Let 1, 2, 3 It is known that the last 13 digits of the product are all exactly 0. May I ask: what is the minimum natural number that appears at the end of multiplication?

Only the multiplication of factor 2 and 5 can get a 0. If the last 13 bits of the product are all exactly 0, then at least 13 factor 2 and 13 factor 5 are needed. There are many factors in factor 2. To get 13 factors, 5, 5, 10, 15, 20, 30, 35, 40, 45, 55 each can get a factor 5, a total of 9, 25, 50 each can get

Some continuous natural numbers 1,2,3 There are 13 continuous zeros at the end of the product of. What is the largest natural number?

If we can generate 0 at the end, we just need to look at 5, 10, 15, 20... These numbers
Five times an even number produces one zero,
10 can produce a 0, 15 can produce a 0, 20 can produce a 0,
If 25 * 4 = 100, two zeros can be generated; if 30 = 100, one zero can be generated; if 50 * 2 = 100, two zeros can be generated,
At this point, there are 12 zeros in total, and further down 55 can produce another 0
Because the question is the largest natural number, and to 60 can produce 14 0
So the largest natural number is 59

All natural numbers except 0 can be regarded as false fractions whose denominator is 1

Yes
For example: 1, 2, 3, 4, 5, 6. Can be changed into 1 / 1, 2 / 1, 3 / 1, 4 / 1, 5 / 1, 6 / 1
A false score of

Let 1, 2, 3 It is known that the last 13 digits of the product are all exactly 0. May I ask: what is the minimum natural number that appears at the end of multiplication?