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 4 of 9.

Ram said:   1 decade ago
k=4 because we need least value of k which is divided 7.

So in equation 90k+4. If k=4 then answer is 364 which is divisible by 7.

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.

Denusan said:   7 years ago
When the leaves of the remainder is not equal to get the value of 4 then 90+4=94.

Is it right thew given is used in it.

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?

Sachin said:   1 decade ago
Given here the answer should be multiple of 7, so simply divide all options by 7.

Then we only got the answer 364.

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.

Nagesh Bhukya said:   6 years ago
K = 4.

Because this value decided by 7 in 90k+4.
90(4)+4=364/7 = 52.
Remainder is 0 .
So k = 4 is correct.
(1)

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.

Sowmya said:   1 decade ago
94 is divisible by 7 and it leaves a reminder of 4 when divided with all other given numbers so y not 94?

Meenakshi said:   1 decade ago
@bhavi,

The trick which was said by krishna was very simple and even though you ask for a short trick ?


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