Fourth of July- Holiday Multiplication - Coloring Squared - Free Printable

Let’s solve this step by step.

We are given a multiplication grid called “Fourth of July Practice Multiplication.” The goal is to fill in the blank squares (which are white) with the correct product of the row number and column number. Then, we color each square based on the key:

- Red: products 1–10
- Yellow: products 31–50
- Light Blue: products 11–20
- Dark Blue: products 21–30
- White squares stay white (but actually, according to the note, *blank squares are white* — meaning they’re meant to be filled in, then colored based on the product)

Wait — let’s read carefully:
“*Blank squares are white” — that probably means the empty cells in the grid are currently white, and after you calculate the product, you color them according to the key. So our job is to compute each missing product and assign its color.

But looking at the grid — it’s already mostly filled! Let me check if there are any blanks.

Actually, scanning the grid… every cell seems to have a number. For example:

Top-left corner: row 10 × col 7 = 70 → written as 70
Row 10, col 8 = 80 → written
Row 10, col 9 = 90 → written
... all the way to row 10, col 64? Wait no — columns go from 7 to 64? That doesn’t make sense for a multiplication table.

Wait — look again.

The rows are labeled on the left: 10, 20, 30, ..., up to 70? No — let's list the row headers:

Leftmost column (row labels):
10
20
30
40
50
60
70

And top row (column headers):
7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64

That’s 58 columns? And 7 rows? That seems odd.

Wait — perhaps the row labels are not multipliers but something else? Let’s test one cell.

Take first row (labeled 10), first column (labeled 7). Cell value is 70 → 10 × 7 = 70 ✔️

Second row (labeled 20), first column (7): 140 → 20 × 7 = 140 ✔️

Third row (30), first column (7): 210 → 30 × 7 = 210 ✔️

So yes — this is a multiplication table where:

- Row header = first factor
- Column header = second factor
- Cell = product

Now, the instruction says: “Practice Multiplication” and gives a coloring key based on the product value.

Also: “*Blank squares are white” — but in the image provided, all squares appear to have numbers. Maybe in the original worksheet, some were blank, but here they’re all filled? Or perhaps we are to verify or use this for coloring?

Looking back at user request: “Solve the problem accurately.”

But what is the actual problem? The image shows a completed multiplication grid with a coloring key. There’s no explicit question like “color the grid” or “find how many red squares”.

Wait — perhaps the task is implied: since it’s a “Practice Multiplication” sheet with a coloring key, maybe the student is supposed to color the grid based on the products. But since we can’t color here, maybe we need to count how many squares fall into each color category?

That would make sense as a solvable problem.

Let me assume that’s the task: Count how many cells in the grid have products in each range:

- Red: 1–10
- Yellow: 31–50
- Light Blue: 11–20
- Dark Blue: 21–30
- (Note: 51+ are not assigned a color? But the key only goes up to 50. Also, 0? Not present.)

Looking at the grid dimensions:

Rows: 7 rows (headers: 10,20,30,40,50,60,70)

Columns: Let’s count the column headers.

From the top row: starts at 7, ends at 64.

List: 7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64

That’s 58 columns.

Total cells = 7 rows × 58 columns = 406 cells.

Now, we need to find for each cell, the product = row_header × col_header, and see which range it falls into.

But note: row headers are multiples of 10: 10,20,30,40,50,60,70

Column headers are integers from 7 to 64 inclusive.

So minimum product: 10 × 7 = 70

Maximum product: 70 × 64 = 4480

All products are at least 70.

Now look at the color key:

- Red: 1–10 → but our smallest product is 70 → so NO red squares
- Light Blue: 11–20 → still less than 70 → none
- Dark Blue: 21–30 → none
- Yellow: 31–50 → none

All products are ≥70, so none fall into any of the colored categories?

That can’t be right. Perhaps I misread the row and column headers.

Wait — let me double-check the grid.

In the image, the first row (after header) has values starting with 70, 80, 90, 100, etc.

