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 5 of 9.

V!cky said:   10 years ago
Divide all given answer by 7 when remainder remains 0 i.e correct answer or assume k.

Take LCM i.e 90 K+4.

Hira said:   10 years ago
Correct answer is 94. Read question carefully.

Anukriti said:   10 years ago
Is it for all cases we put the value of remainder in k; e.g if 7 is remainder then we should keep the value of k = 7? Is it general or anything else?

Anita said:   10 years ago
6+9+15+18 = 48.
48*7 = 336.

7*4 = 28.
Then 336+28 = 364.

364/48 = 7.5833.

Answer = 7.
Remainder = 4.

Is this right?

Abhik Sheek said:   10 years ago
Don't go for calculation first here it is told that multiple of 7. Here is your answer.

SATYA said:   9 years ago
How find the value of k==4?

John saida said:   9 years ago
Sir, is there any shortcut method present for this type of question other then options verification?

Ritu said:   9 years ago
But why to do lcm of 6, 9, 15, 18 here?

Will somebody help me out?

O jung beom said:   9 years ago
No need to be messed up. Here we are given a clue (multiple of 7) it means the no is to be divided by 7. We got only 364 divisible with 7.

Now just make a test by dividing 364 by 6. As it leaves a remainder of 4 so 364 is the correct answer.

Snehesh said:   9 years ago
In the equation,

AX + B,

-> A is the LCM of all the given values.
-> B is the remainder.
-> The first number of the form: 90A+4 is 364. Hence, the ans.


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