The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90$ k $ + 4, which is multiple of 7.
Least value of $ k $ for which [90$ k $ + 4] is divisible by 7 is $ k $ = 4.
$\therefore$ Required number = $\left(90 \times 4\right)$ + 4 = 364.
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