$2340 + 26 = 2366$
$RHS = 2366$

Conclusion:
Since $LHS = RHS$ ($2366 = 2366$), the relationship is verified.

Final Answer: The HCF is 13, the LCM is 182, and the relationship $LCM \times HCF = \text{product of the two numbers}$ is verified.

Given: The number $5005$.

To Find: The prime factorization of $5005$.

Step 1: Understanding Prime Factorization
Prime factorization is the process of expressing a composite number as a product of prime numbers. We will use the method of successive division by the smallest prime numbers ($2, 3, 5, 7, 11, 13, \dots$) until the quotient becomes $1$.

Step 2: Sequential Division

We test the divisibility of $5005$ by prime numbers in ascending order:

1. Check for 2: The last digit of $5005$ is $5$, which is odd. Therefore, it is not divisible by $2$.

2. Check for 3: The sum of the digits is $5 + 0 + 0 + 5 = 10$. Since $10$ is not divisible by $3$, $5005$ is not divisible by $3$.

3. Check for 5: The last digit is $5$. According to the divisibility rule for $5$, any number ending in $0$ or $5$ is divisible by $5$.
$5005 \div 5 = 1001$

4. Check for 7: We divide $1001$ by $7$.
$1001 \div 7 = 143$
Since the remainder is $0$, $7$ is a prime factor.

5. Check for 11: We divide $143$ by $11$.
$143 \div 11 = 13$
Since the remainder is $0$, $11$ is a prime factor.

6. Check for 13: We divide $13$ by $13$.
$13 \div 13 = 1$
Since the quotient is $1$, the process is complete.

Step 3: Tabular Representation of Factors

Step 4: Expressing as a Product
By collecting the divisors used in the successive division, we can write:
$5005 = 5 \times 1001$
$5005 = 5 \times 7 \times 143$
$5005 = 5 \times 7 \times 11 \times 13$

Final Answer: The prime factorization of $5005$ is $5 \times 7 \times 11 \times 13$.

Given: The set of integers $\{12, 15, 21\}$.

To Find: The Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the given integers using the prime factorisation method.

Step 1: Prime Factorisation of the given integers

We express each integer as a product of its prime factors:

For $12$:
$12 = 2 \times 6$
$12 = 2 \times 2 \times 3$
$12 = 2^2 \times 3^1$

For $15$:
$15 = 3 \times 5$
$15 = 3^1 \times 5^1$

For $21$:
$21 = 3 \times 7$
$21 = 3^1 \times 7^1$

Step 2: Determining the HCF

The Highest Common Factor (HCF) is defined as the product of the smallest power of each common prime factor in the numbers.

Looking at the prime factors:
Common prime factor: $3$
Smallest power of $3$ present in all factorisations is $3^1$.

Therefore, $\text{HCF}(12, 15, 21) = 3^1 = 3$.

Step 3: Determining the LCM

The Least Common Multiple (LCM) is defined as the product of the greatest power of each prime factor involved in the numbers.

The prime factors involved are $2, 3, 5,$ and $7$.
Greatest power of $2$ is $2^2$.
Greatest power of $3$ is $3^1$.
Greatest power of $5$ is $5^1$.
Greatest power of $7$ is $7^1$.

Calculating the LCM:
$\text{LCM}(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1$
$\text{LCM}(12, 15, 21) = 4 \times 3 \times 5 \times 7$
$\text{LCM}(12, 15, 21) = 12 \times 5 \times 7$
$\text{LCM}(12, 15, 21) = 60 \times 7$
$\text{LCM}(12, 15, 21) = 420$

Final Answer: The HCF of 12, 15, and 21 is 3, and the LCM is 420.

Given: Two integers, $336$ and $54$.

To Find: The Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the given integers, and to verify the relationship: $LCM \times HCF = \text{Product of the two numbers}$.

Step 1: Prime Factorization of the given numbers

We express each number as a product of its prime factors.

For $336$:
$336 = 2 \times 168$
$168 = 2 \times 84$
$84 = 2 \times 42$
$42 = 2 \times 21$
$21 = 3 \times 7$
Thus, $336 = 2^4 \times 3^1 \times 7^1$

For $54$:
$54 = 2 \times 27$
$27 = 3 \times 9$
$9 = 3 \times 3$
Thus, $54 = 2^1 \times 3^3$

Step 2: Determining HCF and LCM

HCF (Highest Common Factor): The HCF is the product of the smallest power of each common prime factor in the numbers.
Common prime factors are $2$ and $3$.
Smallest power of $2$ is $2^1$.
Smallest power of $3$ is $3^1$.
$HCF = 2^1 \times 3^1 = 6$.

LCM (Least Common Multiple): The LCM is the product of the greatest power of each prime factor involved in the numbers.
Greatest power of $2$ is $2^4$.
Greatest power of $3$ is $3^3$.
Greatest power of $7$ is $7^1$.
$LCM = 2^4 \times 3^3 \times 7^1 = 16 \times 27 \times 7$.
$16 \times 27 = 432$.
$432 \times 7 = 3024$.
$LCM = 3024$.

Step 3: Verification of the relationship

We need to verify if $LCM \times HCF = \text{Product of the two numbers}$.

Left Hand Side (LHS):
$LCM \times HCF = 3024 \times 6$
$3024 \times 6 = 18144$.

Right Hand Side (RHS):
Product of the two numbers $= 336 \times 54$
$336 \times 50 = 16800$
$336 \times 4 = 1344$
$16800 + 1344 = 18144$.

Since $LHS = RHS$ ($18144 = 18144$), the relationship is verified.

