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Aptitude · Q10

Problems On H.C.F And L.C.M

Competitive Exams · Aptitude · question 10

Q10

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A.
74
B.
94
C.
184
D.
364
Answer

Answer: Option D

Solution

Answer: Option D
Solution:
L.C.M. of 6, 9, 15 and 18 is 90
Let required number be 90k + 4, which is multiple of 7
Least value of k for which (90k + 4) is divisible by 7 is k = 4
∴ Required number = (90 x 4) + 4 = 364