But 10×7=70, 10×8=80, ..., 10×10=100, 10×11=110, etc.

All these are above 50.

Similarly, even the smallest possible product in the grid is 10×7=70.

But the coloring key only goes up to 50 for yellow, and nothing above.

This suggests that either:

1. The coloring key is misprinted, or
2. I misinterpreted the grid.

Another possibility: perhaps the row and column headers are not the factors, but something else.

Look at the title: “Fourth of July Practice Multiplication”

And the key has colors for small numbers, but the grid has large numbers.

Perhaps the "practice" is to multiply, and then the coloring is based on the digits or something else? But the key says "1-10 Red", etc., implying the product value.

Maybe the grid is not row×col, but something else.

Let me check a different cell.

Take row labeled 70, column labeled 7: cell value is 490 → 70×7=490 ✔️

Row 70, col 8: 560 → 70×8=560 ✔️

So it is definitely multiplication.

But then all products are between 70 and 4480, so none are in 1-50.

Unless... perhaps the coloring is based on the sum of digits or the last digit or something? But the key doesn't say that.

Another idea: perhaps the "key" is for a different part, or maybe we are to consider the quotient or remainder.

Or perhaps the blank squares are to be filled, and in this version, they are already filled, but originally some were blank, and we need to identify which ones would be which color if calculated.

But still, all products are >50.

Let me calculate the smallest product: min row = 10, min col = 7, product = 70

Largest in the "colored" range is 50, so indeed, no cell qualifies for any color.

But that seems absurd for a worksheet.

Perhaps the row and column headers are not the multipliers, but the indices, and the cell is i*j, but the headers are just labels.

For example, maybe the first row corresponds to multiplier 1, but it's labeled 10? That doesn't make sense.

Another thought: look at the very first cell: row "10", col "7", value "70" — which is 10*7.

But 10 and 7 are both greater than 1, so product is at least 70.

Unless the coloring key is for the factors, not the product.

Let me read the key again: "Key: 1-10 Red, 11-20 Light Blue, 21-30 Dark Blue, 31-50 Yellow, *Blank squares are white"

It doesn't specify what the numbers refer to. In context, it should be the product, but perhaps it's the row or column number.

But the row numbers are 10,20,30,40,50,60,70 — so row 10 would be red (since 10 is in 1-10), row 20 would be light blue? 20 is in 11-20? 20 is included in 11-20? The key says 11-20 Light Blue, so 20 is light blue.

But the key is probably for the product, as it's a multiplication practice.

Perhaps "blank squares" means squares that are empty, and in this image, all are filled, so no blank squares, and we need to color based on product, but since all products >50, they are uncolored or something.

But the key doesn't have a category for >50.

This is confusing.

Let me try to see if there are any products <=50 in the grid.

Is there any cell with value <=50? Looking at the grid, the smallest number visible is 70 in the top-left. All others are larger. For example, row 10, col 7: 70; row 10, col 8: 80; ... row 20, col 7: 140; etc. Even the smallest is 70.

So no cell has product <=50.

Therefore, according to the key, no cell should be colored red, light blue, dark blue, or yellow. All would be white or uncolored.

But the note says "*Blank squares are white", implying that non-blank squares are to be colored, but there are no categories for >50.

Perhaps the key is incomplete, or perhaps for this grid, we only care about certain cells.

Another idea: perhaps the "Fourth of July" theme means we only color cells where the product is related to 4th of July, like 76, 1776, etc., but that seems unlikely.

Or perhaps the coloring is based on the digit sum or something.

Let's calculate the digit sum for a few cells.

For example, 70: 7+0=7 -> red (1-10)
80: 8+0=8 -> red
90: 9+0=9 -> red
100: 1+0+0=1 -> red
110: 1+1+0=2 -> red
...
140: 1+4+0=5 -> red
150: 1+5+0=6 -> red
...
200: 2+0+0=2 -> red
210: 2+1+0=3 -> red
...
70*7=490: 4+9+0=13 -> light blue (11-20)
70*8=560: 5+6+0=11 -> light blue
70*9=630: 6+3+0=9 -> red
etc.

