The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
A. 74
B. 94
C. 184
D. 364
Answer: Option D
Solution(By Examveda Team)
L.C.M. of 6, 9, 15 and 18 is 90Let required number be 90k + 4, which is multiple of 7
Least value of k for which (90k + 4) is divisible by 7 is k = 4
∴ Required number = (90 x 4) + 4 = 364
This Question Belongs to Arithmetic Ability >> Problems On H.C.F And L.C.M
Join The Discussion
Comments ( 4 )
Related Questions on Problems on H.C.F and L.C.M
In all the options only 364 is the multiple of 7..
K ki value kahaa se aa gyee apne aap ???
Let's say, x be the number which when divided by 6,9,15,18 leaves remainder 0. This implies (x+4) must leave remainder 4 when divided by the above numbers.The least number to satisfy the above condition is the LCM of the above numbers which is 90.
So, x=90. x+4=94
But 94 is not a multiple of 7.
With the help of hit and trial method,we can get
x=180,x+4=184 but not a multiple of 7.
x=270,x+4=274 but not a multiple of 7.
x=360,x+4=364.
364 leaves remainder 4 when divided by 6,9,15 and 18.And is also a multiple of 7.
Hence,364 is the least number satisfying all the conditions.
90K+4 is the required no.
if K=1
90*1+4=94 , it is not multiple of 7.
90*2+4=184,it is not multiple of 7
90*3+4=274,it is not multiple of 7
90*4+4=364,it is multiple of 7
so least value will be 364.