ABC26GN0432 · H.C.F. and L.C.M. of Numbers

Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2004 · Marks: · Difficulty:

The sum of two numbers is 462 and their highest common factor is 22. What is the minimum number of pairs that satisfy these conditions?
Answer
Answer (as printed):
Explanation
Let the required numbers be 22a and 22b. Then, 22a + 22b = 462 ⇒ a + b = 21. Now, co-primes with sum 21 are (1, 20), (2, 19), (4, 17), (5, 16), (8, 13) and (10, 11). \ Required numbers are (22 × 1, 22 × 20), (22 × 2, 22 × 19), (22 × 4, 22 × 17), (22 × 5, 22 × 16), (22 × 8, 22 × 13) and (22 × 10, 22 × 11). Clearly, the number of such pairs is 6.

Explanation as extracted from the printed page; notation may be imperfect.

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