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 8 of 9.
Aditya Kulkarni said:
6 years ago
@All.
Here, We have been asked for a least multiple of 7, therefore one of the numbers given in option must be divided by 7.
74, 94 and 184 are not divisible by 7, but 364 is. So, 364 is the answer.
Here, We have been asked for a least multiple of 7, therefore one of the numbers given in option must be divided by 7.
74, 94 and 184 are not divisible by 7, but 364 is. So, 364 is the answer.
(1)
Jaisri said:
6 years ago
6, 9, 15, 18 LCM is 90.
Least number from this is 6. So 90+6= 96 when 96÷7 is 73 then remainder is 4.
Least number from this is 6. So 90+6= 96 when 96÷7 is 73 then remainder is 4.
Tushar said:
6 years ago
L.C.M. of 6, 9, 15 and 18 is 90.
Let the 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.
90K+4 is the required no.
How?
If K=1
90*1+4=94 , it is not multiple of 7.
90*2+4=184, it is not multiple of 7.
90*3+4=274, it is not multiple of 7.
90*4+4=364, it is multiple of 7.
So the least value will be 364.
Let the 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.
90K+4 is the required no.
How?
If K=1
90*1+4=94 , it is not multiple of 7.
90*2+4=184, it is not multiple of 7.
90*3+4=274, it is not multiple of 7.
90*4+4=364, it is multiple of 7.
So the least value will be 364.
(1)
Shalini said:
6 years ago
LCM of 6, 9, 15, 18 is 90.
Such that there is one formula these type of questions ie "LCM*k+remainder "should be divisible by a multiple.
So we want to do "trail and error" method by giving value from the options to the 'k' so that the number is divisible, so we substitute k=4.
Such that there is one formula these type of questions ie "LCM*k+remainder "should be divisible by a multiple.
So we want to do "trail and error" method by giving value from the options to the 'k' so that the number is divisible, so we substitute k=4.
Nagesh Bhukya said:
6 years ago
K = 4.
Because this value decided by 7 in 90k+4.
90(4)+4=364/7 = 52.
Remainder is 0 .
So k = 4 is correct.
Because this value decided by 7 in 90k+4.
90(4)+4=364/7 = 52.
Remainder is 0 .
So k = 4 is correct.
(1)
Choudhary said:
6 years ago
What if the value of k has very large?
(1)
Jambay said:
5 years ago
Why we are taking LCM instead of GCF? Please explain.
(7)
Jeni said:
5 years ago
6 = 2,3.
9 = 3,3.
15 = 3,5.
18 = 2,3,3.
LCM = 7 x 2 x 3 x 3 x 5 = 360.
Add the reminder = 360 + 4 = 364.
9 = 3,3.
15 = 3,5.
18 = 2,3,3.
LCM = 7 x 2 x 3 x 3 x 5 = 360.
Add the reminder = 360 + 4 = 364.
(21)
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?
(4)
SUBHANKAR DAS said:
4 years ago
L.C.M. of 6, 9, 15, and 18 is 90.
Let the required number be 90k + 4, which is a multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4,as it is not divisible if k=1,2,3
So, the Required number = (90 x 4) + 4 = 364.
Let the required number be 90k + 4, which is a multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4,as it is not divisible if k=1,2,3
So, the Required number = (90 x 4) + 4 = 364.
(3)
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