Given: Three pairs of integers: (i) 26 and 91, (ii) 510 and 92, (iii) 336 and 54.

To Find: LCM and HCF for each pair, and verify \(\text{LCM} \times \text{HCF} = \text{Product of the two numbers}\).

Part (i): Find HCF and LCM of 26 and 91

Step-by-step reasoning:

We’ll find the prime factorization of both numbers to determine HCF, then use the relationship to find LCM.

Prime factorization: \(26 = 2 \times 13\), \(91 = 7 \times 13\).
HCF is the product of common prime factors: \(\text{HCF} = 13\).
Using \(a \times b = \text{HCF} \times \text{LCM}\),
we have \(26 \times 91 = 13 \times \text{LCM}\).
So, \(\text{LCM} = \frac{26 \times 91}{13} = \frac{2366}{13} = 182\).
Verification: \(\text{LCM} \times \text{HCF} = 182 \times 13 = 2366\) and \(26 \times 91 = 2366\).
Both are equal.

Result: HCF = 13, LCM = 182. Verification successful: \(182 \times 13 = 2366 = 26 \times 91\).

Part (ii): Find HCF and LCM of 510 and 92

Step-by-step reasoning:

We’ll use prime factorization for HCF. Since numbers are larger, we can also use Euclidean algorithm as an alternative.

Prime factorization: \(510 = 2 \times 3 \times 5 \times 17\), \(92 = 2^2 \times 23\).
HCF is the product of common prime factors: \(\text{HCF} = 2\).
Using \(a \times b = \text{HCF} \times \text{LCM}\),
we have \(510 \times 92 = 2 \times \text{LCM}\).
So, \(\text{LCM} = \frac{510 \times 92}{2} = \frac{46920}{2} = 23460\).
Verification: \(\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920\) and \(510 \times 92 = 46920\).
Both are equal.

Result: HCF = 2, LCM = 23460. Verification successful: \(23460 \times 2 = 46920 = 510 \times 92\).

Part (iii): Find HCF and LCM of 336 and 54

Step-by-step reasoning:

We’ll find prime factorization for HCF. Note that these numbers have common factors beyond just small primes.

Prime factorization: \(336 = 2^4 \times 3 \times 7\), \(54 = 2 \times 3^3\).
HCF is the product of common prime factors with lowest powers: \(\text{HCF} = 2^1 \times 3^1 = 6\).
Using \(a \times b = \text{HCF} \times \text{LCM}\),
we have \(336 \times 54 = 6 \times \text{LCM}\).
So, \(\text{LCM} = \frac{336 \times 54}{6} = \frac{18144}{6} = 3024\).
Verification: \(\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144\) and \(336 \times 54 = 18144\).
Both are equal.

Result: HCF = 6, LCM = 3024. Verification successful: \(3024 \times 6 = 18144 = 336 \times 54\).

Final Answer:

(i) HCF(26,91) = 13, LCM(26,91) = 182; Verification: \(182 × 13 = 2366 = 26 × 91\). (ii) HCF(510,92) = 2, LCM(510,92) = 23460; Verification: \(23460 × 2 = 46920 = 510 × 92\). (iii) HCF(336,54) = 6, LCM(336,54) = 3024; Verification: \(3024 × 6 = 18144 = 336 × 54\).

Key Concepts Used

• Key Concept: For any two positive integers \(a\) and \(b\), the product of the numbers is equal to the product of their Highest Common Factor (HCF) and Least Common Multiple (LCM): \(a × b = HCF(a,b) × LCM(a,b)\). This is derived from the Fundamental Theorem of Arithmetic, which states every integer greater than 1 can be uniquely expressed as a product of primes.

Given: Three sets of integers: (i) 12, 15, 21; (ii) 17, 23, 29; (iii) 8, 9, 25

To Find: LCM and HCF for each set using prime factorisation method

Part (i): Prime Factorisation of 12, 15, and 21

Step-by-step reasoning:

First, we find the prime factors of each number by dividing them by prime numbers starting from 2.

\(12 = 2 \times 2 \times 3 = 2^2 \times 3^1\)
\(15 = 3 \times 5 = 3^1 \times 5^1\)
\(21 = 3 \times 7 = 3^1 \times 7^1\)

Result: Prime factors: \(12: 2^2, 3^1\); \(15: 3^1, 5^1\); \(21: 3^1, 7^1\)

Part (i): Calculate LCM for 12, 15, and 21

Step-by-step reasoning:

LCM is the product of the highest powers of all prime factors present in any of the numbers. The prime factors involved are 2, 3, 5, and 7.

