Доска 14 х 501 раскрашена в шахматном порядке. В её клетках расставлены числа, как показано на рисунке. Найдите разность сумм чисел на чёрных и белых клетках.

Задание 2. Разность сумм чисел на чёрных и белых клетках

Дано:

• Размер доски: 14 x 501.
• Клетки расставлены числа, как показано на рисунке.

Найти: разность сумм чисел на чёрных и белых клетках.

Решение:

Чтобы решить эту задачу, нужно понять, как числа располагаются на доске и как они распределяются по цвету клеток. На рисунке показан небольшой фрагмент доски, но важно не его точное содержимое, а закономерность.

Доска раскрашена в шахматном порядке. Это значит, что соседние клетки (по горизонтали и вертикали) имеют разные цвета.

Рассмотрим, как связаны числа и цвета клеток:

1. Первая строка: 1 (белая), 2 (чёрная), 3 (белая), 4 (чёрная), 5 (белая).
2. Вторая строка: 15 (чёрная), 16 (белая), 17 (чёрная), 18 (белая), 19 (чёрная).
3. Третья строка: 29 (белая), 30 (чёрная), 31 (белая), 32 (чёрная), 33 (белая).

Мы видим, что:

• Белые клетки: 1, 3, 5, 16, 30, 32, 33.
• Чёрные клетки: 2, 4, 15, 17, 18, 29, 31.

Однако, это только часть чисел. Более важный момент в том, что на доске 14x501, и числа идут последовательно. Как расположены числа относительно цвета клетки?

Давайте проанализируем разницу между соседними числами:

• Строка 1: 2-1=1, 3-2=1, 4-3=1, 5-4=1.
• Строка 2: 16-15=1, 17-16=1, 18-17=1, 19-18=1.
• Строка 3: 30-29=1, 31-30=1, 32-31=1, 33-32=1.

Разница между числами в соседних строках (например, между 1 и 15): 15-1 = 14. Это соответствует количеству столбцов на доске.

Ключевой момент:

На доске 14x501. Если мы возьмём любое число и посмотрим на число в клетке, которая находится справа от него, то разница будет 1. Если посмотреть на число под ним, то разница будет 14 (т.е. 14 ячеек между ними).

Рассмотрим пару соседних клеток: одна белая, другая чёрная. Например, клетки с числами 1 (белая) и 2 (чёрная).

Разность между числом в чёрной клетке и числом в белой клетке, расположенных рядом по горизонтали, равна 1 (2 - 1 = 1).

Теперь посмотрим на разность между числом в чёрной клетке и числом в белой клетке, расположенных рядом по вертикали. Например, число 2 (чёрная) и число 16 (белая) в первом столбце. Разность: 16 - 2 = 14. Или число 15 (чёрная) и 29 (белая). Разность: 29 - 15 = 14.

Важно: В шахматном порядке, если клетка (i, j) имеет один цвет, то клетка (i, j+1) имеет другой цвет, и клетка (i+1, j) имеет другой цвет.

Рассмотрим разницу между суммой чисел на чёрных клетках и суммой чисел на белых клетках. Каждая пара соседних клеток (по горизонтали) отличается на 1. Если клетка слева белая, а справа чёрная, то разность (чёрная - белая) = 1. Если клетка слева чёрная, а справа белая, то разность (белая - чёрная) = 1, и (чёрная - белая) = -1.

В каждой строке у нас будет примерно равное количество белых и чёрных клеток (из-за четного количества столбцов 14). Например, в первой строке 1, 3, 5 - белые (3 шт.), 2, 4 - чёрные (2 шт.). Но это не так, у нас 14 столбцов, значит 7 белых и 7 чёрных.

Важное свойство шахматной доски:

На доске с чётным количеством столбцов (14), в каждой строке будет одинаковое количество клеток каждого цвета (14 / 2 = 7). То есть, в каждой строке будет 7 белых и 7 чёрных клеток.

В пределах одной строки, суммируя разницы между соседними клетками:

• Если у нас идут пары (белая, чёрная), то сумма разностей будет: (2-1) + (4-3) + ... = 1 + 1 + ...
• Если у нас идут пары (чёрная, белая), то сумма разностей будет: (3-2) + (5-4) + ... = 1 + 1 + ...

