EXERCISE 1.1
Real Numbers • 7 Questions
Question 1
Hint available
Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Key Idea
Use the division method of prime factorisation – repeatedly divide the given number by the smallest possible prime (2, 3, 5, 7, …) until the quotient becomes 1. The primes used in the divisions, together with their multiplicities, give the required product of prime factors.
Step-by-Step Solution
(i) 140
- Step 1: 140 is even, divide by 2 → $140 \div 2 = 70$.
- Step 2: 70 is even, divide by 2 → $70 \div 2 = 35$.
- Step 3: 35 is not divisible by 2, try 3 (no), try 5 → $35 \div 5 = 7$.
- Step 4: 7 is a prime number.
- Hence $140 = 2 \times 2 \times 5 \times 7 = 2^{2}\,5\,7$.
(ii) 156
- Step 1: 156 is even, divide by 2 → $156 \div 2 = 78$.
- Step 2: 78 is even, divide by 2 → $78 \div 2 = 39$.
- Step 3: 39 is not divisible by 2, try 3 → $39 \div 3 = 13$.
- Step 4: 13 is a prime number.
- Hence $156 = 2 \times 2 \times 3 \times 13 = 2^{2}\,3\,13$.
(iii) 3825
- Step 1: The last digit is 5, so 3825 is divisible by 5 → $3825 \div 5 = 765$.
- Step 2: 765 also ends with 5, divide by 5 again → $765 \div 5 = 153$.
- Step 3: Sum of digits of 153 is $1+5+3=9$, divisible by 3 → $153 \div 3 = 51$.
- Step 4: 51 is divisible by 3 → $51 \div 3 = 17$.
- Step 5: 17 is a prime number.
- Hence $3825 = 5 \times 5 \times 3 \times 3 \times 17 = 5^{2}\,3^{2}\,17$.
(iv) 5005
- Step 1: The number is odd, not divisible by 2. Sum of digits $5+0+0+5=10$ → not divisible by 3. Ends with 5 → not divisible by 5. Try 7: $5005 \div 7 = 715$ (exact).
- Step 2: 715 is odd; sum of digits $7+1+5=13$ → not divisible by 3. Ends with 5 → divisible by 5 → $715 \div 5 = 143$.
- Step 3: 143 is not divisible by 2,3,5. Try 7 → $143 \div 7 = 20.428$ (not integer). Try 11 → $143 \div 11 = 13$.
- Step 4: 13 is prime.
- Hence $5005 = 7 \times 5 \times 11 \times 13 = 5\,7\,11\,13$.
(v) 7429
- Step 1: The number is odd, not divisible by 2. Sum of digits $7+4+2+9 = 22$ → not divisible by 3. Does not end with 5 → not divisible by 5.
- Step 2: Test divisibility by 7: $7429 \div 7 = 1061.285$ (not integer). Test 11: $7429 \div 11 = 675.363$ (no). Test 13: $7429 \div 13 = 571.461$ (no).
- Step 3: Test 17: $7429 \div 17 = 437$ (exact). So 17 is a factor.
- Step 4: Now factor 437. It is odd, sum of digits $4+3+7 = 14$ → not divisible by 3. Ends with 7 → not 5. Test 7: $437 \div 7 = 62.428$ (no). Test 11: $437 \div 11 = 39.727$ (no). Test 13: $437 \div 13 = 33.615$ (no). Test 17: $437 \div 17 = 25.705$ (no). Test 19: $437 \div 19 = 23$ (exact).
- Step 5: 23 is a prime number.
- Hence $7429 = 17 \times 19 \times 23$.
Thus the prime‑factorisations are:
- $140 = 2^{2}\,5\,7$
- $156 = 2^{2}\,3\,13$
- $3825 = 5^{2}\,3^{2}\,17$
- $5005 = 5\,7\,11\,13$
- $7429 = 17\,19\,23$
- Step 1: 140 is even, divide by 2 → $140 \div 2 = 70$.
- Step 2: 70 is even, divide by 2 → $70 \div 2 = 35$.
