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:
86 comments Page 5 of 9.
Vikram Rajput said:
9 years ago
Divide all options by number which is multiple in the question if remainder 0 for any option, that is your answer.
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?
Manikanta said:
9 years ago
@Itsnoble.
111111111 is divisible by 9.
111111111 is divisible by 9.
Triveni said:
9 years ago
You know concept of "Dividend : (Divisor*Quotient) plus Remainder".
So here 800 is divisor "k" is quotient "3" is remainder.
So here 800 is divisor "k" is quotient "3" is remainder.
Vijay said:
9 years ago
Please tell me how you put the value k=2?
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.
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