Сумма чисел на чёрных клетках минус сумма чисел на белых клетках:

Let S_white be the sum of numbers on white cells, and S_black be the sum of numbers on black cells.

Consider a pair of adjacent cells horizontally: (white, black). The difference is black - white = 1.

Consider a pair of adjacent cells horizontally: (black, white). The difference is white - black = 1, so black - white = -1.

In each row of 14 cells, there are 7 white and 7 black cells. The sequence of colors is W, B, W, B, ..., W, B. This means there are 7 pairs of (W, B). The sum of differences (black - white) in a row will be 7 * 1 = 7.

Similarly, if the row starts with Black, it would be B, W, B, W, ..., B, W. There are 7 pairs of (B, W). The sum of differences (black - white) in a row would be 7 * (-1) = -7.

However, we need to consider the transition between rows.

Let's look at the difference between the sum of all black cells and the sum of all white cells.

Consider the first cell (1,1) which is white. Its value is 1.

The cell (1,2) is black, its value is 2. Difference (black - white) = 2 - 1 = 1.

The cell (1,3) is white, its value is 3. Difference (black - white) = 2 - 3 = -1.

The cell (1,4) is black, its value is 4. Difference (black - white) = 4 - 3 = 1.

In the first row (14 columns):

Sum of numbers on white cells = 1 + 3 + 5 + ... + 13. (These are odd numbers). The last odd number in the first row is 13. There are (13-1)/2 + 1 = 7 white cells.

Sum of numbers on black cells = 2 + 4 + 6 + ... + 14. (These are even numbers). The last even number in the first row is 14. There are 14/2 = 7 black cells.

Sum of differences (black - white) for the first row = (2-1) + (4-3) + (6-5) + ... + (14-13) = 1 + 1 + 1 + ... + 1 (7 times) = 7.

Now, consider the second row. The first cell (2,1) is black. Its value is 15.

The cell (2,2) is white, its value is 16. Difference (black - white) = 15 - 16 = -1.

The cell (2,3) is black, its value is 17. Difference (black - white) = 17 - 16 = 1.

In the second row, the color pattern is Black, White, Black, White, ...

Sum of numbers on black cells = 15 + 17 + 19 + ... + (15 + 2*6) = 15 + 17 + 19 + 21 + 23 + 25 + 27. (This is incorrect, as the numbers are incrementing by 1 in the cells, not by 2 between black cells). Let's re-evaluate the picture.

The picture shows:

Row 1: 1, 2, 3, 4, 5 (Colors: W, B, W, B, W)

Row 2: 15, 16, 17, 18, 19 (Colors: B, W, B, W, B)

Row 3: 29, 30, 31, 32, 33 (Colors: W, B, W, B, W)

Let's assume the coloring starts with White in the top-left corner (1,1).

Then the color of cell (i, j) is White if i+j is even, and Black if i+j is odd.

For a 14x501 board:

The total number of cells is 14 * 501 = 7014.

Since the number of columns (14) is even, in each row there are exactly 14/2 = 7 white cells and 7 black cells.

Therefore, the total number of white cells = 501 * 7.

The total number of black cells = 501 * 7.

The total number of cells is even, so there are exactly half white and half black cells: 7014 / 2 = 3507 white cells and 3507 black cells.

Consider any two adjacent cells, one black and one white. Let their values be x and y.

If y = x+1 (horizontal adjacency), and x is white, y is black, then y-x = 1.

If y = x+1 (horizontal adjacency), and x is black, y is white, then y-x = 1, so x-y = -1.

If y = x+14 (vertical adjacency), and x is white, y is black, then y-x = 14.

If y = x+14 (vertical adjacency), and x is black, y is white, then y-x = 14, so x-y = -14.

Let's consider the sum of numbers on all cells. The sum of numbers from 1 to N is N(N+1)/2.

The problem asks for the difference between the sum of numbers on black cells and the sum of numbers on white cells. Let this be D = Sum(Black) - Sum(White).

Consider the first row. Numbers are 1, 2, 3, ..., 14. White cells: 1, 3, ..., 13. Black cells: 2, 4, ..., 14.

Sum(White in row 1) = 1+3+...+13 = 49.

Sum(Black in row 1) = 2+4+...+14 = 56.

Difference (Black - White) in row 1 = 56 - 49 = 7.

Consider the second row. Numbers start from 15. So, 15, 16, ..., 28.

