Question

If two positive integers p and q are written as p=a2×b^3 and q=a^3×b, a and b are prime numbers, then verify: LCM(p,q) × HCF(p,q)=p,q​

Answer 1

Solution :-

• p = a² * b³
• q = a³ * b
• a and b are prime numbers . ( That means no common factor Possible).

LCM(p,q) :-

→ p = a * a * b * b * b

→ q = a * a * a * b

LCM = b * b * b * a * a * a = b³ * a³

HCF(p,q) :-

→ p = a * a * b * b * b

→ q = a * a * a * b

HCF = a * a * b = a² * b

Therefore,

→ LCM(p,q) × HCF(p,q) = p × q

→ (b³ * a³) × (a² * b) = (a² * b³) × (a³ * b)

→ (b³ * b) × (a² * a³) = (a² * a³) × (b³ * b)

→ b⁴ * a⁵ = a⁵ * b⁴

→ a⁵ * b⁴ = a⁵ * b⁴

→ LHS = RHS . (Verified).

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow TWO\: POSITIVE\:INTEGERS:-p\:AND\:q$

$\sf\dashrightarrow p= a^2 \times b^3$

$\sf\dashrightarrow q= a^3 \times b$

$\sf\dashrightarrow a\:and\: b\: are\: prime\:numbers.$

$\red{\text{NOTE:- prime number doesn't has any common factors.}}$

✯.L.C.M,

$\sf\therefore p= \boxed{a}\times \boxed{a} \times \boxed{b}\times b\times b$

$\sf\therefore q= \boxed{ a} \times \boxed{a} \times a \times \boxed{b}$

$\sf\implies L.C.M= a\times a\times b\times b\times b \times a$

$\sf\implies L.C.M= a^3b^3$

$\large{\boxed{\bf{ \star\:\:L.C.M= a^3b^3 \:\: \star}}}$

✯.H.C.F,

$\sf\therefore p= \boxed{a}\times \boxed{a} \times \boxed{b}\times b\times b$

$\sf\therefore q= \boxed{ a} \times \boxed{a} \times a \times \boxed{b}$

$\sf\implies H.C.F= a\times a\times b$

$\sf\implies H.C.F= a^2 b$

$\large{\boxed{\bf{ \star\:\: H.C.F= a^2b\:\: \star}}}$

✯THEREFORE,

WE KNOW

$\sf\therefore H.C.F= a^2b,L.C.M= a^3b^3$

NOW,

✯VERIFICATION,

$\bf{\boxed{\bf{ \star\:\: L.C.M \times H.C.F= PRODUCT\:OF\:THE\:POSITIVE INTEGERS(p \times q)\:\: \star}}}$

$\sf\therefore H.C.F \times L.C.M= p \times q$

$\sf\implies ( a^2b ) \times (a^3b^3) = (a^2b^3)\times (a^3b)$

$\sf\implies a^2 \times b\times a^3\times b^3= a^2\times b^3\times a^3 \times b$

$\sf\implies a^2\times a^3 \times b \times b^3= a^2 \times a^3 \times b\times b^3$

$\sf\therefore a^5b^4= a^5b^4$

$\sf\therefore L.H.S= R.H.S$

HENCE PROVED✔

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