ABC26GN0440 · H.C.F. and L.C.M. of Numbers

Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2008 · Marks: · Difficulty:

Find the smallest number which when increased by 10 is completely divisible by 12, 15, 18, 20 and 24.
Answer
Answer (as printed):
Explanation
Required number = (L.C.M. of 12, 15, 18, 20, 24) − 10 = (2 × 2 × 3 × 5 × 3 × 2) − 10 = 360 − 10 = 350.

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

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