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 4 of 9.
Lali said:
1 decade ago
It clear when we put a value of k 1 then 5,6,7,8, leave the remainder 3 and 6 in case of 9. when k=2. in this case no remainder is left.
Siva said:
1 decade 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.
Anjali said:
7 years ago
Check the 4 options divisibility for 9.
Option B.
1+6+8+3=18 which is divisible by 9.
While other options are not divisible by 9.
Option B.
1+6+8+3=18 which is divisible by 9.
While other options are not divisible by 9.
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.
Priyanka Mishra said:
1 decade ago
@Kranthi you can check from the options by dividing the number by 5, 6, 7, 8 and find if the required remainder is coming or not.
Triveni said:
10 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.
Ramya namani said:
6 years ago
Or simply you can go for option verification.
Check the options whether it is divisible by 9.
Only 1683 is divisible by 9.
Check the options whether it is divisible by 9.
Only 1683 is divisible by 9.
(2)
Rd sharma said:
1 decade ago
In question when number is divided by 9 no reminder. If reminder is 2 then how can solve this question. Please reply anyone.
Itsnoble said:
1 decade 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.
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.
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