Table of Contents
Exercise 1.2 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on the Fundamental Theorem of Arithmetic. Mathematical expressions are rendered using MathJax.
1. Express each of the following numbers as a product of its prime factors.
(i) 140
Divide by smallest prime: \( 140 \div 2 = 70 \)
\( 70 \div 2 = 35 \)
\( 35 \div 5 = 7 \)
\( 7 \div 7 = 1 \)
Prime factors: \( 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7 \)
Product = \( 2^2 \times 5 \times 7 \)
(ii) 156
\( 156 \div 2 = 78 \)
\( 78 \div 2 = 39 \)
\( 39 \div 3 = 13 \)
\( 13 \div 13 = 1 \)
Prime factors: \( 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13 \)
Product = \( 2^2 \times 3 \times 13 \)
(iii) 3825
\( 3825 \div 5 = 765 \)
\( 765 \div 5 = 153 \)
\( 153 \div 3 = 51 \)
\( 51 \div 3 = 17 \)
\( 17 \div 17 = 1 \)
Prime factors: \( 5 \times 5 \times 3 \times 3 \times 17 = 5^2 \times 3^2 \times 17 \)
Product = \( 5^2 \times 3^2 \times 17 \)
(iv) 5005
\( 5005 \div 5 = 1001 \)
\( 1001 \div 7 = 143 \)
\( 143 \div 11 = 13 \)
\( 13 \div 13 = 1 \)
Prime factors: \( 5 \times 7 \times 11 \times 13 \)
Product = \( 5 \times 7 \times 11 \times 13 \)
(v) 7429
\( 7429 \div 17 = 437 \) (17 is a prime factor)
\( 437 \div 19 = 23 \) (19 and 23 are prime)
\( 23 \div 23 = 1 \)
Prime factors: \( 17 \times 19 \times 23 \)
Product = \( 17 \times 19 \times 23 \)
2. Find the LCM and HCF of the following integers by the prime factorization method.
(i) 12, 15 and 21
Prime Factorization:
- \( 12 = 2^2 \times 3 \)
- \( 15 = 3 \times 5 \)
- \( 21 = 3 \times 7 \)
HCF: Lowest power of common factors = \( 3 \)
LCM: Highest power of all factors = \( 2^2 \times 3 \times 5 \times 7 = 420 \)
HCF = 3, LCM = 420
(ii) 17, 23, and 29
Prime Factorization:
- \( 17 = 17 \)
- \( 23 = 23 \)
- \( 29 = 29 \)
HCF: No common factors, so \( 1 \)
LCM: \( 17 \times 23 \times 29 = 11339 \)
HCF = 1, LCM = 11339
(iii) 8, 9, and 25
Prime Factorization:
- \( 8 = 2^3 \)
- \( 9 = 3^2 \)
- \( 25 = 5^2 \)
HCF: No common factors, so \( 1 \)
LCM: \( 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800 \)
HCF = 1, LCM = 1800
(iv) 72 and 108
Prime Factorization:
- \( 72 = 2^3 \times 3^2 \)
- \( 108 = 2^2 \times 3^3 \)
HCF: Lowest power of common factors = \( 2^2 \times 3^2 = 36 \)
LCM: Highest power of all factors = \( 2^3 \times 3^3 = 216 \)
HCF = 36, LCM = 216
(v) 306 and 657
Prime Factorization:
- \( 306 = 2 \times 3 \times 3 \times 17 = 2 \times 3^2 \times 17 \)
- \( 657 = 3 \times 3 \times 73 = 3^2 \times 73 \)
HCF: Lowest power of common factors = \( 3^2 = 9 \)
LCM: Highest power of all factors = \( 2 \times 3^2 \times 17 \times 73 = 22338 \)
HCF = 9, LCM = 22338
3. Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).
For a number to end with 0, it must be divisible by 10, i.e., have factors \( 2 \) and \( 5 \).
\( 6^n = (2 \times 3)^n = 2^n \times 3^n \)
Contains \( 2 \) but no \( 5 \), so not divisible by 10.
Examples: \( 6^1 = 6 \), \( 6^2 = 36 \), \( 6^3 = 216 \) (none end with 0).
Conclusion: \( 6^n \) cannot end with 0 for any natural number \( n \).
4. Explain why \(7 \times 11 \times 13 + 13\) and \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) are composite numbers.
First number: \( 7 \times 11 \times 13 + 13 \)
Factor out 13: \( 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 \)
78 is composite (\( 78 = 2 \times 3 \times 13 \)), so \( 13 \times 78 \) is composite.
Second number: \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \)
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)
\( 5040 + 5 = 5045 \)
Check divisibility: \( 5045 \div 5 = 1009 \) (ends with 5, divisible by 5), and 1009 is prime but \( 5045 = 5 \times 1009 \), a product of primes > 1.
Conclusion: Both are composite.
5. How will you show that \((17 \times 11 \times 2) + (17 \times 11 \times 5)\) is a composite number? Explain.
Factor out common terms: \( (17 \times 11 \times 2) + (17 \times 11 \times 5) = 17 \times 11 \times (2 + 5) \)
\( 2 + 5 = 7 \)
So, \( 17 \times 11 \times 7 \)
Product of three primes \( > 1 \), hence composite.
Calculate: \( 17 \times 11 = 187 \), \( 187 \times 7 = 1309 \)
1309 is composite (e.g., divisible by 7: \( 1309 \div 7 = 187 \)).
Conclusion: It is composite.
6. What is the last digit of \(6^{100}\)?
Last digit depends on the units digit of \(6^n\).
\( 6^1 = 6 \) (ends with 6)
\( 6^2 = 36 \) (ends with 6)
\( 6^3 = 216 \) (ends with 6)
Pattern: Units digit is always 6 for any \( n \).
So, \( 6^{100} \) ends with 6.
Last digit = 6