Oh! This might be it. Perhaps the coloring is based on the sum of the digits of the product, not the product itself.

Let me verify with the key.

If digit sum is 1-10: red
11-20: light blue
21-30: dark blue
31-50: yellow

And blank squares are white, but in this case, all are filled.

This makes sense because otherwise, no cells would be colored.

Moreover, in educational worksheets, sometimes they use digit sums for coloring activities.

Let me confirm with a cell.

Take row 10, col 7: product 70, digit sum 7+0=7 -> red

Row 10, col 8: 80, 8+0=8 -> red

Row 10, col 9: 90, 9+0=9 -> red

Row 10, col 10: 100, 1+0+0=1 -> red

Row 10, col 11: 110, 1+1+0=2 -> red

...

Row 10, col 19: 190, 1+9+0=10 -> red

Row 10, col 20: 200, 2+0+0=2 -> red

Row 10, col 21: 210, 2+1+0=3 -> red

...

When does digit sum exceed 10?

For example, row 10, col 29: 290, 2+9+0=11 -> light blue

Row 10, col 38: 380, 3+8+0=11 -> light blue

Row 10, col 39: 390, 3+9+0=12 -> light blue

Row 10, col 47: 470, 4+7+0=11 -> light blue

Row 10, col 48: 480, 4+8+0=12 -> light blue

Row 10, col 49: 490, 4+9+0=13 -> light blue

Row 10, col 56: 560, 5+6+0=11 -> light blue

Row 10, col 57: 570, 5+7+0=12 -> light blue

Row 10, col 58: 580, 5+8+0=13 -> light blue

Row 10, col 59: 590, 5+9+0=14 -> light blue

Row 10, col 64: 640, 6+4+0=10 -> red

Now for row 20:

20*7=140, 1+4+0=5 -> red

20*8=160, 1+6+0=7 -> red

...

20*19=380, 3+8+0=11 -> light blue

20*29=580, 5+8+0=13 -> light blue

20*38=760, 7+6+0=13 -> light blue

20*47=940, 9+4+0=13 -> light blue

20*49=980, 9+8+0=17 -> light blue

20*56=1120, 1+1+2+0=4 -> red

Digit sum can be small even for large numbers.

For example, 1000: 1+0+0+0=1 -> red

So it's possible.

Now, is there any cell with digit sum >=21?

For example, row 70, col 64: 70*64=4480, digit sum 4+4+8+0=16 -> light blue

Row 70, col 59: 70*59=4130, 4+1+3+0=8 -> red

Row 70, col 63: 70*63=4410, 4+4+1+0=9 -> red

Row 70, col 62: 4340, 4+3+4+0=11 -> light blue

What about row 50, col 59: 50*59=2950, 2+9+5+0=16 -> light blue

Row 60, col 59: 60*59=3540, 3+5+4+0=12 -> light blue

Row 70, col 58: 70*58=4060, 4+0+6+0=10 -> red

Can we get digit sum 21 or more?

For example, 999: 9+9+9=27, but our max product is 70*64=4480, digit sum 4+4+8+0=16

Max possible digit sum for numbers up to 4480: for example, 3999: 3+9+9+9=30, but do we have such a product?

Let's see if any product has high digit sum.

For instance, row 70, col 49: 70*49=3430, 3+4+3+0=10 -> red

Row 70, col 59: 4130, 4+1+3+0=8 -> red

Row 60, col 49: 60*49=2940, 2+9+4+0=15 -> light blue

Row 50, col 49: 2450, 2+4+5+0=11 -> light blue

Row 40, col 49: 1960, 1+9+6+0=16 -> light blue

Row 30, col 49: 1470, 1+4+7+0=12 -> light blue

Row 20, col 49: 980, 9+8+0=17 -> light blue

Row 10, col 49: 490, 4+9+0=13 -> light blue

Now, try row 70, col 39: 70*39=2730, 2+7+3+0=12 -> light blue

Row 70, col 29: 2030, 2+0+3+0=5 -> red

Perhaps row 60, col 39: 60*39=2340, 2+3+4+0=9 -> red

To get high digit sum, need many 9's.

