nerdboy

1. Find the sums of the prime numbers in each interval

Primes from 1–10:

2, 3, 5, 7
Sum: (2 + 3 + 5 + 7 = 17)

Primes from 11–20:

11, 13, 17, 19
Sum: (11 + 13 + 17 + 19 = 60)

Primes from 21–30:

23, 29
Sum: (23 + 29 = 52)

Primes from 31–40:

31, 37
Sum: (31 + 37 = 68)

Primes from 41–50:

41, 43, 47
Sum: (41 + 43 + 47 = 131)

2. We want:

\( [ P = 17 \times 60 \times 52 \times 68 \times 131 \times 997 ] \), and we need the remainder when P is divided by 97.

3. Reduce each number modulo 97

17 is already < 97, so 17 mod 97 = 17

60 mod 97 = 60

52 mod 97 = 52

68 mod 97 = 68

131 mod 97 = 131 - 97 = 34

997 ÷ 97 = 10 remainder 27, so 997 mod 97 = 27

So:

\([ P \equiv 17 \times 60 \times 52 \times 68 \times 34 \times 27 \pmod{97} ]\).

4. Compute the product step by step mod 97. 

Start multiplying step by step:

Calculate each step modulo 97.

Step 1: 17 × 60 = 1020

1020 ÷ 97 = 10 × 97 = 970, remainder 50
So 1020 mod 97 = 50

Step 2: 50 × 52 = 2600

2600 ÷ 97 = 26 × 97 = 2522, remainder 78
So 2600 mod 97 = 78

Step 3: 78 × 68 = 5304

5304 ÷ 97 = 54 × 97 = 5238, remainder 66
So 5304 mod 97 = 66

Step 4: 66 × 34 = 2244

2244 ÷ 97 = 23 × 97 = 2231, remainder 13
So 2244 mod 97 = 13

Step 5: 13 × 27 = 351

351 ÷ 97 = 3 × 97 = 291, remainder 60
So 351 mod 97 = 60.

Final answer: \(\boxed{60}\)

1. Center bar:

Middle row, cols 2-4 = positions (3,2)-(3,3)-(3,4)

This can be rotated to vertical or reflected, but always stays at the center.

2. Off-center:

Top row, cols 1-3: (1,1)-(1,2)-(1,3)

Top row, cols 2-4: (1,2)-(1,3)-(1,4)

Top row, cols 3-5: (1,3)-(1,4)-(1,5)

But these are all equally distant from their accompanying edge; by symmetry, place (1,1)-(1,2)-(1,3) is same as (1,3)-(1,4)-(1,5) under reflection.

So for each "run of 3" not at the center, there are only two types by distance from the edge: edge and one-step-in.

So per orientation, horizontal and vertical, that gives:

Center

Edge

One in from edge

But rotations mix horizontals and verticals. So these three types exhaust the possibility for straight bars:

Case 1: Centered three-in-a-row (middle row & col)

Case 2: Edge three-in-a-row (either on top/left or bottom/right edge)

Case 3: Offset three-in-a-row (one away from edge, but not center)

Centered bar: ((3,2)-(3,3)-(3,4)) (also ((2,3)-(3,3)-(4,3)), rotated)

Edge bar: ((1,1)-(1,2)-(1,3)) (can be rotated/reflected to any edge)

Offset bar: ((1,2)-(1,3)-(1,4))

So 3 distinct orbits for straight bars.

B. Diagonal Bars:

Now, consider the diagonal three-in-a-row:

((1,1)-(2,2)-(3,3))

((2,2)-(3,3)-(4,4))

((3,3)-(4,4)-(5,5))

Similar situation on other main diagonal.

Only possibilities for "bar on a diagonal":

Main diagonal, centered

Central run: ((2,2)-(3,3)-(4,4))

Main diagonal, edge

Edge run: ((1,1)-(2,2)-(3,3))

((3,3)-(4,4)-(5,5))

The two runs at the ends of the diagonal are symmetric by 180° rotation.

Similarly, diagonals above/below the main diagonal, but only have length 3.

Also, shorter diagonals: length 3, only one way to pick three in a row.

But are these different up to symmetry? For a 5x5, all "sloping" bars at the edge are the same under symmetry.

Diagonal (main): Centered and edge.

And diagonal bars not on the true main diagonal or true anti-diagonal can be rotated to each other.

So, just as before:

Diagonal centered

Diagonal edge

So, in total, for "three-in-a-row" on a diagonal, we only get two types: centered and edge.

In conclusion, there are 5 distinct types of three-in-a-row colorings (orbits under symmetry):

Final Answer;

\(\boxed{5}\)

There are 5 distinct colorings of three squares in a row (of any direction) in a 5 by 5 grid, up to rotation and reflection.

Let's call this "spadesuit" a function called g, so, A g B=a^2/y

which would conclude that a can be \(\frac{9 - 3\sqrt{13}}{2}\), and \(\frac{9 + 3\sqrt{13}}{2}\).

She needs 450 so 450-(95+24)

                              450-(119)=331

331/5=66.2 but, you need round up another jar to make it a whole number 

In conclusion, the solution to this question is 67.

You cannot find a specific number due to variables being the sidelengths, but, it would be this if you can calculate the area of said triangle.

R = (a*b*c)/(4*(area of triangle))

The system of equations is:

p - 2q = 3 q - 2r = -2 p + r = 9

We can solve this system using Gaussian Elimination.

First, we can add the first and second equations to get:

p - 2q + q - 2r = 1

This simplifies to:

-q - 2r = 1

Now, we can subtract the third equation from this equation to get:

-q - 2r - (p + r) = 1 - 9

This simplifies to:

-2q = -8

Therefore, q = 4.

Now, we can substitute this value into the second equation to get:

4 - 2r = -2

This simplifies to:

2r = 6

Therefore, r = 3.

Finally, we can substitute these values into the first equation to get:

p - 2(4) = 3

This simplifies to:

p = 5

Therefore, the ordered triple that satisfies the system of equations is (5, 4, 3).