Let $x$ and $y$ be two distinct prime numbers and $p = x^2 y^3$, $q = xy^4$, $r = x^5 y^2$. Find the HCF and LCM of $p$, $q$ and $r$. Further check if $\text{HCF}(p, q, r) \times \text{LCM}(p, q, r) = p \times q \times r$ or not.
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
Given: $p = x^2y^3$, $q = xy^4$, $r = x^5y^2$, where $x$ and $y$ are distinct primes.
HCF = product of smallest powers of common prime factors:
$$\text{HCF}(p,q,r) = x^1 \times y^2 = xy^2$$
LCM = product of greatest powers of all prime factors:
$$\text{LCM}(p,q,r) = x^5 \times y^4 = x^5y^4$$
Checking whether HCF × LCM = p × q × r:
$$\text{HCF} \times \text{LCM} = xy^2 \times x^5y^4 = x^6y^6$$
$$p \times q \times r = x^2y^3 \times xy^4 \times x^5y^2 = x^8y^9$$
Since $x^6y^6 \neq x^8y^9$,
$$\boxed{\text{HCF}(p,q,r) \times \text{LCM}(p,q,r) \neq p \times q \times r}$$
Source: Chapter 1, Section 1.2 (The Fundamental Theorem of Arithmetic)
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Explanation
- HCF takes the lowest power of each common prime factor across all three numbers. Both $x$ and $y$ appear in all three, so take $x^1$ (lowest of 2,1,5) and $y^2$ (lowest of 3,4,2).
- LCM takes the highest power of every prime factor: $x^5$ and $y^4$.
- The identity HCF × LCM = product of numbers holds only for two numbers, not three. The textbook explicitly states this (see Remark after Example 4). Examiners expect you to demonstrate this with calculation and conclude clearly.