Solution : Let us assume, to the contrary, that \(3 + 2\sqrt{5}\) is an irrational number. Now, let \(3 + 2\sqrt{5}\) = \(a\over b\), where a and b are coprime and b \(ne\) 0. So, \(2\sqrt{5}\) = \(a\over b\) – 3 or \(\sqrt{5}\) = \(a\over 2b\) – \(3\over 2\) Since a and b are integers, …
Solution : Suppose that \(\sqrt{5}\) is an irrational number. Then \(\sqrt{5}\) can be expressed in the form \(p\over q\) where p, q are integers and have no common factor, q \(ne\) 0. \(\sqrt{5}\) = \(p\over q\) Squaring both sides, we get 5 = \(p^2\over q^2\) \(\implies\) \(p^2\) = \(5q^2\) …
Prove that \(\sqrt{5}\) is an irrational number by contradiction method. Read More »
Solution : They will be again at the starting point at least common multiples of 18 and 12 minutes. To find the L.C.M of 18 and 12, we have : 18 = \(2 \times 3\times 3\) and 12 = \(2 \times 2 \times 3\) L.C.M of 18 and 12 = \(2 \times 2 \times 3 …
Solution : We have, \(7 \times 11 \times 13\) + 13 = 1001 + 13 =1014 1014 = \(2 \times 3 \times 13 \times 13\) So, it is the product of more than two prime numbers. 2, 3 and 13. Hence, it is a composite number. \(7 \times 6 \times 5 \times 4 \times 3 …
Solution : If the number \(6^n\) ends with the digit zero. Then it is divisible by 5. Therefore, the prime factors of \(6^n\) contains the prime number 5. This is not possible because the only primes in the factors of \(6^n\) are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees …
Check whether \(6^n\) can end with the digit 0 or any n \(\in\) N. Read More »
Solution : We have, H.C.F. (306, 657) = 9. We know that, Product of L.C.M and H.C.F = Product of two numbers. \(\implies\) L.C.M \(\times\) 9 = \(306 \times 657\) \(\implies\) L.C.M = \(306 \times 657 \over 9\) = 22338 Hence, L.C.M is 22338.
Question : Find the H.C.F and L.C.M of the following integers by applying prime factorisation method. (i) 12, 15, 21 (ii) 17, 23, 29 (iii) 8, 9 and 25 Solution : (i) 12, 15, 21 12 = \(2 \times 2 \times 3\) 15 = \(3 \times 5\) 21 = \(3 \times 7\) Here 3 is a …
Question : Find the L.C.M and H.C.F of the following pairs of integers and verify : L.C.M \(\times\) H.C.F = Product of the two numbers (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 Solution : (i) 26 and 91 26 = 2 \(\times\) 13 and 91 …
Find the L.C.M and H.C.F of the following pairs of integers and verify : Read More »
Question : Express each number as a product of its prime factors : (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Solution : (i) We use the division method as shown below : \(\therefore\) 140 = \(2 \times 2 \times 5 \times 7\) = \(2^2 \times 5 \times 7\) (ii) We use the division …
Express each number as a product of its prime factors : Read More »
Solution : Let m be any positive integer. Then it is of the form 3m, 3m + 1 or 3m + 2. Now, we have to prove that the cube of these can be rewritten in the form 9q, 9q + 1 or 9q + 8. Now, \((3m)^3\) = \(27m^3\) = \(9(m^3)\) = 9q, where …
