If the product of the HCF and LCM of two distinct numbers is the cube of one of the numbers, then which of the following statements is/are correct? I. The difference of the numbers is an even number. II. One of the numbers is a perfect square. Select the answer using the code given below.
- (a) I only
- (b) II only
- (c) Both I and II ✓ UPSC's answer
- (d) Neither I nor II
Why the answer is (c)
• Let the two distinct numbers be $a$ and $b$. The fundamental property of HCF and LCM states that $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.
• The problem states that this product is the cube of one of the numbers, so $a \times b = a^3$ (assuming $a$ is the number whose cube is formed).
• Dividing both sides by $a$ (since $a \neq 0$), we get $b = a^2$. This means one number is the square of the other.
• Statement II is correct because $b$ is explicitly defined as $a^2$, which is a perfect square.
• To check Statement I, let $a = 2k$ (an even number). Then $b = (2k)^2 = 4k^2$. The difference $b - a = 4k^2 - 2k = 2k(2k - 1)$, which is always even.
• If $a = 2k + 1$ (an odd number), then $b = (2k + 1)^2 = 4k^2 + 4k + 1$. The difference $b - a = (4k^2 + 4k + 1) - (2k + 1) = 4k^2 + 2k = 2k(2k + 1)$, which is also always even.
• Since the difference is even in all cases, Statement I is also correct, making option (c) the right answer.
Why the other options are wrong
- (a) I only
- This option is incorrect because Statement II is also true, as one number is necessarily the square of the other.
- (b) II only
- This option is incorrect because Statement I is also true, as the difference between a number and its square is always even.
- (d) Neither I nor II
- This option is incorrect because both statements I and II are mathematically valid under the given condition.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2026, held on 24 May 2026. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.