The sum of four prime numbers of two digits is 204. The sum of the first and last i.e, `(p_1 + p_4)` is same as the sum of second number `p_2` and third number `p_3` (i.e., `p_2 + p_3`). Again `p_3 - p_2 = 2 (p_2 - p_1) = 2 (p_4 - p_3)`. The average number of all the four numbers is not a prime number, but a product of two prime numbers one of these prime numbers (out of four) is :

Options

A. a. 23
B. b. 89
C. c. 71
D. d. can't be determined

Answer

To solve the problem step by step, we need to find four two-digit prime numbers \( p_1, p_2, p_3, p_4 \) that satisfy the following conditions:

1. The sum of the four prime numbers is 204:
\[ p_1 + p_2 + p_3 + p_4 = 204 \]

2. The sum of the first and last primes is equal to the sum of the second and third primes:
\[ p_1 + p_4 = p_2 + p_3 \]

3. The differences between the primes are given by:
\[ p_3 - p_2 = 2(p_2 - p_1) = 2(p_4 - p_3) \]

4. The average of the four numbers is not a prime number but a product of two prime numbers.

### Step 1: Set Up the Equations

From the first equation, we can express \( p_4 \) in terms of \( p_1, p_2, \) and \( p_3 \):
\[ p_4 = 204 - p_1 - p_2 - p_3 \]

### Step 2: Substitute into the Second Equation

Using the second equation:
\[ p_1 + (204 - p_1 - p_2 - p_3) = p_2 + p_3 \]
This simplifies to:
\[ 204 - p_2 - p_3 = p_2 + p_3 \]
\[ 204 = 2p_2 + 2p_3 \]
\[ 102 = p_2 + p_3 \]

### Step 3: Express \( p_3 \) in Terms of \( p_2 \)

From \( p_2 + p_3 = 102 \), we can express \( p_3 \):
\[ p_3 = 102 - p_2 \]

### Step 4: Substitute \( p_3 \) into the Difference Equations

Now, substituting \( p_3 \) into the difference equations:
1. \( p_3 - p_2 = 102 - 2p_2 \)
2. \( 2(p_2 - p_1) = 2(p_4 - p_3) \)

### Step 5: Solve for \( p_1 \) and \( p_4 \)

From the first difference equation:
\[ 102 - 2p_2 = 2(p_2 - p_1) \]
This gives:
\[ 102 - 2p_2 = 2p_2 - 2p_1 \]
Rearranging gives:
\[ 102 = 4p_2 - 2p_1 \]
\[ p_1 = 2p_2 - 51 \]

For \( p_4 \):
\[ p_4 = 204 - p_1 - p_2 - p_3 \]
Substituting \( p_3 \):
\[ p_4 = 204 - p_1 - p_2 - (102 - p_2) = 102 - p_1 \]

### Step 6: Finding Prime Numbers

Now we have expressions for \( p_1, p_2, p_3, \) and \( p_4 \):
- \( p_1 = 2p_2 - 51 \)
- \( p_3 = 102 - p_2 \)
- \( p_4 = 102 - p_1 \)

We need to find \( p_2 \) such that all \( p_1, p_2, p_3, p_4 \) are prime numbers.

### Step 7: Testing Values for \( p_2 \)

Let’s test values for \( p_2 \) (which must be a two-digit prime):
- If \( p_2 = 41 \):
- \( p_1 = 2(41) - 51 = 31 \)
- \( p_3 = 102 - 41 = 61 \)
- \( p_4 = 102 - 31 = 71 \)

Now we have the primes: \( 31, 41, 61, 71 \).

### Step 8: Check the Conditions

1. Sum: \( 31 + 41 + 61 + 71 = 204 \) (satisfied)
2. \( p_1 + p_4 = 31 + 71 = 102 \) and \( p_2 + p_3 = 41 + 61 = 102 \) (satisfied)
3. Differences:
- \( p_3 - p_2 = 61 - 41 = 20 \)
- \( 2(p_2 - p_1) = 2(41 - 31) = 20 \)
- \( 2(p_4 - p_3) = 2(71 - 61) = 20 \) (satisfied)

### Step 9: Average Calculation

The average of the four numbers:
\[ \text{Average} = \frac{204}{4} = 51 \]
51 is not a prime number but can be expressed as a product of two primes \( 3 \times 17 \).

### Conclusion

One of the prime numbers is \( 71 \).