Final Answer: The HCF is 6, the LCM is 3024, and the relationship $LCM \times HCF = \text{Product of the two numbers}$ is verified.

Given:

Time taken by Sonia to complete one round = $18$ minutes.

Time taken by Ravi to complete one round = $12$ minutes.

Both start from the same point, at the same time, and move in the same direction.

To Find:

The time after which they will meet again at the starting point.

Step 1: Understanding the Mathematical Requirement

Since Sonia and Ravi are moving in a circular path and we need to find the earliest time they meet at the starting point, we are looking for the smallest time interval that is a multiple of both $18$ and $12$. This is defined as the Least Common Multiple (LCM) of the two time durations.

Step 2: Prime Factorization of the Given Numbers

To find the LCM, we first express each number as a product of its prime factors.

For $18$:

$18 = 2 \times 9 = 2 \times 3 \times 3 = 2^1 \times 3^2$

For $12$:

$12 = 2 \times 6 = 2 \times 2 \times 3 = 2^2 \times 3^1$

Step 3: Calculating the Least Common Multiple (LCM)

The LCM of two or more numbers is the product of the highest power of each prime factor involved in the numbers.

Prime factors involved: $2$ and $3$.

Highest power of $2$: $2^2$ (from $12$).

Highest power of $3$: $3^2$ (from $18$).

Therefore, $\text{LCM}(18, 12) = 2^2 \times 3^2$.

$\text{LCM}(18, 12) = 4 \times 9 = 36$.

Step 4: Verification

Sonia completes rounds at: $18, 36, 54, \dots$ minutes.

Ravi completes rounds at: $12, 24, 36, 48, \dots$ minutes.

The first common multiple is $36$.

Final Answer: They will meet again at the starting point after 36 minutes.

Given: Two numerical expressions:
1. $A = 7 \times 11 \times 13 + 13$
2. $B = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

To Prove: Both $A$ and $B$ are composite numbers.

Theoretical Background:
According to the Fundamental Theorem of Arithmetic, every composite number can be expressed (factorized) as a product of primes. A composite number is defined as a positive integer greater than $1$ that has at least one divisor other than $1$ and itself. In other words, if a number can be expressed as a product of two or more factors where each factor is greater than $1$, it is a composite number.

Step 1: Analyzing the first expression ($A$)
Let $A = 7 \times 11 \times 13 + 13$.
We observe that $13$ is a common factor in both terms of the expression.
Factoring out $13$:
$A = 13 \times (7 \times 11 \times 1 + 1)$ [Using the Distributive Property: $ab + ac = a(b+c)$]
$A = 13 \times (77 + 1)$
$A = 13 \times 78$
Since $78$ can be further factorized as $2 \times 3 \times 13$, we have:
$A = 13 \times 2 \times 3 \times 13 = 2 \times 3 \times 13^2$
Since $A$ is expressed as a product of prime factors ($2, 3, 13$), it satisfies the definition of a composite number.

Step 2: Analyzing the second expression ($B$)
Let $B = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$.
We observe that $5$ is a common factor in both terms of the expression.
Factoring out $5$:
$B = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$ [Using the Distributive Property]
Calculating the product inside the parentheses:
$7 \times 6 = 42$
$42 \times 4 = 168$
$168 \times 3 = 504$
$504 \times 2 = 1008$
$B = 5 \times (1008 + 1)$
$B = 5 \times 1009$
Since $1009$ is a prime number, $B$ is expressed as the product of two factors ($5$ and $1009$), both of which are greater than $1$. Therefore, $B$ is a composite number.

Conclusion:
Since both expressions $A$ and $B$ can be written as a product of factors other than $1$ and the numbers themselves, they are confirmed to be composite numbers.

Final Answer: Both $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers because they possess factors other than $1$ and themselves.

Given: A number of the form $6^n$, where $n$ is a natural number ($n \in \mathbb{N}$).

To Find: Whether there exists any natural number $n$ such that $6^n$ ends with the digit $0$.

Step 1: Understanding the condition for a number to end with the digit 0.

Any positive integer that ends with the digit $0$ must be divisible by $10$. In terms of prime factorization, if a number is divisible by $10$, it must also be divisible by the prime factors of $10$. Since $10 = 2 \times 5$, any number ending in $0$ must have both $2$ and $5$ as prime factors in its prime factorization.

Step 2: Prime factorization of the base.

The given number is $6^n$. First, we find the prime factorization of the base, which is $6$.
$6 = 2 \times 3$.

Step 3: Expressing $6^n$ in terms of its prime factors.

Substituting the prime factorization of $6$ into the expression $6^n$:
$6^n = (2 \times 3)^n$
Using the exponent rule $(a \times b)^n = a^n \times b^n$, we get:
$6^n = 2^n \times 3^n$

Step 4: Analyzing the Fundamental Theorem of Arithmetic.

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

In the prime factorization of $6^n = 2^n \times 3^n$, the only prime factors present are $2$ and $3$.

Step 5: Conclusion based on the analysis.

For $6^n$ to end with the digit $0$, its prime factorization must contain the prime factor $5$. However, we have shown that the prime factorization of $6^n$ is $2^n \times 3^n$. Since $5$ is not a prime factor of $6^n$ for any value of $n$, it is impossible for $6^n$ to be divisible by $10$.

Therefore, there is no natural number $n$ for which $6^n$ ends with the digit $0$.

Final Answer: No, $6^n$ cannot end with the digit 0 for any natural number $n$ because its prime factorization does not contain the prime factor 5.