Cell (2,1) is black (since 2+1=3 is odd). Its value is 15.

Cell (2,2) is white (since 2+2=4 is even). Its value is 16.

Cell (2,3) is black (since 2+3=5 is odd). Its value is 17.

Sum(Black in row 2) = 15 + 17 + ... + 27. (This is an arithmetic progression with 7 terms, first term 15, last term 27). Sum = (7/2) * (15 + 27) = (7/2) * 42 = 7 * 21 = 147.

Sum(White in row 2) = 16 + 18 + ... + 28. (This is an arithmetic progression with 7 terms, first term 16, last term 28). Sum = (7/2) * (16 + 28) = (7/2) * 44 = 7 * 22 = 154.

Difference (Black - White) in row 2 = 147 - 154 = -7.

Consider the third row. Numbers start from 29. So, 29, 30, ..., 42.

Cell (3,1) is white (since 3+1=4 is even). Its value is 29.

Cell (3,2) is black (since 3+2=5 is odd). Its value is 30.

Sum(White in row 3) = 29 + 31 + ... + 41. (7 terms). Sum = (7/2) * (29 + 41) = (7/2) * 70 = 7 * 35 = 245.

Sum(Black in row 3) = 30 + 32 + ... + 42. (7 terms). Sum = (7/2) * (30 + 42) = (7/2) * 72 = 7 * 36 = 252.

Difference (Black - White) in row 3 = 252 - 245 = 7.

We see a pattern in the differences for each row: +7, -7, +7, -7, ...

The sign of the difference depends on whether the first cell of the row is white or black.

If the first cell (i,1) is White (i+1 is even, so i is odd), the difference is +7.

If the first cell (i,1) is Black (i+1 is odd, so i is even), the difference is -7.

There are 501 rows.

Number of odd rows (1, 3, ..., 501) = (501 - 1) / 2 + 1 = 500 / 2 + 1 = 250 + 1 = 251.

Number of even rows (2, 4, ..., 500) = (500 - 2) / 2 + 1 = 498 / 2 + 1 = 249 + 1 = 250.

Total difference = (Number of odd rows * +7) + (Number of even rows * -7)

Total difference = (251 * 7) + (250 * -7) = 251 * 7 - 250 * 7 = (251 - 250) * 7 = 1 * 7 = 7.

So the difference between the sum of numbers on black cells and the sum of numbers on white cells is 7.

Let's verify this logic.

Consider a 2x2 board:

1 W, 2 B

3 B, 4 W

Sum(White) = 1 + 4 = 5

Sum(Black) = 2 + 3 = 5

Difference = 0.

Our row calculation: Row 1 (1,2): W, B. Diff = 2-1 = 1. Row 2 (3,4): B, W. Diff = 3-4 = -1. Total diff = 1 + (-1) = 0.

Consider a 2x3 board:

1 W, 2 B, 3 W

4 B, 5 W, 6 B

Sum(White) = 1 + 3 + 5 = 9

Sum(Black) = 2 + 4 + 6 = 12

Difference = 12 - 9 = 3.

Our row calculation: Row 1 (1,2,3): W, B, W. Diff = (2-1) + (2-3) = 1 - 1 = 0.

Row 2 (4,5,6): B, W, B. Diff = (4-5) + (4-6) = -1 - 2 = -3. This is not matching. The numbers are incrementing by 1 per cell.

Let's restart the row logic.

For a row of length N, let the numbers be k, k+1, ..., k+N-1.

If N is even (like 14):

Row starts with W: W, B, W, B, ..., W, B

Sum(W) = k + (k+2) + ... + (k+N-2)

Sum(B) = (k+1) + (k+3) + ... + (k+N-1)

Sum(B) - Sum(W) = [(k+1)-k] + [(k+3)-(k+2)] + ... + [(k+N-1)-(k+N-2)] = 1 + 1 + ... + 1 (N/2 times) = N/2.

In our case, N=14. So, Sum(B) - Sum(W) for a row starting with White = 14/2 = 7.

Row starts with B: B, W, B, W, ..., B, W

Sum(B) = k + (k+2) + ... + (k+N-2)

Sum(W) = (k+1) + (k+3) + ... + (k+N-1)

Sum(B) - Sum(W) = [k-(k+1)] + [(k+2)-(k+3)] + ... + [(k+N-2)-(k+N-1)] = -1 + -1 + ... + -1 (N/2 times) = -N/2.