- Step 3: 35 is not divisible by 2, try 3 (no), try 5 → $35 \div 5 = 7$.
- Step 4: 7 is a prime number.
- Hence $140 = 2 \times 2 \times 5 \times 7 = 2^{2}\,5\,7$.
(ii) 156
- Step 1: 156 is even, divide by 2 → $156 \div 2 = 78$.
- Step 2: 78 is even, divide by 2 → $78 \div 2 = 39$.
- Step 3: 39 is not divisible by 2, try 3 → $39 \div 3 = 13$.
- Step 4: 13 is a prime number.
- Hence $156 = 2 \times 2 \times 3 \times 13 = 2^{2}\,3\,13$.
(iii) 3825
- Step 1: The last digit is 5, so 3825 is divisible by 5 → $3825 \div 5 = 765$.
- Step 2: 765 also ends with 5, divide by 5 again → $765 \div 5 = 153$.
- Step 3: Sum of digits of 153 is $1+5+3=9$, divisible by 3 → $153 \div 3 = 51$.
- Step 4: 51 is divisible by 3 → $51 \div 3 = 17$.
- Step 5: 17 is a prime number.
- Hence $3825 = 5 \times 5 \times 3 \times 3 \times 17 = 5^{2}\,3^{2}\,17$.
(iv) 5005
- Step 1: The number is odd, not divisible by 2. Sum of digits $5+0+0+5=10$ → not divisible by 3. Ends with 5 → not divisible by 5. Try 7: $5005 \div 7 = 715$ (exact).
- Step 2: 715 is odd; sum of digits $7+1+5=13$ → not divisible by 3. Ends with 5 → divisible by 5 → $715 \div 5 = 143$.
- Step 3: 143 is not divisible by 2,3,5. Try 7 → $143 \div 7 = 20.428$ (not integer). Try 11 → $143 \div 11 = 13$.
- Step 4: 13 is prime.
- Hence $5005 = 7 \times 5 \times 11 \times 13 = 5\,7\,11\,13$.
(v) 7429
- Step 1: The number is odd, not divisible by 2. Sum of digits $7+4+2+9 = 22$ → not divisible by 3. Does not end with 5 → not divisible by 5.
- Step 2: Test divisibility by 7: $7429 \div 7 = 1061.285$ (not integer). Test 11: $7429 \div 11 = 675.363$ (no). Test 13: $7429 \div 13 = 571.461$ (no).
- Step 3: Test 17: $7429 \div 17 = 437$ (exact). So 17 is a factor.
- Step 4: Now factor 437. It is odd, sum of digits $4+3+7 = 14$ → not divisible by 3. Ends with 7 → not 5. Test 7: $437 \div 7 = 62.428$ (no). Test 11: $437 \div 11 = 39.727$ (no). Test 13: $437 \div 13 = 33.615$ (no). Test 17: $437 \div 17 = 25.705$ (no). Test 19: $437 \div 19 = 23$ (exact).
- Step 5: 23 is a prime number.
- Hence $7429 = 17 \times 19 \times 23$.
Thus the prime‑factorisations are:
- $140 = 2^{2}\,5\,7$
- $156 = 2^{2}\,3\,13$
- $3825 = 5^{2}\,3^{2}\,17$
- $5005 = 5\,7\,11\,13$
- $7429 = 17\,19\,23$
Question 2
Hint available
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Key Idea
Use prime factorisation to obtain the Highest Common Factor (HCF) as the product of common prime factors with the smallest exponents, and the Least Common Multiple (LCM) as the product of all prime factors taken with the highest exponents. Then verify the relation \(\text{LCM}\times\text{HCF}=\text{product of the two numbers}\).
Step-by-Step Solution
Given two positive integers, we have to find their HCF and LCM and then check the identity \(\text{LCM}\times\text{HCF}=ab\) where \(a\) and \(b\) are the numbers.
(i) 26 and 91
Step 1: Prime factorise each number.
\[26 = 2\times13\]
\[91 = 7\times13\]
Step 2: Identify common prime factors with the smallest powers. The only common factor is \(13\). Hence, \(\text{HCF}=13\).
