If two positive integers a and b are written as a = x^3y^2 and b = xy^3; x, y are prime numbers, then HCF (a, b) is

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 : i) 12,15 and 21 12=2×2×3 15=5×3 21=7×3 From the above ,HCF(12,15,21)=3and LCM(12,15,21)=420 ii)17,23,and 29 17=17×1 23=23×1 29=29×1 From the above ,HCF(17,23,29)=1and LCM(17,23,29)=11339 iii)8,9 and 25 8=2×2×2 9=3×3 25=5×5 From the above ,HCF(8,9,25)=1and LCM(8,9,25)=1800

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

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.

Let, a = 2p+1, p∈N and b = 2q+1 , q∈N\(\cup\){0} \(\because\) a > b ⇒ 2p+1 > 2q+1 ⇒ p > q \(\frac{a+b}{2}\) = \(\frac{2p+1+2q+1}{2}\) = p+q+1 \(\frac{a-b}{2}\) = \(\frac{2p+1-(2q+1)}{2}\) = p-q Case-I: \(\frac{a+b}{2}\) is odd which implies p+q+1 is odd \(\Rightarrow\) p+q is even (\(\because\) odd - 1 = even) \(\Rightarrow\) p+q-2q...

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...

Given integers are 468 and 222 where 468 > 222. By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i) Since remainder ≠ 0, apply division lemma...

The correct answer is a2 b2

(d) Third number = (436 × 5) – (344 × 2 + 554 × 2) = 2180 – 1796 = 384