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 2 of 9.
Sana said:
8 years ago
Very clear and effective explanation. Thanks a lot @Suchita.
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.
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.
Bhagirath said:
8 years ago
Lcm of 5,6,7, and8=840.
840÷9=93,leaves remainder =3.
multiply all the values=5*6*7*8=1680.
and leaves remainder are added =1680+3=1683.
840÷9=93,leaves remainder =3.
multiply all the values=5*6*7*8=1680.
and leaves remainder are added =1680+3=1683.
Nitesh said:
8 years ago
Thanks @Suchita.
Kiruthikabaskaran said:
8 years ago
Super shorcut method. Thank you so much @Suchita.
Akshay said:
8 years ago
How did you get that the req number is of the form 840k+ 3?
SHIVA SHANKAR said:
8 years ago
Please clarifly my doubt after find out LCM for the numbers 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3.
How K = 2 comes please clarify friends.
Required number is of the form 840k + 3.
How K = 2 comes please clarify friends.
Sandeep said:
8 years ago
Find the least number which when divided by 16,18,20,25 leaves 4 as a remainder in each case but when divided by 7 leaves no remainder.
a) 17004
b) 18000
c) 18002
d) 18004
Can anyone explain the answer?
a) 17004
b) 18000
c) 18002
d) 18004
Can anyone explain the answer?
Manikanta said:
9 years ago
@Itsnoble.
111111111 is divisible by 9.
111111111 is divisible by 9.
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