How many integer solutions of equation b c 13 satisfy 2,3 b o,c 3?

1. The stupidest way is to count one by one, if you don't want to count, you have to find a pattern.

2. The situation of B is relatively easy to do, and the solution of B is that there are only 4 different solutions.

3. Let's take a look at the case of b=0 and the solution of a and c. Since b = 0, therefore, a+c = 13, and because a 2, c 3, it can be assumed that a = 2 + x, c = 3 + y, x and y are both non-negative integers; There are 9 cases of non-negative integer solutions with (2+x)+(3+y)=13, x+y=8, and x+y=8.

4. According to the method of step 3 above, calculate the situation of B respectively. Therefore, in the end, there are 9 + 8 + 7 + 6 = 30 integer solutions.

When b = 3, a = 2, c = 8 or a = 3, c = 7 or a = 4, c = 6 or a = 5, c = 5 or a = 6, c = 4 or a = 7, c = 3

b = 2 at a = 2, c = 9 or a = 3, c = 8 or a = 4, c = 7 or a = 5, c = 6 or a = 6, c = 5 or a = 7, c = 4 or a = 8, c = 3

b = 1 at a = 2, c = 10 or a = 3, c = 9 or a = 4, c = 8 or a = 5, c = 7 or a = 6, c = 6 or a = 7, c = 5 or a = 8, c = 4 or a = 9, c = 3

b = 0 at a = 2, c = 11 or a = 3, c = 10 or a = 4, c = 9 or a = 5, c = 8 or a = 6, c = 7 or a = 7, c = 6 or a = 8, c = 5 or a = 9, c = 4 or a = 10, c = 3

Use the liquid shed partition method.

12 elements of the same beam.

There are 11 empty.

Insert 3 rubber key partitions.

c(11,3)=11*10*9/3!=11*5*3=165

with the partition method.

12 identical elements.

There are 11 empty.

Insert 3 partitions.

c(11,3)=11*10*9/3!=11*5*3=165

with the partition method.

12 identical elements.

There are 11 empty.

Insert 3 partitions.

c(11,3)=11*10*9/3!=11*5*3=165 hope it can help you!

There are 4 combinations.

There are 4x3 groups in the number of combinations.

There are 4x3 groups in the number of combinations.

There are 4 combinations.

There are 4x3 groups in the number of combinations.

There are 4x3 groups in the number of combinations.

There is 1 set of combinations

In the end, 56 sets were added.

Solution: You might as well set a b c, then there is.

abc=2(a+b+c)≤2(c+c+c)＝6c

So there is: 1< = ab 6

ab=1 i.e. a=b=1 is obviously unsolvable.

ab=2, i.e. a=1b=2 is obviously unsolvable.

ab=3, i.e., a=1b=3, 3c=2(4+c), c=8 (1, 3, 8).

ab=4=1*4=2*2 a=1b=4 c=5 （1 4 5）

a=2b=2 c=4 （ 2 2 4）

ab=5 a=1b=5 c=4 rounded (a, b, c).

ab=6=1*6=2*3 c has no positive integer solution.

Because abc is equivalent to (145) 145 154 415 451 541 514 6 pairs.

138) 138 183 381 318 813 831 6 pairs.

224) 224 242 422 3 pairs.

6+6+3=15 groups.

Solution: a==1

Let b==3 c==8 3!==6 (species a==2.)

Let b==2 c==4 3!/2!==3 (species) a ==3

Let b==2 c==4 3!==6 (species) 6+3+6 ==15 choose d

One. ｜ab｜=0,｜a+b｜=1；

a=0 Nianmengb =1, so there are two groups;

b = 0 a = 1, so there are two groups.

Two. ｜ab｜=1,｜a+b｜=0；

a = b a high defeat = 1; So there are two groups;

A total of six groups of finger tremors are as follows:

a=0,b=1

a=0,b=-1

a=1,b=0

a=-1,b=0

a=-1,b=1

a=1,b=-1

9a+6b+4c=80

5a+2b+4（a+b+c）=80

5a+2b+4x15=80

5a+2b=80-60

5a+2b=20

Since they are all positive integers, a=2 b=5