Let a, b, c, d, e be natural numbers in an arithmetic progression such that a + b + c + d + e is the cube of an integer and b + c + d is square of an integer. The least possible value of the number of digits of c is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem step by step, we will denote the five natural numbers in an arithmetic progression as follows:

Let:
- \( a = c - 2d \)
- \( b = c - d \)
- \( c = c \)
- \( d = c + d \)
- \( e = c + 2d \)

### Step 1: Sum of the five numbers
The sum of these five numbers is:
\[ a + b + c + d + e = (c - 2d) + (c - d) + c + (c + d) + (c + 2d) \]
Simplifying this, we get:
\[ = 5c \]
According to the problem, this sum is the cube of an integer, so we can write:
\[ 5c = \alpha^3 \quad \text{(1)} \]
where \( \alpha \) is some integer.

### Step 2: Sum of the middle three numbers
Now, we calculate the sum of the middle three numbers:
\[ b + c + d = (c - d) + c + (c + d) = 3c \]
According to the problem, this sum is the square of an integer, so we can write:
\[ 3c = \beta^2 \quad \text{(2)} \]
where \( \beta \) is some integer.

### Step 3: Equating the two equations
From equations (1) and (2), we can express \( c \) in terms of \( \alpha \) and \( \beta \):
From (1):
\[ c = \frac{\alpha^3}{5} \]
From (2):
\[ c = \frac{\beta^2}{3} \]
Setting these equal gives:
\[ \frac{\alpha^3}{5} = \frac{\beta^2}{3} \]
Cross-multiplying yields:
\[ 3\alpha^3 = 5\beta^2 \quad \text{(3)} \]

### Step 4: Finding integer solutions
To find integer solutions for \( \alpha \) and \( \beta \), we can express \( \beta^2 \) in terms of \( \alpha \):
\[ \beta^2 = \frac{3\alpha^3}{5} \]
For \( \beta^2 \) to be an integer, \( 3\alpha^3 \) must be divisible by 5. This implies \( \alpha \) must be a multiple of 5. Let \( \alpha = 5k \) for some integer \( k \).

Substituting this back into equation (3):
\[ 3(5k)^3 = 5\beta^2 \]
This simplifies to:
\[ 375k^3 = 5\beta^2 \]
Dividing both sides by 5 gives:
\[ 75k^3 = \beta^2 \]

### Step 5: Finding the least value of \( c \)
To find the least possible value of \( c \), we need to find the smallest integer \( k \) such that \( 75k^3 \) is a perfect square. The prime factorization of 75 is \( 3 \times 5^2 \). For \( 75k^3 \) to be a perfect square, \( k \) must contain at least one factor of 3. Let \( k = 3m \).

Substituting \( k \):
\[ 75(3m)^3 = 75 \cdot 27m^3 = 2025m^3 \]
To find \( c \):
\[ c = \frac{\alpha^3}{5} = \frac{(5k)^3}{5} = 25k^3 = 25(3m)^3 = 675m^3 \]
The smallest value occurs when \( m = 1 \):
\[ c = 675 \]

### Step 6: Number of digits in \( c \)
The number of digits in \( 675 \) is \( 3 \).

### Final Answer
The least possible value of the number of digits of \( c \) is:
\[ \boxed{3} \]