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 6 of 9.
Phanindra said:
1 decade ago
Isn't it the least common multiple of 6, 7, 9, 15 and 18 with a remainder 4?
Ananya said:
8 years ago
Hey guys, what will we take in the right-hand side? Please help me out.
Sonu said:
2 decades ago
How can you find the value of k=4 i.e.
How we can divide (90k + 4)by 7.
How we can divide (90k + 4)by 7.
Myk said:
1 decade ago
a/b ve get quocent q and reminder r
there for a=bq+r that why 90k+4
there for a=bq+r that why 90k+4
Ritu said:
9 years ago
But why to do lcm of 6, 9, 15, 18 here?
Will somebody help me out?
Will somebody help me out?
Akash said:
4 years ago
@Jeni.
Can you please explain how came 7 when calculating the LCM?
Can you please explain how came 7 when calculating the LCM?
(3)
Bhavi said:
1 decade ago
Is there any short trick to solve these types of questions???
Ram said:
1 decade ago
I'm confused to find LCM and HCF. Can anyone please explain?
Ananth said:
1 decade ago
Shortest Way => Only one option divisible by 7 is "364".
Mani said:
1 decade ago
How can u choose the value of k=4 ? It can be 2 also know.
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