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.

Ram said:   1 decade ago
I'm confused to find LCM and HCF. Can anyone please explain?

Ujwal said:   1 decade ago
@Mohsin.

Factors of 6 = 2*3.
Factors of 9 = 3*3.
Factors of 15= 3*3*5.
Factors of 18 = 3*2*3.
LCM = 3*2*3*5= 90.

Mohsin said:   1 decade ago
How did you get 90? Can anyone please explain.

Akshara said:   1 decade ago
So basically this is kind of a trial an error method question you just need to find the value of k such that 90k+4/7 gives 0.

Anonymous said:   1 decade ago
But in this question, the least number is 4. However, what if the number is too large. Will we have to try with every number in place of k and keep checking if it is divisible on not?

Dipak lal said:   1 decade ago
L.C.M. of 6, 9, 15 and 18 is 90.

Take k =1,2,3,... till reminder is 00.

Let required number be (90 k + 4 )/7.

if k=1 then 90(1)+4/7 = 94/7 =13.42 = reminder is not 00.

if k=2 then 90(2)+4/7 = 184/7 =26.28 = reminder is not 00.

if k=3 then 90(3)+4/7 = 274/7 =39.14 = reminder is not 00.

if k=4 then 90(4)+4/7 = 364/7 =52.00 = reminder is 00.

OK now you take k = 4 because remainder 0.
Solution is 90k + 4 = 90*4 + 4 = 364.


Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

Myk said:   1 decade ago
a/b ve get quocent q and reminder r
there for a=bq+r that why 90k+4

Jennifer said:   1 decade ago
Let required number be 90k + 4, which is multiple of 7.

How we are taking 90k+4??


Post your comments here:

Your comments will be displayed after verification.