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 2 of 9.
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)
Jambay said:
5 years ago
Why we are taking LCM instead of GCF? Please explain.
(7)
Choudhary said:
5 years ago
What if the value of k has very large?
(1)
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)
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.
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)
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.
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)
Deepak said:
6 years ago
Thanks all.
Deb said:
6 years ago
It should be 94 which is divisible by 7 as well as leaves reminder 4 for other number.
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