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. 14)
14.
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
1677
1683
2523
3363
Answer: Option
Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

Discussion:
86 comments Page 5 of 9.

SHANKAR said:   6 years ago
Simply we can solve this by;

1+6+8+3=18
1+6+7+7=21
Same as other;

18 is divided by 9 and then B is correct.

Shubham shah said:   8 years ago
Answer is just simple divide all option values by 9 which is exactly divisible is your answer. Just simple.

Shruti said:   8 years ago
@Suchita.

You are truly right this really works!

Must say this is the best Sol for this question. Thanks.

Bodla aravind said:   6 years ago
Let (840k + 3) = 1683 [ 1683 taken from Option B].
Therefore, k = (1683-3)/840.
k = 1680/840,
k= 2.

Mehar said:   1 decade ago
multiply all the values ==> 5*6*7*8=1680.
these are leaving 3 as a reminder so,1680+3=1683.

Swati padewal said:   5 years ago
Simply we can solve this by,
5*6*7*8 = 1680.

And then after 1680+3 = 1683
The answer is 1683.
(12)

Sunita said:   9 years ago
If there is no option is given, and the question is asking sum of that no what will we do?

Bujji said:   1 decade ago
Simply divide the options with 9 and which option whose remainder is 0 it is the answer.

Rupesh said:   2 decades ago
All thing I understand but how this value of k is taken 2 can som one explain me.

Ritvik said:   10 years ago
Best possible way to add the digits in the option and check divisibility with 9.


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