If p , q , r are prime numbers and α, β, γ are positive integers such that L.C.M. of α, β, γ is p 3 q 2 r and greatest common divisor of α, β, γ is pqr then the number of possible triplets (α, β, γ) will be

Question

If p , q , r are prime numbers and α, β, &gamma; are positive integers such that L.C.M. of α, β, &gamma; is p 3 q 2 r and greatest common divisor of α, β, &gamma; is pqr then the number of possible triplets (α, β, &gamma;) will be (A) 6 (B) 12 (C) 72 (D) 96

Tardigrade Admin · Accepted Answer

Θ LCM of α, β, &gamma;=p3 q2 r HCF = pqr ∴ α =pm1 qn1 r β=pm2 qn2 r &gamma; =pm3 qn3 r minimum of ( m 1, m 2, m 3)=1 maximum of ( m 1, m 2, m 3)=3 ∴ Number of possibilities for m 1, m 2, m 3=12 and minimum of ( n 1, n 2, n 3)=1 and maximum ( n 2, n 2, n 3)=2 ∴ Number of possibilities =6 ∴ Total number of ordered triplets =12 × 6=72

Q. If p,q,r are prime numbers and α,β,γ are positive integers such that L.C.M. of α,β,γ is p3q2r and greatest common divisor of α,β,γ is pqr then the number of possible triplets (α,β,γ) will be

6

12

72

96

Q. If $p, q, r$ are prime numbers and $α, β, γ$ are positive integers such that L.C.M. of $α, β, γ$ is $p^{3} q^{2} r$ and greatest common divisor of $α, β, γ$ is pqr then the number of possible triplets $(α, β, γ)$ will be