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:
74
94
184
364
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.

Navnath said:   4 years ago
As sample as that.

If the number is multiple of 7 then Obviously it should divide by 7.
Then check all the options by dividing 7.

Look only one option that is 364 is completely divided by 7.
(22)

Bhavesh said:   4 years ago
90K+4 is the required number.

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.
(61)

Sagar Khare said:   4 years ago
@ALL.

Guys here you use divisibility rules.

We have to find 7 multiple so apply the simple divisibility rule of 7 on all the answers.
(23)

Codi said:   3 years ago
Thanks for explaining @Rahul.
(2)

Nakum pragnesh said:   3 years ago
Or which option is divisible by 7 is your answer
(23)

MMTbelieve youtube said:   2 years ago
We got till lcm 90 ,
Then N=90k+4 should be divisible by 4 then we will get n.
So, put k=1,2,3,4 on 4 the ans 364 is divisible by 7.
So N = 364.
(15)

Sherin lazar said:   5 months ago
Thank you all for explaining answer.
(1)

Sherin lazar said:   5 months ago
Thank you everyone for explaining the session clearly.
(1)


Post your comments here:

Your comments will be displayed after verification.