Given five consecutive natural numbers, if twice the smallest number is 40 less than the sum of these five numbers, what is the largest number among them?
五个连续自然数，如果其中最小的数的两倍比这五个数的和小40，那么其中最大的数是多少？
<gadget id="calculator">40 - 1 - 2 - 3 - 4</gadget>
<output>30</output>

<gadget id="calculator">1 - 2</gadget>
<output>-1</output>

<gadget id="calculator">1 + 1 + 1 + 1 + (-1)</gadget>
<output>3</output>

<gadget id="calculator">30 / 3</gadget>
<output>10</output>

<gadget id="calculator">10 + 4</gadget>
<output>14</output>

<result>14</result>