2. Next, satisfy the condition that the number at100th after the decimal point is 3. We can try one by one.

Suppose the cycle starts at 7, with * *12 bits, 12×8=96, and the last four bits are: 7, 3, 8, 2, which does not meet the condition.

Suppose the cycle starts from 3, and the cycle part * *11,1 1×9=99, plus the first number 7, the last number is 0, which does not meet the condition.

Suppose the cycle starts at 8, and the cycle part is * * 10, 10×9=90, plus the first two digits 7,3, * * 92, and the last eight digits are 823 16934, which does not meet the condition.

Suppose the loop starts from 2, and the loop part has ***9 bits, 10×9=90, plus the first three bits 7, 3, 8 and * * 93, and the last seven bits are 23 16934, which does not meet the condition.

Suppose the cycle starts from 3, the cycle part is ***8 bits, 12×8=96, plus the first four bits 7,382, *** 100, and the last bit is 0, which does not meet the condition.

Suppose the loop starts from 1, and the loop part has * * 7 bits, 13×7=9 1, plus the first five bits 73823, * * 96, and the last four bits are 1693, which satisfies the condition.

So this cyclic decimal is 0.73823 (1) 69345 (0) on1and 0.