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The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Answer & Solution
Correct Answer:
Option
D
L.C.M. of 6, 9, 15 and 18 is 90
Let 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
Let 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
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LoginSo, 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.
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.