For example, 1998: 1+9+9+8=27, but is 1998 in the grid? 1998 / 10 = 199.8, not integer, so not in row 10.

1998 / 20 = 99.9, not integer.

1998 / 30 = 66.6, not.

1998 / 40 = 49.95, not.

1998 / 50 = 39.96, not.

1998 / 60 = 33.3, not.

1998 / 70 ≈ 28.54, not.

So not in grid.

Another number: 2997: 2+9+9+7=27

2997 / 10 = 299.7, not

/20=149.85, not

/30=99.9, not

/40=74.925, not

/50=59.94, not

/60=49.95, not

/70≈42.814, not

Not in grid.

3996: 3+9+9+6=27

/10=399.6, not

/20=199.8, not

/30=133.2, not

/40=99.9, not

/50=79.92, not

/60=66.6, not

/70≈57.085, not

Not in grid.

4480 is max, digit sum 16.

What about 3899? 3+8+9+9=29, but 3899 / 70 ≈ 55.7, not integer.

Perhaps 2999: 2+9+9+9=29, /70≈42.84, not.

Maybe smaller numbers with high digit sum.

For example, 99: 9+9=18, but 99 is not in grid since min product 70, and 99/10=9.9, not integer column.

Columns start at 7, so for row 10, col 9.9 not integer.

Col must be integer, so product must be multiple of row header.

Since row headers are 10,20,30,40,50,60,70, all products are multiples of 10.

Oh! Important point: all row headers are multiples of 10, so all products are multiples of 10.

Therefore, every product ends with 0.

So the last digit is always 0.

Therefore, the digit sum is the sum of the other digits.

For example, 70: 7+0=7

100:1+0+0=1

140:1+4+0=5

190:1+9+0=10

200:2+0+0=2

290:2+9+0=11

380:3+8+0=11

470:4+7+0=11

560:5+6+0=11

640:6+4+0=10

700:7+0+0=7

770:7+7+0=14

840:8+4+0=12

910:9+1+0=10

980:9+8+0=17

1050:1+0+5+0=6

etc.

Since the number ends with 0, digit sum is sum of digits except the last 0.

Now, what is the maximum possible digit sum for a multiple of 10 between 70 and 4480.

For example, 3990: 3+9+9+0=21 -> dark blue (21-30)

Is 3990 in the grid? 3990 / 10 = 399, but columns only up to 64, so no.

3990 / 20 = 199.5, not integer.

3990 / 30 = 133, but 133 > 64, not in columns.

Similarly, 2990: 2+9+9+0=20 -> light blue (since 20 is in 11-20? Key says 11-20 Light Blue, so 20 is included.

2990 / 10 = 299 >64, not in grid.

2990 / 20 = 149.5, not.

2990 / 30 ≈99.666, not.

2990 / 40 = 74.75, not.

2990 / 50 = 59.8, not.

2990 / 60 ≈49.833, not.

2990 / 70 ≈42.714, not.

Not in grid.

What about 1990: 1+9+9+0=19 -> light blue

1990 / 10 = 199 >64, no

/20=99.5, no

/30≈66.333, no

/40=49.75, no

/50=39.8, no

/60≈33.166, no

/70≈28.428, no

Not in grid.

990: 9+9+0=18 -> light blue

990 / 10 = 99 >64, no

/20=49.5, no

/30=33, and 33 is within 7 to 64? Yes, col 33 is there.

Row 30, col 33: 30*33=990, digit sum 9+9+0=18 -> light blue

Yes! So 990 is in the grid.

Similarly, 890: 8+9+0=17 -> light blue

890 / 10 = 89 >64, no

/20=44.5, no

/30≈29.666, no

/40=22.25, no

/50=17.8, no

/60≈14.833, no

/70≈12.714, no

Not in grid.

