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The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

  • A.74
  • B.97
  • C.184
  • D.364

Answer: 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,
\(= (90 \times 4) + 4\)
\(= 364\)

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