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.

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.
(21)

Mansi said:   7 years ago
Answer = 7.
Remainder = 4,
7*4 =28.
which answer divide by 28 that is the correct answer is it right?

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.

Narendra said:   9 years ago
Guys , here it is asked the least multiple so the answer is 94.

This will satisfy all the conditions.

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

Krishan rajwar said:   1 decade ago
Hey. Out of all the given option only 364 is the multiple of 7. Question is solved there itself.

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

Deb said:   6 years ago
It should be 94 which is divisible by 7 as well as leaves reminder 4 for other number.

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

How we are taking 90k+4??

Smile said:   6 years ago
I didn't understand this problem.

Please anyone explain this problem clearly.


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