790: 7+9+0=16 -> light blue

790 / 10 = 79 >64, no

/20=39.5, no

/30≈26.333, no

/40=19.75, no

/50=15.8, no

/60≈13.166, no

/70≈11.285, no

Not.

690: 6+9+0=15 -> light blue

690 / 10 = 69 >64, no (col max 64)

/20=34.5, no

/30=23, and 23 is in columns? Columns from 7 to 64, yes, col 23.

Row 30, col 23: 30*23=690, digit sum 6+9+0=15 -> light blue

Yes.

Similarly, 590: 5+9+0=14 -> light blue

590 / 10 = 59, col 59 is there.

Row 10, col 59: 10*59=590, digit sum 5+9+0=14 -> light blue

Yes.

490: 4+9+0=13 -> light blue, row 10, col 49 or row 70, col 7, etc.

Now, can we get digit sum 21 or more?

For example, 3890: 3+8+9+0=20 -> light blue (since 20<=20)

3890 / 10 = 389 >64, no

/20=194.5, no

/30≈129.666, no

/40=97.25, no

/50=77.8, no

/60≈64.833, close but 64.833 not integer, and col 64 is max.

3890 / 60 = 64.833, not integer.

3880: 3+8+8+0=19 -> light blue

3880 / 60 ≈64.666, not integer.

3870: 3+8+7+0=18 -> light blue

3870 / 60 = 64.5, not.

3860: 3+8+6+0=17 -> light blue

3860 / 60 ≈64.333, not.

3850: 3+8+5+0=16 -> light blue

3850 / 60 ≈64.166, not.

3840: 3+8+4+0=15 -> light blue

3840 / 60 = 64, yes! Col 64.

Row 60, col 64: 60*64=3840, digit sum 3+8+4+0=15 -> light blue

Still 15.

What about 4790: 4+7+9+0=20 -> light blue

4790 / 10 = 479 >64, no

/20=239.5, no

/30≈159.666, no

/40=119.75, no

/50=95.8, no

/60≈79.833, no

/70≈68.428, no

Not in grid.

4890: 4+8+9+0=21 -> dark blue

4890 / 10 = 489 >64, no

/20=244.5, no

/30=163, >64, no

/40=122.25, no

/50=97.8, no

/60=81.5, no

/70≈69.857, no

Not in grid.

3990: 3+9+9+0=21 -> dark blue

As before, not in grid.

2990: 2+9+9+0=20 -> light blue, not in grid.

1990: 1+9+9+0=19 -> light blue, not in grid.

990: 9+9+0=18 -> light blue, in grid as row 30, col 33.

Now, is there any product with digit sum >=21?

Let me try row 70, col 57: 70*57=3990, digit sum 3+9+9+0=21 -> dark blue

Col 57 is within 7 to 64? Yes, 57 is there.

Row 70, col 57: 70*57.

Calculate: 70*50=3500, 70*7=490, total 3500+490=3990, yes.

Digit sum: 3+9+9+0=21 -> dark blue (21-30)

Perfect.

Similarly, row 70, col 58: 70*58=4060, 4+0+6+0=10 -> red

Row 70, col 59: 4130, 4+1+3+0=8 -> red

Row 70, col 60: 4200, 4+2+0+0=6 -> red

Row 70, col 61: 4270, 4+2+7+0=13 -> light blue

Row 70, col 62: 4340, 4+3+4+0=11 -> light blue

Row 70, col 63: 4410, 4+4+1+0=9 -> red

Row 70, col 64: 4480, 4+4+8+0=16 -> light blue

Now, row 60, col 64: 3840, 3+8+4+0=15 -> light blue

Row 50, col 64: 3200, 3+2+0+0=5 -> red

Etc.

So yes, there are cells with digit sum in all ranges.

