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 7 of 9.

Emu said:   1 decade ago
suchita it"s working.........

Priya said:   1 decade ago
nice Suchita ......it was nice

Suchita said:   1 decade ago
hey.....just add the digits of given options as
1+6+8+3=18 the answer is that which can be divided by 9.

remaining numbers can't divide by 9.

Sundar said:   1 decade ago
Let (840k + 3) = 1683     [ 1683 taken from Option B]

Therefore, k = (1683-3)/840

k = 1680/840

k = 2.

Note: (840k + 3) should be divisible by 9 if we substitute k = 2.

Chinmaya said:   1 decade ago
How you will get 2 ?

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.

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.

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.


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