In our case, N=14. So, Sum(B) - Sum(W) for a row starting with Black = -14/2 = -7.

This confirms our previous row calculation.

Now, we need to consider the starting numbers for each row.

Row 1 starts with 1. First cell (1,1) is White. So, the difference for Row 1 is +7.

Row 2 starts with 15. First cell (2,1) is Black (since 2+1=3 is odd). So, the difference for Row 2 is -7.

Row 3 starts with 29. First cell (3,1) is White (since 3+1=4 is even). So, the difference for Row 3 is +7.

The pattern of row differences is: +7, -7, +7, -7, ...

There are 501 rows.

Number of rows with difference +7 (odd-numbered rows: 1, 3, ..., 501): 251 rows.

Number of rows with difference -7 (even-numbered rows: 2, 4, ..., 500): 250 rows.

Total difference = (251 rows * +7) + (250 rows * -7) = 251 * 7 - 250 * 7 = (251 - 250) * 7 = 1 * 7 = 7.

The difference between the sum of numbers on black cells and the sum of numbers on white cells is 7.

Let's double check with the picture provided.

Row 1: 1 W, 2 B, 3 W, 4 B, 5 W

Sum(W) = 1+3+5 = 9

Sum(B) = 2+4 = 6

Diff (B-W) = 6-9 = -3. Our formula gave N/2 = 5/2 = 2.5. This formula is for even N.

For N=5, odd number of columns. The number of white and black cells is not equal. W,B,W,B,W. 3 White, 2 Black.

Let's assume the image is just an illustration and the problem refers to a 14x501 board.

The key is that the coloring is standard chess pattern and numbers are sequential.

Let's assume the top-left cell (1,1) is white.

Cell (i,j) is white if i+j is even, and black if i+j is odd.

Consider a block of 2x2 cells:

i, i+1

i+N, i+N+1

If i is White (i+j even), then i+1 is Black, i+N is Black (if N is even), i+N+1 is White.

Example: 1 W, 2 B

3 B, 4 W (N=2)

Sum(W) = 1+4=5. Sum(B) = 2+3=5. Diff = 0.

Example: 1 W, 2 B, 3 W

4 B, 5 W, 6 B (N=3)

Sum(W) = 1+3+5 = 9. Sum(B) = 2+4+6 = 12. Diff = 3.

Example: 1 W, 2 B, 3 W, 4 B

5 B, 6 W, 7 B, 8 W (N=4)

Sum(W) = 1+3+6+8 = 18. Sum(B) = 2+4+5+7 = 18. Diff = 0.

This shows that for N=even, the difference is 0 for a block of 2xN.

Here we have 14x501.

Since N=14 is even, we can consider the board as 501 blocks of 2x14 (stacked vertically). However, the number of rows is 501, which is odd.

Let's think about pairs of cells (white, black) that differ by 1.

Sum(Black) - Sum(White)

Each pair of horizontal adjacent cells (cell value X and X+1) will contribute either +1 or -1 to the total difference, depending on their colors.

Let's consider cells (i, j) and (i, j+1).

If (i,j) is white and (i, j+1) is black, then (j+1) - j = 1. Contribution is +1.

If (i,j) is black and (i, j+1) is white, then (j+1) - j = 1. Contribution is -1.

In each row of 14 columns, there are 13 such pairs.

Row 1 (W, B, W, B, ..., W, B): Pairs are (W,B), (B,W), (W,B), ... (W,B). There are 7 (W,B) pairs and 6 (B,W) pairs. Sum of contributions = 7*(+1) + 6*(-1) = 7-6 = 1.

This contradicts our previous calculation of 7 for the row difference. My understanding of row difference is likely flawed.

Let's re-evaluate row calculation.

Row 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

Colors: W, B, W, B, W, B, W, B, W, B, W, B, W, B

Sum(W) = 1+3+5+7+9+11+13 = 49

Sum(B) = 2+4+6+8+10+12+14 = 56

Sum(B) - Sum(W) = 56 - 49 = 7.

Row 2: 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28

Colors: B, W, B, W, B, W, B, W, B, W, B, W, B, W

Sum(B) = 15+17+19+21+23+25+27 = 147

Sum(W) = 16+18+20+22+24+26+28 = 154

Sum(B) - Sum(W) = 147 - 154 = -7.

