1 20 1 1 1 1 1 1 are different natural numbers

There is such a method: 1 a = 1 ()+1 (), first find the divisor of a set to b and c, and then multiply (b+c) by the numerator and denominator of 1 a respectively (so that the magnitude of a does not change), and get (b+c) [a*(b+c)], that is.

b [a*(b+c)]+c [a*(b+c)] Because b,c is the divisor of a, b [a*(b+c)] and c [a*(b+c)] can be of the form 1 (). For example, the divisor of 1 20 and 20 is 1, 2, 4, 5, 10, and 20, and any two numbers such as 2 and 4 are taken, so 1 20 = (2 + 4) [20 * (2 + 4)] = 2 [20 * (2 + 4)] + 4 [20 * (2 + 4)] = 1 60 + 1 30

If 1 20==1 ( 1 ( then the numerator denominator of the sum multiplied by the divisors of three 20s is (2+4+5) [20*(2+4+5)]=2 [20*(2+4+5)]+4 [20*(2+4+5)]+5 [20*(2+4+5)]+5 [20*(2+4+5)].

In summary, 1 20 = 1 60 + 1 30 = 1 110 + 1 55 + 1 44 (PS: this is just one of the answers).

It's so tiring to type, and the landlord can add some points.

This question belongs to the split question of the score of the second volume of the fifth grade.

The factors of 20 are 1, 2, 4, 5, 10, and 20, a total of 6.

Then the numerator and denominator expand the multiple of the sum of these 6 factors at the same time, that is, expand the auspiciousness by 1 + 2 + 4 + 5 + 10 + 20 = 42 times:

If you want to learn this method systematically, I will guide you!

Write three at the front and see how many more.

For example, if n m is left, try a different a, convert the hidden rotten n m to an am and then split an=b+c, so that b and c are also the divisor of am when the stove hood leaks.

For example, dig into what I wrote earlier.

And 1 8 = 5 40 = 1 40 + 4 40 = 1 40 + 1 10

Summary. 24 and 552

1 23 = 1 ()1 () Fill in different natural numbers.

24 and 552

Process. Did you get it, pro.

Because. 1/18=n/(18n)=1/(18n)+(n-1)/(18n)

For n-1 (18n) = 1 (natural number), then n-1 must be an approximation of 18. The approximate number of 18 is 1, 2, 3, 6, 9, 18then n=2,3,4,7,10,19

The denominator of 1 18n is 36, 54, 72, 126, 180, 342, and the denominator of 1 () obtained after 18n is 36, 27, 24, 21, 20, 19, respectively

The two denominators of the first group are the same, and the other five groups all meet the requirements of the ruler shirt question. So the answer is.

The number in the regimental calendar = 15 multiplied by the number of the number of missing paragraphs in the empty search!