Step 3: For LCM, take the highest power of each prime appearing in either factorisation.
\[\text{LCM}=2\times7\times13=182\]
Step 4: Verify the relation.
\[\text{LCM}\times\text{HCF}=182\times13=2366\]
\[26\times91=2366\]
Thus, the relation holds.
(ii) 510 and 92
Step 1: Prime factorisation.
\[510 = 2\times3\times5\times17\]
\[92 = 2^{2}\times23\]
Step 2: Common prime factor with smallest exponent is \(2\). Hence, \(\text{HCF}=2\).
Step 3: LCM uses the highest powers of all primes: \(2^{2},3,5,17,23\).
\[\text{LCM}=2^{2}\times3\times5\times17\times23=4\times3\times5\times17\times23=23460\]
Step 4: Verification.
\[\text{LCM}\times\text{HCF}=23460\times2=46920\]
\[510\times92=46920\]
The identity is satisfied.
(iii) 336 and 54
Step 1: Prime factorisation.
\[336 = 2^{4}\times3\times7\]
\[54 = 2\times3^{3}\]
Step 2: Common primes with smallest exponents are \(2^{1}\) and \(3^{1}\). Hence, \(\text{HCF}=2\times3=6\).
Step 3: LCM takes the highest powers: \(2^{4},3^{3},7\).
\[\text{LCM}=2^{4}\times3^{3}\times7=16\times27\times7=3024\]
Step 4: Verification.
\[\text{LCM}\times\text{HCF}=3024\times6=18144\]
\[336\times54=18144\]
Thus, the required relation holds for all three pairs.
(i) 26 and 91
Step 1: Prime factorise each number.
\[26 = 2\times13\]
\[91 = 7\times13\]
Step 2: Identify common prime factors with the smallest powers. The only common factor is \(13\). Hence, \(\text{HCF}=13\).
Step 3: For LCM, take the highest power of each prime appearing in either factorisation.
\[\text{LCM}=2\times7\times13=182\]
Step 4: Verify the relation.
\[\text{LCM}\times\text{HCF}=182\times13=2366\]
\[26\times91=2366\]
Thus, the relation holds.
(ii) 510 and 92
Step 1: Prime factorisation.
\[510 = 2\times3\times5\times17\]
\[92 = 2^{2}\times23\]
Step 2: Common prime factor with smallest exponent is \(2\). Hence, \(\text{HCF}=2\).
Step 3: LCM uses the highest powers of all primes: \(2^{2},3,5,17,23\).
\[\text{LCM}=2^{2}\times3\times5\times17\times23=4\times3\times5\times17\times23=23460\]
Step 4: Verification.
\[\text{LCM}\times\text{HCF}=23460\times2=46920\]
\[510\times92=46920\]
The identity is satisfied.
(iii) 336 and 54
Step 1: Prime factorisation.
\[336 = 2^{4}\times3\times7\]
\[54 = 2\times3^{3}\]
Step 2: Common primes with smallest exponents are \(2^{1}\) and \(3^{1}\). Hence, \(\text{HCF}=2\times3=6\).
Step 3: LCM takes the highest powers: \(2^{4},3^{3},7\).
\[\text{LCM}=2^{4}\times3^{3}\times7=16\times27\times7=3024\]
Step 4: Verification.
\[\text{LCM}\times\text{HCF}=3024\times6=18144\]
\[336\times54=18144\]
Thus, the required relation holds for all three pairs.
Question 3
Hint available
Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Key Idea
Prime factorisation method: HCF is the product of the lowest powers of common prime factors; LCM is the product of the highest powers of all prime factors appearing in the numbers.
Step-by-Step Solution
Given: Three sets of integers.
To Find: HCF and LCM of each set using prime factorisation.
(i) Numbers: 12, 15, 21
Step 1: Write prime factorisation.
$$12 = 2^{2}\times 3, \quad 15 = 3\times 5, \quad 21 = 3\times 7.$$
Step 2: Identify common prime factors. Only \(3\) is common to all three numbers.
