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:
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:
87 comments Page 8 of 9.
Rohit said:
10 years ago
Please help me how we get k=2?
Vinay said:
10 years ago
Yes if there was no option for how to find k. I can also go with @Mukul.
Truptti said:
10 years ago
How get 840?
Mukul said:
10 years ago
What if there was no option for and, then how to find k?
Itsnoble said:
10 years ago
@Suchita.
What if we have 111111111? Sum equal 9 but not divisible by 9.
An exception but nice work in this case.
What if we have 111111111? Sum equal 9 but not divisible by 9.
An exception but nice work in this case.
Siva said:
10 years ago
Finding the answer with the choose options is the best method but finding the value of k has some formula. That is what should know.
Ritvik said:
1 decade ago
Best possible way to add the digits in the option and check divisibility with 9.
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?
Ayush said:
1 decade ago
L.C.M. Of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3.
We know that it is divisible by 9.
So, 840k+3 = 9x. (x being the quotient).
Now, k is found by just putting values starting from 1, 2, 3.
If we get the lowest value which when used in k makes equation divisible by 9, it will surely be 2.
Hope it helped.
Required number is of the form 840k + 3.
We know that it is divisible by 9.
So, 840k+3 = 9x. (x being the quotient).
Now, k is found by just putting values starting from 1, 2, 3.
If we get the lowest value which when used in k makes equation divisible by 9, it will surely be 2.
Hope it helped.
Mohit said:
1 decade ago
For those who giving method of dividing options by 9 and leaving 0 as remainder. During actual tests question makers just pull out trick like they give options which are leaving 0 remainder after dividing by 9 and also these options are satisfying other conditions.
So examples like this must be solve by given method of solving equation, this is the fastest and reliable method.
So examples like this must be solve by given method of solving equation, this is the fastest and reliable method.
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