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 3 of 9.
Sai kowsalya.D said:
7 years ago
K=4 because 4 is d Perfect no that suits to d equation i.e (90k+4).
If we substitute k=4 then it becomes 364 hence 364 is divisible by 7 i.e 52 hence k=4.
If we substitute k=4 then it becomes 364 hence 364 is divisible by 7 i.e 52 hence k=4.
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.
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.
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?
Srihari said:
1 decade ago
It can not be 1, 2, 3 because if we substitute these numbers the result which is not divisible by 7. Let's check out. The only least number is 4.
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.
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.
(13)
Prince said:
8 years ago
@Priya it is so easy.
6 = 2 * 3
9 = 3 * 3
15 = 3 * 5
18 = 2 * 3 * 3.
Take one number from common multiple and multiply rest digits.
2*3*3*5.
6 = 2 * 3
9 = 3 * 3
15 = 3 * 5
18 = 2 * 3 * 3.
Take one number from common multiple and multiply rest digits.
2*3*3*5.
Srishti said:
1 decade ago
Here is a simple logic.
First find the LCM of the no.s which comes 90.
Now find the no. among the options greater than 90 and divisible by 7.
First find the LCM of the no.s which comes 90.
Now find the no. among the options greater than 90 and divisible by 7.
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.
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)
Shruti said:
8 years ago
You don't need to take anything on right-hand side. Just see whether the values of k when multiplied and added are a multiple of 7.
Sravanthi said:
1 decade ago
It is given in the question that it should be least multiple of 7 which leaves remainder 4. So why don't it be 94. Please clarify.
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