Step 3: HCF = product of lowest powers of common primes = \(3^{1}=3\).
Step 4: For LCM, take the highest power of each prime appearing in any number: \(2^{2}, 3^{1}, 5^{1}, 7^{1}\).
Step 5: LCM = \(2^{2}\times 3\times 5\times 7 = 4\times 3\times 5\times 7 = 420\).
(ii) Numbers: 17, 23, 29
Step 1: Prime factorisation (each number is prime).
$$17 = 17, \quad 23 = 23, \quad 29 = 29.$$
Step 2: No common prime factor ⇒ HCF = 1.
Step 3: LCM = product of the highest powers (each appears to power 1).
$$\text{LCM}=17\times 23\times 29 = 11339.$$
(iii) Numbers: 8, 9, 25
Step 1: Prime factorisation.
$$8 = 2^{3}, \quad 9 = 3^{2}, \quad 25 = 5^{2}.$$
Step 2: No common prime factor ⇒ HCF = 1.
Step 3: LCM = product of highest powers: \(2^{3}, 3^{2}, 5^{2}\).
$$\text{LCM}=2^{3}\times 3^{2}\times 5^{2}=8\times 9\times 25=1800.$$
Conclusion: The HCF and LCM for each set are obtained as shown above.
To Find: HCF and LCM of each set using prime factorisation.
(i) Numbers: 12, 15, 21
Step 1: Write prime factorisation.
$$12 = 2^{2}\times 3, \quad 15 = 3\times 5, \quad 21 = 3\times 7.$$
Step 2: Identify common prime factors. Only \(3\) is common to all three numbers.
Step 3: HCF = product of lowest powers of common primes = \(3^{1}=3\).
Step 4: For LCM, take the highest power of each prime appearing in any number: \(2^{2}, 3^{1}, 5^{1}, 7^{1}\).
Step 5: LCM = \(2^{2}\times 3\times 5\times 7 = 4\times 3\times 5\times 7 = 420\).
(ii) Numbers: 17, 23, 29
Step 1: Prime factorisation (each number is prime).
$$17 = 17, \quad 23 = 23, \quad 29 = 29.$$
Step 2: No common prime factor ⇒ HCF = 1.
Step 3: LCM = product of the highest powers (each appears to power 1).
$$\text{LCM}=17\times 23\times 29 = 11339.$$
(iii) Numbers: 8, 9, 25
Step 1: Prime factorisation.
$$8 = 2^{3}, \quad 9 = 3^{2}, \quad 25 = 5^{2}.$$
Step 2: No common prime factor ⇒ HCF = 1.
Step 3: LCM = product of highest powers: \(2^{3}, 3^{2}, 5^{2}\).
$$\text{LCM}=2^{3}\times 3^{2}\times 5^{2}=8\times 9\times 25=1800.$$
Conclusion: The HCF and LCM for each set are obtained as shown above.
Question 4
Hint available
Given that HCF (306, 657) = 9, find LCM (306, 657).
Key Idea
For any two positive integers a and b, the product of their Highest Common Factor (HCF) and Lowest Common Multiple (LCM) equals the product of the numbers themselves: \(\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\).
Step-by-Step Solution
Given: \(\text{HCF}(306,657) = 9\).
To Find: \(\text{LCM}(306,657)\).
Step 1: Write down the relation between HCF and LCM.
\[\text{HCF}(306,657) \times \text{LCM}(306,657) = 306 \times 657\]
Step 2: Substitute the known HCF value.
\[9 \times \text{LCM}(306,657) = 306 \times 657\]
Step 3: Compute the product of the two numbers.
\[306 \times 657 = 201042\]
Step 4: Solve for LCM.
\[\text{LCM}(306,657) = \frac{201042}{9}\]
Step 5: Perform the division.
\[\frac{201042}{9} = 22338\]
Conclusion: The LCM of 306 and 657 is 22338.
To Find: \(\text{LCM}(306,657)\).
