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 2 of 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?

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.

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.

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?

Manikanta said:   9 years ago
@Itsnoble.

111111111 is divisible by 9.

Rupam Gupta said:   8 years ago
Explanation of the answer is very helpful, Thank You all.


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