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:
86 comments Page 1 of 9.
Rupesh said:
2 decades ago
All thing I understand but how this value of k is taken 2 can som one explain me.
Chinmaya said:
1 decade ago
How you will get 2 ?
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.
Therefore, k = (1683-3)/840
k = 1680/840
k = 2.
Note: (840k + 3) should be divisible by 9 if we substitute k = 2.
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.
1+6+8+3=18 the answer is that which can be divided by 9.
remaining numbers can't divide by 9.
Priya said:
1 decade ago
nice Suchita ......it was nice
Emu said:
1 decade ago
suchita it"s working.........
Mehar said:
1 decade ago
multiply all the values ==> 5*6*7*8=1680.
these are leaving 3 as a reminder so,1680+3=1683.
these are leaving 3 as a reminder so,1680+3=1683.
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.
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.
Sakshi said:
1 decade ago
nice Mehar.
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.
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