For example, digit sum 21: row 70, col 57: 3990, sum 21

Is there higher? Row 70, col 56: 70*56=3920, 3+9+2+0=14 -> light blue

Row 70, col 55: 3850, 3+8+5+0=16 -> light blue

Row 70, col 54: 3780, 3+7+8+0=18 -> light blue

Row 70, col 53: 3710, 3+7+1+0=11 -> light blue

Row 70, col 52: 3640, 3+6+4+0=13 -> light blue

Row 70, col 51: 3570, 3+5+7+0=15 -> light blue

Row 70, col 50: 3500, 3+5+0+0=8 -> red

So only col 57 gives 21 for row 70.

What about row 60, col 64: 3840, sum 15

Row 50, col 64: 3200, sum 5

Row 40, col 64: 2560, 2+5+6+0=13 -> light blue

Row 30, col 64: 1920, 1+9+2+0=12 -> light blue

Row 20, col 64: 1280, 1+2+8+0=11 -> light blue

Row 10, col 64: 640, 6+4+0=10 -> red

Now, another cell with high digit sum: row 70, col 48: 70*48=3360, 3+3+6+0=12 -> light blue

Row 70, col 47: 3290, 3+2+9+0=14 -> light blue

Row 70, col 46: 3220, 3+2+2+0=7 -> red

Etc.

Row 60, col 57: 60*57=3420, 3+4+2+0=9 -> red

Row 50, col 57: 2850, 2+8+5+0=15 -> light blue

Row 40, col 57: 2280, 2+2+8+0=12 -> light blue

Row 30, col 57: 1710, 1+7+1+0=9 -> red

Row 20, col 57: 1140, 1+1+4+0=6 -> red

Row 10, col 57: 570, 5+7+0=12 -> light blue

Now, is there a cell with digit sum 31 or more? Probably not, since max is 4480 sum 16, or 3990 sum 21, and 4890 not in grid.

What about 4990? 4+9+9+0=22, but 4990 / 70 ≈71.285, col 71>64, not in grid.

So likely, digit sums range from 1 to 21 or so.

Specifically, we have at least one with 21.

Now, for the problem, since the user said "solve the problem", and given that the grid is filled, perhaps the task is to count how many cells fall into each color category based on digit sum of the product.

That seems reasonable.

So let's define:

For each cell, product P = row_header * col_header

Then digit_sum = sum of digits of P

Then:

- If digit_sum between 1 and 10 inclusive: Red

- 11 to 20 inclusive: Light Blue

- 21 to 30 inclusive: Dark Blue

- 31 to 50 inclusive: Yellow

- Else: no color, or white, but since all are filled, and key has no >50, perhaps they are not colored, but in this case, max digit sum is 21, so no yellow.

Digit sum 21 is dark blue, 22+ would be dark blue or yellow, but we may not have any.

First, let's confirm if there are any with digit sum >21.

For example, row 70, col 57: 3990, sum 21

Row 70, col 67: but col 67>64, not in grid.

Row 60, col 64: 3840, sum 15

Row 50, col 64: 3200, sum 5

What about row 70, col 49: 3430, 3+4+3+0=10 -> red

Row 70, col 39: 2730, 2+7+3+0=12 -> light blue

Row 60, col 59: 3540, 3+5+4+0=12 -> light blue

Row 50, col 59: 2950, 2+9+5+0=16 -> light blue

Row 40, col 59: 2360, 2+3+6+0=11 -> light blue

Row 30, col 59: 1770, 1+7+7+0=15 -> light blue

Row 20, col 59: 1180, 1+1+8+0=10 -> red

Row 10, col 59: 590, 5+9+0=14 -> light blue

Now, row 70, col 58: 4060, 4+0+6+0=10 -> red

Row 70, col 56: 3920, 3+9+2+0=14 -> light blue

Row 70, col 55: 3850, 3+8+5+0=16 -> light blue

Row 70, col 54: 3780, 3+7+8+0=18 -> light blue

Row 70, col 53: 3710, 3+7+1+0=11 -> light blue

Row 70, col 52: 3640, 3+6+4+0=13 -> light blue

Row 70, col 51: 3570, 3+5+7+0=15 -> light blue

Row 70, col 50: 3500, 3+5+0+0=8 -> red

So for row 70, only col 57 has digit sum 21.