Row 3: 29, 30, 31, ..., 42

Colors: W, B, W, B, ..., W, B

Sum(W) = 29+31+...+41 = 245

Sum(B) = 30+32+...+42 = 252

Sum(B) - Sum(W) = 252 - 245 = 7.

The pattern of row differences is indeed +7, -7, +7, -7, ...

There are 501 rows.

Odd-numbered rows (1, 3, ..., 501) have a difference of +7.

Even-numbered rows (2, 4, ..., 500) have a difference of -7.

Number of odd rows = 251.

Number of even rows = 250.

Total Difference = (251 * 7) + (250 * -7) = 1757 - 1750 = 7.

The result is 7.

The image provided is just an illustration and might not represent the full 14x501 board or the exact coloring/numbering scheme for the first few cells, but the problem states the board is colored in a chessboard pattern and numbers are arranged sequentially.

The core logic relies on the fact that in rows with an even number of columns (14), the difference between the sum of black cells and white cells alternates between +7 and -7. With an odd number of rows (501), there's one more row contributing +7 than -7, leading to a net difference of +7.

Final check of logic:

The difference between any two horizontally adjacent cells is 1. If the pair is (W, B), the contribution to Sum(B) - Sum(W) is +1. If the pair is (B, W), the contribution is -1.

In a row of 14 columns, we have 7 pairs of (W, B) and 6 pairs of (B, W) if it starts with W.

Example Row 1 (1-14): W B W B W B W B W B W B W B

Pairs: (1,2) W-B, (2,3) B-W, (3,4) W-B, (4,5) B-W, ... (13,14) W-B.

Let's be precise about which number is assigned to which color.

Cell (i,j) value: V(i,j) = (i-1)*14 + j.

Cell (i,j) color: White if i+j is even, Black if i+j is odd.

We want to calculate $$\sum_{i,j \text{ black}} V(i,j) - \sum_{i,j \text{ white}} V(i,j)$$.

Consider the difference between adjacent cells: V(i,j+1) - V(i,j) = 1.

If (i,j) is White and (i,j+1) is Black: V(i,j+1) - V(i,j) = 1. Contribution to (Sum B - Sum W) = +1.

If (i,j) is Black and (i,j+1) is White: V(i,j+1) - V(i,j) = 1. Contribution to (Sum B - Sum W) = -1.

In a row of 14 columns:

If row starts with White (i is odd): W B W B W B W B W B W B W B. Pairs are (W,B), (B,W), (W,B), ... (B,W), (W,B). There are 7 (W,B) pairs and 6 (B,W) pairs. Total contribution from horizontal pairs in this row = 7*(+1) + 6*(-1) = 1.

If row starts with Black (i is even): B W B W B W B W B W B W B W. Pairs are (B,W), (W,B), (B,W), ... (W,B), (B,W). There are 7 (B,W) pairs and 6 (W,B) pairs. Total contribution from horizontal pairs in this row = 7*(-1) + 6*(+1) = -1.

Now consider vertical pairs: V(i+1,j) - V(i,j) = 14.

If (i,j) is White and (i+1,j) is Black: V(i+1,j) - V(i,j) = 14. Contribution = +14.

If (i,j) is Black and (i+1,j) is White: V(i+1,j) - V(i,j) = 14. Contribution = -14.

This approach becomes complicated with the total sum.

Let's stick to the row-wise sum of differences.

Difference for Row i: D_i = Sum(Black cells in row i) - Sum(White cells in row i).

If row i starts with White (i is odd): D_i = +7.

If row i starts with Black (i is even): D_i = -7.

Total difference = $$\sum_{i=1}^{501} D_i$$.

Number of odd rows (i=1, 3, ..., 501) = 251.

Number of even rows (i=2, 4, ..., 500) = 250.

Total difference = (251 * 7) + (250 * -7) = 1757 - 1750 = 7.

The logic seems solid for a board with sequential numbering and standard chessboard coloring.

The provided image shows numbers starting from 1, 15, 29. This pattern matches the logic for the first few cells if the board were smaller.

Row 1 starts with 1 (W). Row 2 starts with 15 (B). Row 3 starts with 29 (W).

This confirms the coloring and the starting numbers are consistent with our assumption.

The difference is the sum of the differences for each row.

Ответ: 7.