mathemerize

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, …

Prove that \(3 + 2\sqrt{5}\) is irrational. Read More »

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 …

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 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 ? Read More »

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 …

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. Read More »

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 …

Find the H.C.F and L.C.M of the following integers by applying prime factorisation method. Read More »

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 …

Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m or 9m + 1 or 9m + 8. Read More »