\(\text{LCM} = 2^{\text{max}(2)} \times 3^{\text{max}(1,1,1)} \times 5^{\text{max}(1)} \times 7^{\text{max}(1)}\)
Highest powers: \(2^2\), \(3^1\), \(5^1\), \(7^1\)
\(\text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7\)

Result: \(\text{LCM} = 420\)

Part (i): Calculate HCF for 12, 15, and 21

Step-by-step reasoning:

HCF is the product of the lowest powers of common prime factors. The only common prime factor among all three numbers is 3.

\(\text{HCF} = 3^{\text{min}(1,1,1)}\)
Lowest power: \(3^1\)

Result: \(\text{HCF} = 3\)

Part (ii): Prime Factorisation of 17, 23, and 29

Step-by-step reasoning:

These numbers are all prime numbers themselves, so their prime factorisation is straightforward.

\(17 = 17^1\) (prime)
\(23 = 23^1\) (prime)
\(29 = 29^1\) (prime)

Result: Prime factors: \(17: 17^1\); \(23: 23^1\); \(29: 29^1\)

Part (ii): Calculate LCM for 17, 23, and 29

Step-by-step reasoning:

Since all numbers are distinct primes with no common factors, LCM is simply the product of all numbers.

\(\text{LCM} = 17^1 \times 23^1 \times 29^1 = 17 \times 23 \times 29\)

Result: \(\text{LCM} = 11339\)

Part (ii): Calculate HCF for 17, 23, and 29

Step-by-step reasoning:

HCF requires common prime factors. Since these are distinct primes with no common factors, HCF is 1.

\(\text{HCF} = \text{Product of lowest powers of common prime factors}\)
No common prime factors → HCF = \(1\)

Result: \(\text{HCF} = 1\)

Part (iii): Prime Factorisation of 8, 9, and 25

Step-by-step reasoning:

Find prime factors by dividing each number by primes. Note that these are composite numbers with no common primes.

\(8 = 2 \times 2 \times 2 = 2^3\)
\(9 = 3 \times 3 = 3^2\)
\(25 = 5 \times 5 = 5^2\)

Result: Prime factors: \(8: 2^3\); \(9: 3^2\); \(25: 5^2\)

Part (iii): Calculate LCM for 8,9,and25

Step-by-step reasoning:

LCM takes the highest power of each prime factor from all numbers. The primes are distinct:2,3,and5.

\(\text{LCM} = 2^{\text{max}(3)} \times 3^{\text{max}(2)} \times 5^{\text{max}(2)}\)
Highest powers:\(2^3,\,3^2,\,5^2\)
\(\text{LCM}=8\times9\times25=8\times225=1800\)

Result: \(\text{LCM}=1800\)

Part(iii):Calculate HCF for8,9,and25

Step-by-step reasoning:

HCF requires common prime factors.Since these numbers have no common primes,HCF is1.

No common prime factors→HCF=\(1\)

Result: \(\text{HCF}=1\)

Final Answer:

(i) LCM=420,HCF=3;(ii)LCM=11339,HCF=1;(iii)LCM=1800,HCF=1

Key Concepts Used

• Key Concept: Prime factorisation method:LCM=product of highest powers of all primes;HCF=product of lowest powers of common primes.If no common primes,HCF=1.

Given: \(6^n\) where \(n\) is any natural number (positive integer)

To Find: Whether \(6^n\) can end with digit 0 for any natural number \(n\)

Understanding what ‘ending with digit 0’ means

Step-by-step reasoning:

A number ending with digit 0 is divisible by 10. For example: 10, 20, 30, 100, etc. all end with 0 and are divisible by 10. Mathematically, if a number ends with digit 0, it can be written as \(10 \times k\) where \(k\) is some integer.

If \(6^n\) ends with digit 0 ⇒ \(6^n\) is divisible by 10 ⇒ \(10 \mid 6^n\)

Result: \(6^n\) must be divisible by 10

Prime factorization of 10

Step-by-step reasoning:

To understand divisibility by 10, we need to examine the prime factors of 10. According to the Fundamental Theorem of Arithmetic, every composite number can be expressed as a product of primes in a unique way (up to the order of factors).

\(10 = 2 \times 5\)

Result: 10 has prime factors: 2 and 5

Condition for divisibility by 10

Step-by-step reasoning:

For a number to be divisible by 10, it must contain both prime factors 2 and 5 in its prime factorization. This is because if a number is divisible by 10, it must be divisible by both 2 and 5 (since 10 = 2 × 5).

