Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2009 · Marks: · Difficulty:
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is
(a)1
(b)2
(c)3
(d)4
Answer
Answer (as printed): B
Explanation
Let the numbers be 13a and 13b. Then, 13a × 13b = 2028 ⇒ ab = 12. Now, co-primes with product 12 are (1, 12) and (3, 4). So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4). Clearly, there are 2 such pairs.
Explanation as extracted from the printed page; notation may be imperfect.