Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 10)
10.
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Answer: Option
Explanation:
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.
Discussion:
88 comments Page 9 of 9.
Sakthivel C.M said:
1 decade ago
How did you find K ?
Koti reddy said:
1 decade ago
How to come k=4 ?
Manasa said:
1 decade ago
Thank you sundar.
Sweet said:
7 years ago
How come 90k+ 4?
Habibali@mail.com said:
1 decade ago
Thankx Rahul...
Anonymous said:
1 decade ago
How k=4 comes?
Akhil said:
1 decade ago
Thank rahul.
Deepak said:
6 years ago
Thanks all.
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