There is a class of natural numbers. The product of all divisors (including 1 and itself) of each number has exactly 8 consecutive 0s at the end. What is the smallest natural number that satisfies this condition.
有一类自然数，每个数所有约数(包括1和它自己)的乘积末尾刚好有8个连续的0，满足这种条件的最小自然数是多少．
<gadget id="calculator">1 + 59</gadget>
<output>60</output>

<gadget id="calculator">59 / 2</gadget>
<output>59/2 = around 29.5</output>

<gadget id="calculator">60 * (59/2)</gadget>
<output>1_770</output>

<result>1_770</result>