The product of two numbers is 2028 and their H.C.F. is 13. How many distinct pairs of such positive integers are possible?

Introduction / Context:This problem combines the ideas of highest common factor (H.C.F.), factorization, and counting possible pairs of numbers. It checks whether you can use the relationship between H.C.F., co-prime factors, and the product of two numbers to count how many pairs of numbers satisfy the given conditions. This is a popular style of question in quantitative aptitude sections.

Given Data / Assumptions:

• The product of the two positive integers is 2028.
• The H.C.F. (greatest common divisor) of the two numbers is 13.
• We must find how many such distinct pairs of numbers exist.

Concept / Approach:If two numbers have H.C.F. 13, we can write them as 13a and 13b, where a and b are co-prime (their H.C.F. is 1). The product of the numbers is then 13a * 13b = 13^2 * ab. This product is given as 2028, which allows us to find ab. Once ab is known, we look for co-prime factor pairs (a, b) whose product equals that value. Each such pair gives one possible pair of original numbers 13a and 13b.

Step-by-Step Solution:Step 1: Let the numbers be 13a and 13b with H.C.F. 13, so H.C.F.(a, b) = 1.Step 2: Product = 13a * 13b = 169ab.Step 3: Given product is 2028, so 169ab = 2028.Step 4: Compute ab = 2028 / 169 = 12.Step 5: We now need all co-prime positive integer pairs (a, b) such that ab = 12.Step 6: Factor pairs of 12 are (1, 12), (2, 6), (3, 4) and their reverses.Step 7: Check co-primeness: gcd(1, 12) = 1 (valid), gcd(2, 6) = 2 (invalid), gcd(3, 4) = 1 (valid).Step 8: So valid co-prime pairs are (1, 12) and (3, 4). Each pair corresponds to numbers (13, 156) and (39, 52).Step 9: As we count pairs without ordering, {13, 156} and {39, 52} give 2 distinct pairs.

Verification / Alternative check:For pair (13, 156): product = 13 * 156 = 2028; H.C.F. is 13.For pair (39, 52): product = 39 * 52 = 2028; H.C.F. is also 13. No other co-prime factor pairs of 12 exist, so there are exactly 2 valid pairs.

Why Other Options Are Wrong:Option a (1) undercounts the pairs by missing one valid combination. Option c (3) and option d (4) overcount, usually due to treating ordered pairs (a, b) and (b, a) as different or including non co-prime pairs like (2, 6). The question asks for the number of such pairs of numbers, which naturally refers to distinct unordered pairs of positive integers.

Common Pitfalls:A common error is forgetting that a and b must be co-prime. Some candidates take all factor pairs of 12 without checking gcd, and others treat (a, b) and (b, a) as separate pairs. Always carefully check co-prime conditions and interpret whether the question refers to ordered or unordered pairs. Most exam questions consider pairs like (13, 156) and (156, 13) as the same pair.

Final Answer:The number of such pairs is 2.