Is there another? Row 60, col 64: 3840, sum 15

Row 50, col 64: 3200, sum 5

What about row 40, col 64: 2560, 2+5+6+0=13

Row 30, col 64: 1920, 1+9+2+0=12

Row 20, col 64: 1280, 1+2+8+0=11

Row 10, col 64: 640, 6+4+0=10

All less than 21.

Row 70, col 47: 3290, 3+2+9+0=14

Row 70, col 46: 3220, 3+2+2+0=7

Row 70, col 45: 3150, 3+1+5+0=9

Row 70, col 44: 3080, 3+0+8+0=11

Row 70, col 43: 3010, 3+0+1+0=4

Row 70, col 42: 2940, 2+9+4+0=15

Row 70, col 41: 2870, 2+8+7+0=17

Row 70, col 40: 2800, 2+8+0+0=10

Row 70, col 39: 2730, 2+7+3+0=12

Etc.

So only one cell with digit sum 21: row 70, col 57.

Now, are there any with digit sum 22 or more? Seems not, as per earlier calculation.

For example, 4890 not in grid, 3990 is 21, 4990 not, etc.

So dark blue only for digit sum 21-30, and we have only one cell with 21, none with 22+.

Yellow is 31-50, no cells.

Now, for the count, we need to go through all 7*58=406 cells and categorize based on digit sum of product.

This is tedious, but perhaps we can write a small program or find a pattern.

Since this is a text-based response, I need to do it manually or find a smart way.

Notice that all products are multiples of 10, so last digit is 0, so digit sum is sum of digits of P/10.

Let Q = P / 10 = (row_header / 10) * col_header

Row headers are 10,20,30,40,50,60,70, so row_header / 10 = 1,2,3,4,5,6,7

Let R = row_header / 10, so R = 1,2,3,4,5,6,7

C = col_header, from 7 to 64 inclusive.

Then P = 10 * R * C

Digit sum of P is the same as digit sum of R*C, because multiplying by 10 adds a 0 at the end, which doesn't change the digit sum (since 0 adds nothing).

Is that true?

For example, R*C = 7*7=49, digit sum 4+9=13

P=490, digit sum 4+9+0=13, same.

R*C=10*7=70, but R=10? No, R is 1 to 7.

R=1, C=7, R*C=7, digit sum 7

P=70, digit sum 7+0=7, same.

R=1, C=10, R*C=10, digit sum 1+0=1

P=100, digit sum 1+0+0=1, same.

R=3, C=33, R*C=99, digit sum 9+9=18

P=990, digit sum 9+9+0=18, same.

R=7, C=57, R*C=399, digit sum 3+9+9=21

P=3990, digit sum 3+9+9+0=21, same.

Yes! Because P = 10 * (R*C), and since R*C is an integer, P is R*C followed by a 0, so the digit sum of P is exactly the digit sum of R*C.

Therefore, we can work with S = R * C, where R=1,2,3,4,5,6,7 and C=7 to 64, and find digit sum of S, then assign color based on that digit sum.

And S ranges from 1*7=7 to 7*64=448.

Digit sum of S, not of P, but since it's the same, we can use S.

So now, the problem reduces to: for each R in {1,2,3,4,5,6,7}, and each C in {7,8,9,...,64}, compute S = R * C, then digit_sum(S), then categorize.

And count how many for each color.

Since R is small, we can do it row by row.

Let me define the ranges:

- Red: digit_sum 1-10

- Light Blue: 11-20

- Dark Blue: 21-30

- Yellow: 31-50

But since max S=448, digit sum max is 4+4+8=16 for 448, or 399 is 3+9+9=21, as before.

448: 4+4+8=16

399:3+9+9=21

449 not possible, max 7*64=448.

