Let a and b be two positive integers such that a = p³q⁴ and b = p²q³, where p and q are prime numbers. If HCF(a,b) = pmqⁿ and LCM(a,b) = prq^s, then (m n)(r s)=?





To find the HCF of a and b, we need to find the highest power of each prime factor that divides both a and b.

The prime factors of a are p and q. The highest power of p that divides a is p³, and the highest power of q that divides a is q⁴.

The prime factors of b are also p and q. The highest power of p that divides b is p², and the highest power of q that divides b is q³.

Therefore, the HCF of a and b is pmq² (the highest power of p and q that divides both a and b).



To find the LCM of a and b, we need to find the lowest multiple of a and b that is divisible by all the prime factors.

The prime factors of a are p and q. The highest power of p that divides a is p³, and the highest power of q that divides a is q⁴.

The prime factors of b are also p and q. The highest power of p that divides b is p², and the highest power of q that divides b is q³.

Therefore, the LCM of a and b is prq⁴ (the product of the highest powers of p and q that divide either a or b).



We have HCF(a,b) = pmq² and LCM(a,b) = prq⁴.

Let's express pmq² and prq⁴ in terms of p, q, and their powers:

pmq² = p^mqⁿ and prq⁴ = p^rq^s

Therefore, we need to find (m n)(r s) given p^mqⁿ and p^rq^s.

Since HCF(a,b) = pmq², we know that the highest power of p that divides both a and b is p^m, and the highest power of q that divides both a and b is qⁿ.

Similarly, since LCM(a,b) = prq⁴

1. if question is asking about (m n) (r s) then, answer will be 72.
2. if (m+n)(r+s) is exist there then, your answer will be 35.