Use Euclid’s division algorithm to find HCF of 441, 567, 693.

Solution : i) 26 and 91 26=2×13×1(expressing as product of it’s prime factors) 91=7×13×1(expressing as product of it’s prime factors) So, LCM(26,91)=2×7×13×1=182 HCF(26,91)=13×1=13 Verification: LCM×HCF=13×182=2366 Product of 26 and 91 =2366 Therefore,LCM×HCF=Product of the two numbers . i) 510 and...

Solution: According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) ×...

Solution: Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively. So, 1251 – 1 = 1250 is exactly divisible by the required number, 9377 – 2 = 9375...

Correct answer is option (C) 0 ≤ r < b

Solution: Since the denominator is not of the form 2m x 5n, the rational number has a nonterminating repeating decimal expansion.

396 = 1 × 231 + 165 231 = 1 × 165 + 66 165 = 2 × 66 + 33 66 = 2 × 33 + 0 ∴ The HCF of 231, 396...

Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420 Since the remainder 420 ≠ 0, we apply the division...

(e) (7)3/√?= 14-7=7 √? = 73/7= 49 ?=492= 2401

(d) ? = 4123 divide (2.3)2 – 446 = 4123/(2.3 x 2.3) - 446 779 – 446 = 333 (approx)

(a) Let the required number be x x2 – (9)3 = 567 x2 = 567 + 729 = 1296  x = √1296 = √(36)2 = 36