On the calendar, circle 3 × 3 numbers arbitrarily with a square, then the sum of these nine numbers may be A.80 B.98 C.108 D.206. Who can tell me the solution of this type of problem? It's best to have a formula

On the calendar, circle 3 × 3 numbers arbitrarily with a square, then the sum of these nine numbers may be A.80 B.98 C.108 D.206. Who can tell me the solution of this type of problem? It's best to have a formula

The calendar is arranged by seven days of a week. So if the middle number in the 3 * 3 square circle is x, then the number in the square should be X
X-8 X-7 X-6
X-1 X X+1
X+6 X+7 X+8
Add these nine numbers to get 9x, so the sum of these nine numbers should be a multiple of 9
So choose C

On the calendar, circle 9 numbers with a square (1). If the 9 numbers are 108, you can find out the numbers of the 9 days by square multiplication
emergency

Let the middle one be x, and we can know the nine numbers
x-7-1 x-7 x-7+1
x-1 x x+1
x+7-1 x+7 x+7+1
The result is: 9x = 108, so x = 12

Can you circle a square in the calendar so that the sum of the four numbers in the square is 78? If so, what are the dates of these four days? If not, please give reasons

Suppose we can find such a square, the number in the upper left corner is x, then the other three numbers should be x + 1, x + 7, x + 8; then we should have: x + X + 1 + X + 7 + X + 8 = 78, that is: 4x = 62, the solution is: x = 15.5, the calendar can't have a decimal part, so the assumption is not true, we can't find such a square, so we can't circle such a square