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 8 of 9.

Anand said:   1 decade ago
The solution of finding the option which is divisible by 9 works only in this specific case of provided options. In case, when more than 1 option is divisible by 9, the method to find out the value of k needs to be understood.

Gogol said:   1 decade ago
@SUCHITA

If one of the option becomes for example..2538, then the sum of 2+5+3+8= 18 will also be divided by 9 leaving no remainder.

So, in this case there is no profit from your logic.

M.V.KRISHNA/PALVONCHA said:   1 decade ago
Excellent suchita.

Simanta said:   1 decade ago
Nice suchita !

Shariq said:   1 decade ago
Just divide all options by 9.

Saurabh agrawal said:   1 decade ago
Suchita's trick is great. Very short and productive.

Gaurav said:   1 decade ago
You can find out whether any num is divisible by nine, 3, 4 by using divisibily test. Jus try google search for divisibility test.

Sakshi said:   1 decade ago
nice Mehar.

Uttam said:   1 decade ago
Just try this,

Sum the digits of each option given. i.e

[A].1+6+7+7=21 [B].1+6+8+3=18


[C].2+5+2+3=12 [D].3+3+6+3=15

Now only 18 is divisible by 9

So answer should be 1683 that is option B.

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.


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