Find the LCM and HCF of the following integers by applying the prime factorization method. i) 12,15 and 21 ii) 17,23 and 29 iii) 8,9 and 25

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 : We know, HCF×LCM=Product of two numbers i.e 9×LCM=306×657 LCM=306×657/9=22338

Solution: LCM x HCF = Product of two numbers => LCM x 18 = 306 x 1314 => LCM = (306 x 1314)/18 = 22338

Solution: 117 = 65 x 1 + 52 65 = 52 x 1 + 13 65 = 13 x 5 + 0 => HCF(65, 117) = 13 Since, HCF(65, 117) = 65m - 117 => 13 = 65m - 117 => 65m...

Solution: x = p2q3 and y = p3q HCF(x, y) = p2 x q LCM(x, y) = p3 x q3 LCM(x, y) = (p x q2) x HCF(x, y) => LCM is a multiple of HCF.

Correct answer is (B) xy2 

By applying Euclid’s division lemma (i) 54 = 32 × 1 + 22 Since remainder ≠ 0, apply division lemma on division of 32 and remainder 22. 32 = 22 × 1 +...

(i) 963 and 657 By applying Euclid’s division lemma 963 = 657 × 1 + 306 …(i) Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306 657 = 306...

The correct answer is 6p

The correct answer is a2 b2