The greatest common divisor of two numbers is 37, and the sum of two numbers is 444, and such a set

If the major common divisor of two numbers is 37, then the sum of two numbers 444 should be 37 and multiples, so there are 5 groups of numbers with a common divisor of 37 and a sum of 444, that is, 1*37 and 11*37

2*37 and 10*37

3*37 and 9*37

4*37 and 7*37

5*37 and 6*37

Among them, the greatest common divisor of 2*37 and 10*37 and 3*37 and 9*37 is not 37, which are 2*37 and 3*37, respectively

In this way, there are 3 groups of natural numbers that meet the criteria, namely:

37 and 407

148 and 259

175 and 222

a b)=37 a+b=444

The greatest common divisor is 37, so the numbers a and b are multiples of 37, and their sum is also a multiple of 37.

2,10; 3,9; 4,8 ；6,6 these 4 groups, their greatest common divisor will not be 37 (e.g., 2,10 The greatest common divisor is 2*37).

Eligible 1*37,11*37=407;

1、a=37 b=407

2、a=185 b=259

A ......37……B......407

A ......74……B......370

A ......111……B......333

A ......148……B......296

A ......185……B......259

A ......222……B......222

A ......259……B......185

A ......296……B......148

A ......333……B......111

A ......370……B......74

A ......407……B......37

444 37 = 12, 12 split into 2 coprime numbers, there are 1+11 and 5+7,2 groups.

So 11*37+1*37=444

11 can only be disassembled into 4 + 7 to be full of chains, feasting on the full shed of silver.

A spike is 36*7=252

The other 36*4=144

The 4 numbers are: 247,247,247,6*247

The biggest hunger of the 4 socks was (1482) limb slippage.

15 1 14 2 13 4 11 7 8

Therefore, there are a total of 4 groups of two numbers that are bent high like this.

Let the greatest common divisor be d

The two books are DA and DB

Then (a,b) = 1

The least common multiple is DAB

So d(ab+1)=143 = 11*13, if d=11, then ab=12 can be taken as (3,4), if d=13, then ab=10 can be taken as (2,5), so the two composite numbers are (33,44) or (26,65).

Let the greatest common divisor of the two composite numbers be n, then the two composite numbers are an and bn respectively (a and b are coprimous), and their least common multiple is abn

The equation n+abn=143

n(1+ab)=11x13

If n = 11, ab = 12, a = 3, b = 4. These two numbers are 33 and 44;

If n=13, ab=10, a=2, b=5. These two numbers are 26 and 65, respectively.

The answer is 130 and 13

The product of the greatest common divisor and the least common multiple is the product of the two numbers", which means that the two numbers are multiples, 143 13 11

Divide 11 into 10+1 with 13 10 130 and 13 1 13 so the answer is 130 and 13

I tried, and the score of 13 didn't fit the topic

The least common multiple must be an integer multiple of the greatest common divisor, and their sum is, of course, an integer multiple of the greatest common divisor.

The greatest common divisor is 13

The least common multiple is 143-13=130

These two composite numbers are 26 and 65

The least common multiple must be an integer multiple of the greatest common divisor, and their sum must also be an integer multiple of the greatest common divisor.

Because: 143 = 11 13

Then, the greatest common divisor is equal to 11 or 13

When the greatest common divisor is equal to 11:

Least common multiple = 143 - 11 = 132 = 11 12 Then, the product of these two composite numbers = least common multiple and greatest common divisor = 11 11 12

That is, these two composite numbers are 33 and 44

When the greatest common divisor = 13:

Least common multiple = 143 - 13 = 130 Then, the product of these two composite numbers = least common multiple and greatest common divisor = 13 130

That is, these two composite numbers are 26 and 65

Let the two composite numbers be a and b respectively.

a=mc，b=nc

m,n coplasmic).

i.e.: c+mnc = 143

c（1+mn)=11x13

c=11;1+mn=13；Or:

c=13；1+mn=11

c=11；mn=12；Or:

c=13；mn=10

So: c=11, m=4, n=3;Or:

c=13；m=5；n=2

That is, the two composite numbers are: a=4x11=44;b = 3x11 = 33 or: a = 13 x 5 = 65;b=2x13=26So: these two composite numbers are (44) and (33); or (65) and (26).