Step 1: Write down the relation between HCF and LCM.
\[\text{HCF}(306,657) \times \text{LCM}(306,657) = 306 \times 657\]
Step 2: Substitute the known HCF value.
\[9 \times \text{LCM}(306,657) = 306 \times 657\]
Step 3: Compute the product of the two numbers.
\[306 \times 657 = 201042\]
Step 4: Solve for LCM.
\[\text{LCM}(306,657) = \frac{201042}{9}\]
Step 5: Perform the division.
\[\frac{201042}{9} = 22338\]
Conclusion: The LCM of 306 and 657 is 22338.
Question 5
Hint available
Check whether 6 n can end with the digit 0 for any natural number n.
Key Idea
A number ends with digit 0 iff it is divisible by 10 = 2 × 5. Use prime factorisation of 6^n.
Step-by-Step Solution
Given: $n \in \mathbb{N}$ (natural number).\
To Find: Whether there exists an $n$ such that the decimal representation of $6^{n}$ ends with 0.\
Step 1: Write the prime factorisation of $6^{n}$.
$$6^{n} = (2 \times 3)^{n} = 2^{n} \cdot 3^{n}.$$\
Step 2: For a number to end with 0 it must be a multiple of 10.
$$10 = 2 \times 5.$$\
Hence a number ending with 0 must contain both the prime factors 2 and 5 in its factorisation.\
Step 3: Examine the prime factors of $6^{n}$.
- $6^{n}$ contains the factor $2^{n}$ (so the factor 2 is present).\
- $6^{n}$ contains only the prime factor 3 besides 2; there is no factor 5 in $6^{n}$ for any $n$ because the only primes appearing are 2 and 3.
Step 4: Since the factor 5 never appears in $6^{n}$, $6^{n}$ can never be divisible by 10.
Therefore $6^{n}$ can never end with the digit 0 for any natural number $n$.\
Conclusion: No natural number $n$ makes $6^{n}$ end with 0.
To Find: Whether there exists an $n$ such that the decimal representation of $6^{n}$ ends with 0.\
Step 1: Write the prime factorisation of $6^{n}$.
$$6^{n} = (2 \times 3)^{n} = 2^{n} \cdot 3^{n}.$$\
Step 2: For a number to end with 0 it must be a multiple of 10.
$$10 = 2 \times 5.$$\
Hence a number ending with 0 must contain both the prime factors 2 and 5 in its factorisation.\
Step 3: Examine the prime factors of $6^{n}$.
- $6^{n}$ contains the factor $2^{n}$ (so the factor 2 is present).\
- $6^{n}$ contains only the prime factor 3 besides 2; there is no factor 5 in $6^{n}$ for any $n$ because the only primes appearing are 2 and 3.
Step 4: Since the factor 5 never appears in $6^{n}$, $6^{n}$ can never be divisible by 10.
Therefore $6^{n}$ can never end with the digit 0 for any natural number $n$.\
Conclusion: No natural number $n$ makes $6^{n}$ end with 0.
Question 6
Hint available
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Key Idea
A natural number greater than 1 is composite if it can be written as a product of two integers, each greater than 1. Hence, showing a factorisation of the given numbers into such a product proves they are composite.
Step-by-Step Solution
Given:
1) $N_1 = 7 \times 11 \times 13 + 13$
2) $N_2 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
To Show: Both $N_1$ and $N_2$ are composite numbers.
Step 1 – Factor out the common factor in each expression.
- For $N_1$, the common factor is $13$:
$$N_1 = 13\bigl(7 \times 11 + 1\bigr).$$
- For $N_2$, the common factor is $5$:
$$N_2 = 5\bigl(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1\bigr).$$
Step 2 – Simplify the brackets.
- $7 \times 11 = 77$, therefore
$$N_1 = 13\bigl(77 + 1\bigr) = 13 \times 78.$$
- $7 \times 6 \times 4 \times 3 \times 2 \times 1 = 1008$, therefore
$$N_2 = 5\bigl(1008 + 1\bigr) = 5 \times 1009.$$
Step 3 – Observe that each factor is greater than 1.
