I would like to ask these numbers 1-9 add up to 24, nine empty, horizontal and vertical should be equal to 24 Horizontal is five empty, vertical is also five empty, the middle one is shared, how can we get 24?

The number in the middle is three
The vertical ones are 8, 7, 3, 4, 2
The horizontal ones are 5, 1, 3, 9 and 6

The sum of 1-9 is equal to 24

1
two
forty-five thousand three hundred and sixty-seven
eight
nine

Use eight eights to make five numbers. The sum of five numbers equals 1000. Can you do that?

888+88+8+8+8=1000

Add the nine numbers 2.4.6.8.10.12.14.16 and 18 to make the sum of each horizontal line, each vertical line and each diagonal line 30

16、2、12
6、10、14
8、18、4

Fill in the blanks with appropriate fractions so that the sum of the three numbers in horizontal, vertical and oblique rows equals 1
() () 2 / 15
1 / 3 ()
（）（）（）

(4 / 15) (3 / 5) 2 / 15
1 / 5 1 / 3 (7 / 15)
（ 8/15 ）（ 1/15 ）（ 2/5 ）

1-33 numbers, choose 6 numbers to add up to 114, how many groups of such numbers? It's better to list them
The numbers are not before and after, and cannot be repeated

This is too much. The core of 19995 group is how to make all combinations of 3 out of 33 numbers_ Click()Dim a(1 To 6) As Integerk = 6: duoshao = 0For i = 1 To 6: a(i) = i: NextDohe = 0For i = 1 To 6: he = he + a(i):...

Any six numbers from 1 to 33 add up to 22 and 23
Don't take into account the decimal 'fraction' of these miscellaneous 7 miscellaneous 8``
It's the 33 numbers of nature 01 02 03 04.31 32 33
You'd better arrange it for me. The sum is 24 25 26 28 29.30... All the way to 183
It's best

1+2+3+4+5+7=22
1+2+3+4+5+8=23
1+2+3+4+5+9=24
1+2+3+4+5+10=25
1 + 2 + 3 + 4 + 5 + 11 = 26 (keep the first five numbers unchanged, and add 1 to the sixth number each time until 33)
Numbers with sum of 22-48)
.
1+2+3+4+5+32=47
1+2+3+4+5+33=48
(then add the fifth number one by one until the fifth number is 32. And becomes 75.)
1+2+3+4+6+33=49
1+2+3+4+7+33=50
1+2+3+4+8+33=51
.
1+2+3+4+32+33=75
(then again, keep the numbers of the first, second, third, fifth and sixth unchanged, and add one to the fourth digit one by one)
1+2+3+5+32+33=76
1+2+3+6+32+33=77
.
1+2+3+31+32+33=102
(then, change the third number)
1+2+4+31+32+33=103
.
1+2+30+31+32+33=129
(then, change only the second number)
1+3+30+31+32+33=130
1+4+30+31+32+33=131
.
1+29+30+31+32+33=156
Last (change the first number only)
2+29+30+31+32+33=157
3+29+30+31+32+33=158
.
28+29+30+31+32+33=183
The results can be obtained
(we should think more and find out the rules. It's easy.)

How to calculate the sum of 6 numbers from 1 to 33 equal to 103
It's better to have a method of calculation

103 / 6 equals 17 and 1
Then you make the average of five numbers 17 and one number 18

1-33 numbers, choose 6 numbers to add up to 111, how many groups of such numbers? It's better to list them

There are 18549 groups of such numbers, just a few groups: 1 + 2 + 12 + 31 + 32 + 33, 1 + 2 + 13 + 30 + 32 + 33, 1 + 2 + 14 + 29 + 32 + 33, 1 + 2 + 14 + 30 + 31 + 33, 1 + 2 + 15 + 28 + 32 + 33, 1 + 2 + 15 + 29 + 31 + 33, 1 + 2 + 15 + 30 + 31 + 32, 1 + 2 + 16 + 27 + 32 + 33, 1 + 2 + 16 + 28 + 31 + 33, 1 + 2 + 16 + 29 + 30 + 33, 1 + 2 + 16

How many combinations can there be if the sum of 6 numbers from 1 to 18 is equal to 33? (not permutation)
Write each group down, or there will be no marks
It can't be repeated

18,5,4,3,2,1；17,6,4,3,2,1；16,7,4,3,2,1；16,6,5,3,2,1；15,8,4,3,2,1；15,7,5,3,2,1；15,6,5,4,2,1；14,9,4,3,2,1；14,8,5,3,2,1；14,7,6,3,2,1；14,6,5,4,3,1；13,10,4,3,2,1；13,9,5,3,2,1；13,8,6,3,2,1；13,...

I would like to ask these numbers 1-9 add up to 24, nine empty, horizontal and vertical should be equal to 24 Horizontal is five empty, vertical is also five empty, the middle one is shared, how can we get 24?