$$2418-3\times[((7\times13)-52+7]+[(18\times12)\div9]$$
$38$
$$248-3\left(91-52+7+\frac{18\times 12}{9}\right)$$
$$248-3\left(39+7+\frac{18\times 12}{9}\right)$$
$$248-3\left(46+\frac{18\times 12}{9}\right)$$
$$248-3\left(46+\frac{216}{9}\right)$$
$$248-3\left(46+24\right)$$
$$248-3\times 70$$
$$248-210$$
$$38$$
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$2\times 19$