- In $N_1 = 13 \times 78$, both $13$ and $78$ are integers $>1$.
- In $N_2 = 5 \times 1009$, both $5$ and $1009$ are integers $>1$.
Step 4 – Conclude using the definition of a composite number.
Since each number can be expressed as a product of two integers each exceeding 1, both $N_1$ and $N_2$ satisfy the definition of a composite number.
Conclusion:
- $7 \times 11 \times 13 + 13 = 13 \times 78$ is composite.
- $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times 1009$ is composite.
1) $N_1 = 7 \times 11 \times 13 + 13$
2) $N_2 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
To Show: Both $N_1$ and $N_2$ are composite numbers.
Step 1 – Factor out the common factor in each expression.
- For $N_1$, the common factor is $13$:
$$N_1 = 13\bigl(7 \times 11 + 1\bigr).$$
- For $N_2$, the common factor is $5$:
$$N_2 = 5\bigl(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1\bigr).$$
Step 2 – Simplify the brackets.
- $7 \times 11 = 77$, therefore
$$N_1 = 13\bigl(77 + 1\bigr) = 13 \times 78.$$
- $7 \times 6 \times 4 \times 3 \times 2 \times 1 = 1008$, therefore
$$N_2 = 5\bigl(1008 + 1\bigr) = 5 \times 1009.$$
Step 3 – Observe that each factor is greater than 1.
- In $N_1 = 13 \times 78$, both $13$ and $78$ are integers $>1$.
- In $N_2 = 5 \times 1009$, both $5$ and $1009$ are integers $>1$.
Step 4 – Conclude using the definition of a composite number.
Since each number can be expressed as a product of two integers each exceeding 1, both $N_1$ and $N_2$ satisfy the definition of a composite number.
Conclusion:
- $7 \times 11 \times 13 + 13 = 13 \times 78$ is composite.
- $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times 1009$ is composite.
Question 7
Hint available
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the 6 same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Key Idea
The time at which two moving objects meet at the starting point again is the Least Common Multiple (LCM) of their individual times for one complete round.
Step-by-Step Solution
Given:
- Time taken by Sonia to complete one round = $18$ minutes.
- Time taken by Ravi to complete one round = $12$ minutes.
- Both start together from the same point and move in the same direction.
To Find: Time after which they will meet again at the starting point.
Step 1: Write the prime factorisation of the two times.
$$\begin{aligned}
18 &= 2 \times 3^2,\\
12 &= 2^2 \times 3.
\end{aligned}$$
Step 2: Determine the Least Common Multiple (LCM) using the highest powers of all prime factors.
$$\text{LCM}=2^2 \times 3^2 = 4 \times 9 = 36\text{ minutes}.$$
Step 3: Interpret the result.
- After $36$ minutes, Sonia will have completed $\dfrac{36}{18}=2$ rounds.
- After $36$ minutes, Ravi will have completed $\dfrac{36}{12}=3$ rounds.
Both have returned to the starting point simultaneously.
Conclusion: The two will meet again at the starting point after $36$ minutes.
- Time taken by Sonia to complete one round = $18$ minutes.
- Time taken by Ravi to complete one round = $12$ minutes.
- Both start together from the same point and move in the same direction.
To Find: Time after which they will meet again at the starting point.
Step 1: Write the prime factorisation of the two times.
$$\begin{aligned}
18 &= 2 \times 3^2,\\
12 &= 2^2 \times 3.
\end{aligned}$$
Step 2: Determine the Least Common Multiple (LCM) using the highest powers of all prime factors.
$$\text{LCM}=2^2 \times 3^2 = 4 \times 9 = 36\text{ minutes}.$$
Step 3: Interpret the result.
- After $36$ minutes, Sonia will have completed $\dfrac{36}{18}=2$ rounds.
- After $36$ minutes, Ravi will have completed $\dfrac{36}{12}=3$ rounds.
Both have returned to the starting point simultaneously.
Conclusion: The two will meet again at the starting point after $36$ minutes.