RRB Technician Grade I 13 March 2026 Paper

Concept:When HCF is $19$, the numbers can be written as $19a$ and $19b$, where $a$ and $b$ are co-prime. The product $a \times b$ equals $\frac{\text{LCM}}{\text{HCF}}$.Explanation:Let the two numbers be $19a$ and $19b$, where $\text{HCF}(a, b) = 1$.Using the relation $\text{LCM} = \text{HCF} \times a \times b$, we get:$$760 = 19 \times a \times b$$$$a \times b = \frac{760}{19} = 40$$Now find all co-prime factor pairs of $40$.The possible pairs are $(1, 40)$ and $(5, 8)$.Checking each pair:For $(1, 40)$: numbers are $19 \times 1 = 19$ and $19 \times 40 = 760$. Here, $19$ is a two-digit number, so this pair is invalid.For $(5, 8)$: numbers are $19 \times 5 = 95$ and $19 \times 8 = 152$. Here, $95$ is a two-digit number, so this pair is also invalid.Since every possible pair gives at least one two-digit number, no valid pair of three-digit numbers exists.Answer:The number of such pairs is $0$.Hence, the correct option is A. 0.