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.

Kannan said:   1 decade ago
In the below eqn.
Least value of k for which (840k + 3) is divisible by 9.

K can be K=0,1,2...

Sub K=1,
(840*1)+3 =-- Not divisible by 9.

Sub K=2,
(840*2)+3 = 187-- Divisible by 9.

Sub K=3,
(840*3)+3 =-- Not Divisible by 9.

Hence take lease no. (i.e 2).
(1)

Neelima said:   6 years ago
@All.

Hai I have a doubt, everyone is explaining from 840k+3 but here the doubt is we got lcm 840 if 5 6 7 8 divide the 840 answer is 0.

So, 840 + 3=843 now we will get remainder 3 but here 5 is divisor 840 is dividend so how you come 840k+3? Explain it.

Ayush said:   1 decade ago
To find value of k, see the following:

840k+3 = 0 (mod9).
30k+3 = 0 (mod9) [840/9=30 (mod9)].
30k = 6 (mod9).
3k = 6 (mod9) [30/9=3 (mod9)].

Dividing both sides by 3, we get

k = 2 (mod9).

So, the value of k = 2.

HOPE THIS ONE ALSO HELPED YOU.

SURESH BINWAL said:   9 years ago
Any number whose digits sum is nine (9) then only the number is divisible by nine.

taking k = 1..means 843 , sum will be 8+4+3 = 15 not divisible by nine.

k = 2 is 1643 & sum of digit is 1+6+4+3 = 18 means 1+8 = 9 hence least number is k=2.

Samiul sk said:   9 years ago
Find the least number which when divided by 12, leaves a remainder of , when divided by 15, leaves a remainder of 10, when divided by 16, leaves a remainder of 11.

(a)115 (b)235 (c)247 (d)475

Solve and find the answer.

Feroz said:   1 decade ago
we get LCM =840 since 843 is divided by 3
when we divide it by 9 it leaves a remainder of 6
since we need a least no. divided by three which also leaves 3 remainder
we put value of k=1, 2, 3.... till we get the required no.
since

Anand said:   1 decade ago
The solution of finding the option which is divisible by 9 works only in this specific case of provided options. In case, when more than 1 option is divisible by 9, the method to find out the value of k needs to be understood.

Munna Mudassir said:   8 years ago
Another shortcut which will work in this case is checking thew divisibility by 9.
only 1683 is divisible by 9 hence the answer is 1683. In case there is more numbers divisible by you have to check it as my previous answer.

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?

Uttam said:   1 decade ago
Just try this,

Sum the digits of each option given. i.e

[A].1+6+7+7=21 [B].1+6+8+3=18


[C].2+5+2+3=12 [D].3+3+6+3=15

Now only 18 is divisible by 9

So answer should be 1683 that is option B.


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