7*64=448, digit sum 4+4+8=16

6*64=384, 3+8+4=15

5*64=320, 3+2+0=5

4*64=256, 2+5+6=13

3*64=192, 1+9+2=12

2*64=128, 1+2+8=11

1*64=64, 6+4=10

So max digit sum is 21 for S=399 (R=7,C=57)

Min is for S=7, digit sum 7

Or S=10, digit sum 1, etc.

Now, let's go row by row for R=1 to 7.

First, R=1:

S = 1 * C = C, for C=7 to 64

So S ranges from 7 to 64.

Digit sum of S:

We need to find for each C from 7 to 64, digit sum of C, then categorize.

C from 7 to 64.

List the digit sums:

C=7:7 -> red

8:8->red

9:9->red

10:1+0=1->red

11:1+1=2->red

12:1+2=3->red

13:1+3=4->red

14:1+4=5->red

15:1+5=6->red

16:1+6=7->red

17:1+7=8->red

18:1+8=9->red

19:1+9=10->red

20:2+0=2->red

21:2+1=3->red

22:2+2=4->red

23:2+3=5->red

24:2+4=6->red

25:2+5=7->red

26:2+6=8->red

27:2+7=9->red

28:2+8=10->red

29:2+9=11->light blue

30:3+0=3->red

31:3+1=4->red

32:3+2=5->red

33:3+3=6->red

34:3+4=7->red

35:3+5=8->red

36:3+6=9->red

37:3+7=10->red

38:3+8=11->light blue

39:3+9=12->light blue

40:4+0=4->red

41:4+1=5->red

42:4+2=6->red

43:4+3=7->red

44:4+4=8->red

45:4+5=9->red

46:4+6=10->red

47:4+7=11->light blue

48:4+8=12->light blue

49:4+9=13->light blue

50:5+0=5->red

51:5+1=6->red

52:5+2=7->red

53:5+3=8->red

54:5+4=9->red

55:5+5=10->red

56:5+6=11->light blue

57:5+7=12->light blue

58:5+8=13->light blue

59:5+9=14->light blue

60:6+0=6->red

61:6+1=7->red

62:6+2=8->red

63:6+3=9->red

64:6+4=10->red

Now, count for R=1:

Red: when digit sum <=10

From above, most are red, except when digit sum >=11.

From list:

Light blue when: C=29(11),38(11),39(12),47(11),48(12),49(13),56(11),57(12),58(13),59(14)

That's 10 values.

C=29,38,39,47,48,49,56,57,58,59 — yes 10.

Total C from 7 to 64 inclusive: number of values = 64-7+1=58

So red = 58 - 10 = 48? But let's verify.

C=7 to 28: all red? C=7 to 19: all red as per above.

C=20 to 28: all red (digit sum 2 to 10)

C=29: light blue

C=30 to 37: all red (3 to 10)

C=38,39: light blue

C=40 to 46: all red (4 to 10)

C=47,48,49: light blue

C=50 to 55: all red (5 to 10)

C=56,57,58,59: light blue

C=60 to 64: all red (6 to 10)

So light blue at: 29,38,39,47,48,49,56,57,58,59 — that's 10 values.

No dark blue or yellow for R=1, since max digit sum is 14<21.

So for R=1: 48 red, 10 light blue, 0 dark blue, 0 yellow.

But 48+10=58, good.

Now R=2:

S=2*C, C=7 to 64

S from 14 to 128

Digit sum of S.

We need to compute for each C, S=2*C, then digit sum.

Since S is even, but anyway.

Let me make a table or find a way.

Note that S=2C, so as C increases, S increases by 2.

Digit sum changes irregularly.

We can list or find when digit sum >=11, etc.

Since it's systematic, perhaps calculate the digit sum for each S.

But to save time, I can group.

Note that for S from 14 to 128, digit sum can be calculated.

Let me list C, S=2C, digit sum.

C=7, S=14, 1+4=5 -> red

C=8, S=16,1+6=7->red

C=9, S=18,1+8=9->red

C=10, S=20,2+0=2->red

C=11, S=22,2