\(10 \mid 6^n\) ⇒ \(2 \mid 6^n\) and \(5 \mid 6^n\)

Result: \(6^n\) must be divisible by both 2 and 5

Prime factorization of 6

Step-by-step reasoning:

Now let’s examine the prime factors of the base number 6. We need to understand what happens when we raise 6 to any power \(n\).

\(6 = 2 \times 3\)

Result: 6 has prime factors: 2 and 3

Prime factorization of \(6^n\)

Step-by-step reasoning:

When we raise a number to a power, we multiply all its prime factors by that exponent. According to the laws of exponents: \((a \times b)^n = a^n \times b^n\).

\(6^n = (2 \times 3)^n = 2^n \times 3^n\)

Result: \(6^n = 2^n \times 3^n\)

Checking for prime factor 5 in \(6^n\)

Step-by-step reasoning:

Now we examine the prime factorization of \(6^n\). From step 5, we see that \(6^n = 2^n \times 3^n\). Notice that this expression contains only prime factors 2 and 3. There is no factor of 5 in this expression.

\(6^n = 2^n \times 3^n\) contains only primes: \(2\) and \(3\)

Result: \(6^n\) does not contain prime factor 5

Conclusion about divisibility by 5

Step-by-step reasoning:

Since \(6^n\) does not contain the prime factor 5 in its prime factorization, it cannot be divisible by 5. Remember from step 3: for \(6^n\) to be divisible by 10, it must be divisible by both 2 and 5.

\(5 \nmid 6^n\) because \(6^n = 2^n \times 3^n\) has no factor of \(5\)

Result: \(6^n\) is not divisible by 5

Final reasoning

Step-by-step reasoning:

Since \(6^n\) is not divisible by 5 (it lacks the prime factor 5), it cannot be divisible by 10. Therefore, \(6^n\) cannot end with digit 0.

\(6^n\) not divisible by \(5\) ⇒ \(6^n\) not divisible by \(10\) ⇒ \(6^n\) cannot end with digit \(0\)

Result: \(6^n\) cannot end with digit 0 for any natural number \(n\)

Final Answer:

\(6^n\) cannot end with the digit \(0\) for any natural number \(n\).

Key Concepts Used

• Ending with digit zero condition: A number ends with digit zero if and only if it is divisible by \(10\). Since \(10 = 2 \times 5\), for a number to end with zero, its prime factorization must contain both prime factors \(2\) and \(5\). This is a fundamental application of the Fundamental Theorem of Arithmetic.

Given: Two expressions: \(7 \times 11 \times 13 + 13\) and \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\).

To Find: Prove that both expressions represent composite numbers.

Analyze the First Expression

Step-by-step reasoning:

We start with the expression \(7 \times 11 \times 13 + 13\). Notice that \(13\) is a common factor in both terms. By factoring out \(13\), we can rewrite the expression in a product form, which will help identify its factors.

\(7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1)\)

Result: The expression simplifies to \(13 \times (77 + 1) = 13 \times 78\).

Factorize Further to Show Compositeness

Step-by-step reasoning:

Now, \(13 \times 78\) is clearly a product of two integers greater than 1. To confirm it’s composite, we factorize \(78\) into its prime factors, showing that the entire expression has multiple factors.

\(78 = 2 \times 3 \times 13\).
Thus, \(13 \times 78 = 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2\).

Result: The expression equals \(2 \times 3 \times 13^2\), which has factors: \(1, 2, 3, 13, 26, 39, 78, 169, 338, 507, 1014\). Since it has more than two factors, it is composite.

Analyze the Second Expression

Step-by-step reasoning:

Now consider \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\). Here, \(5\) is a common factor. Factor out \(5\) to simplify the expression.

\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)\)

Result: The expression simplifies to \(5 \times (1008 + 1) = 5 \times 1009\), since \(7 \times 6 \times 4 \times 3 \times 2 = 1008\).

Check if \(1009\) is Prime or Composite

Step-by-step reasoning:

To determine if \(5 \times 1009\) is composite, we need to check whether \(1009\) is prime or composite. If \(1009\) is prime, then \(5 \times 1009\) has exactly two prime factors, making it composite because it has more than two divisors. We test divisibility by primes up to \(\sqrt{1009}\).

\(\sqrt{1009} \approx 31.76\).
Test primes: Not divisible by \(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31\).

Result: \(1009\) is prime. Thus, \(5 \times 1009\) is a product of two primes.

