Four consecutive natural numbers a, b, c, d are multiples of 2, 3, 4, 5 in turn (the multiple is greater than 1), then what is the minimum value of a+b+c+d.
四个连续的自然数a，b，c，d，依次是2，3，4，5的倍数(倍数大于1)，则a+b+c+d最小是多少．
<gadget id="calculator">3 * 4 * 5</gadget>
<output>60</output>

<gadget id="calculator">60 + 2</gadget>
<output>62</output>

<gadget id="calculator">60 + 3</gadget>
<output>63</output>

<gadget id="calculator">60 + 4</gadget>
<output>64</output>

<gadget id="calculator">60 + 5</gadget>
<output>65</output>

<gadget id="calculator">62 + 63 + 64 + 65</gadget>
<output>254</output>

<result>254</result>