Conclude Compositeness for Second Expression

Step-by-step reasoning:

Since \(5\) and \(1009\) are both primes greater than \(1\), their product \(5 \times 1009\) has divisors: \(1, 5, 1009,\) and \(5045\). This satisfies the definition of a composite number.

Divisors of \(5 \times 1009\): \(1, 5, 1009, 5045\).

Result: The expression has more than two factors, so it is composite.

Final Answer:

Both expressions are composite numbers. For \(7 \times 11 \times 13 + 13\), factorization gives \(2 \times 3 \times 13^2\), which has multiple factors. For \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\), factorization gives \(5 \times 1009\), where \(1009\) is prime, resulting in a product of two primes with divisors beyond \(1\) and itself.

Key Concepts Used

• Key Concept: A number is composite if it has more than two distinct positive divisors. By factoring expressions to reveal products of integers greater than \(1\), we can prove compositeness. This method applies the Fundamental Theorem of Arithmetic indirectly.

Given: Sonia’s time for one round = 18 minutes, Ravi’s time for one round = 12 minutes, Both start together from the same point.

To Find: Time when they meet again at the starting point.

Identify Individual Lap Times

Step-by-step reasoning:

First, note the time each person takes to complete one full round of the circular path. These times are given directly in the problem.

Sonia’s time: \( T_s = 18 \text{ minutes} \)
Ravi’s time: \( T_r = 12 \text{ minutes} \)

Result: \( T_s = 18 \), \( T_r = 12 \)

Understand the Meeting Condition

Step-by-step reasoning:

For Sonia and Ravi to meet again at the starting point, both must have completed an integer number of laps. Let Sonia complete \( n_s \) laps and Ravi complete \( n_r \) laps in the meeting time \( T \). Since they start together, the time taken is the same for both. The distance covered in terms of laps is related to time by their speeds. Alternatively, we can think in terms of time: \( T \) must be a multiple of both 18 and 12 minutes.

Let meeting time = \( T \) minutes.
Then, \( T = n_s \times 18 \) and \( T = n_r \times 12 \),
where \( n_s \) and \( n_r \) are positive integers.
Thus, \( T \) is a common multiple of 18 and 12.

Result: \( T \) is a common multiple of 18 and 12.

Find the Least Common Multiple (LCM)

Step-by-step reasoning:

To find the earliest time they meet, we need the smallest positive common multiple, i.e., the LCM of 18 and 12. We can find LCM using prime factorization or division method.

Prime factorization:
\( 18 = 2 \times 3^2 \)
\( 12 = 2^2 \times 3 \)
LCM = product of highest powers of all prime factors: \( 2^2 \times 3^2 = 4 \times 9 = 36 \).
Alternatively,
using division method:
Divide by common factors:
2 | 18, 12
3 | 9, 6
| 3, 2
LCM = \( 2 \times 3 \times 3 \times 2 = 36 \).

Result: \( \text{LCM}(18, 12) = 36 \)

Interpret the LCM as Meeting Time

Step-by-step reasoning:

The LCM, 36 minutes, is the smallest time that is a multiple of both 18 and 12. At this time, Sonia completes exactly \( \frac{36}{18} = 2 \) laps and Ravi completes exactly \( \frac{36}{12} = 3 \) laps. Both return to the starting point simultaneously.

\( T = 36 \text{ minutes} \)
Sonia’s laps: \( n_s = \frac{T}{18} = \frac{36}{18} = 2 \)
Ravi’s laps: \( n_r = \frac{T}{12} = \frac{36}{12} = 3 \)

Result: \( T = 36 \), Sonia completes 2 laps, Ravi completes 3 laps.

Verify the Solution

Step-by-step reasoning:

Check that at 36 minutes, both are at the starting point. Sonia takes 18 minutes per lap, so in 36 minutes she completes \( 36 / 18 = 2 \) laps, ending at start. Ravi takes 12 minutes per lap, so in 36 minutes he completes \( 36 / 12 = 3 \) laps, also ending at start. Thus, they meet.

Sonia: \( 2 \times 18 = 36 \),
position: start.
Ravi: \( 3 \times 12 = 36 \),
position: start.

Result: Verification confirms they meet at starting point after 36 minutes.

Final Answer:

Sonia and Ravi will meet again at the starting point after \(\mathbf{36}\) minutes.

Key Concepts Used

• : In problems involving periodic events or motions on a circular track, the LCM of the time periods gives the first time when all events coincide or objects meet at the starting point. Here, LCM(18, 12) = 36 minutes is used because both Sonia and Ravi complete integer laps in this time, ensuring they are together at the start.