If (a-7)(b-10)(c-12) = 1000, the least possible value of (a+b+c) equals :

Options

A. 59
B. 29
C. 14
D. any integer less than 1000

Answer

To solve the problem, we need to find the least possible value of \( a + b + c \) given the equation:

\[ (a - 7)(b - 10)(c - 12) = 1000 \]

### Step 1: Define new variables
Let:
- \( x = a - 7 \)
- \( y = b - 10 \)
- \( z = c - 12 \)

Then the equation becomes:
\[ xyz = 1000 \]

### Step 2: Express \( a + b + c \) in terms of \( x, y, z \)
Now, we can express \( a + b + c \) as:
\[ a + b + c = (x + 7) + (y + 10) + (z + 12) = x + y + z + 29 \]

### Step 3: Minimize \( x + y + z \)
To minimize \( a + b + c \), we need to minimize \( x + y + z \). Since \( xyz = 1000 \), we can use the AM-GM inequality which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean.

Applying AM-GM:
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]
\[ \frac{x + y + z}{3} \geq \sqrt[3]{1000} \]
\[ \frac{x + y + z}{3} \geq 10 \]
\[ x + y + z \geq 30 \]

### Step 4: Calculate the minimum value of \( a + b + c \)
Now substituting back into the equation for \( a + b + c \):
\[ a + b + c = x + y + z + 29 \geq 30 + 29 = 59 \]

### Step 5: Check if the minimum can be achieved
To check if this minimum can be achieved, we need \( x, y, z \) to be equal (since AM-GM achieves equality when all terms are equal).

Let:
\[ x = y = z \]

Then:
\[ x^3 = 1000 \implies x = 10 \]

So:
\[ x = 10, \quad y = 10, \quad z = 10 \]

### Step 6: Calculate \( a, b, c \)
Now substituting back to find \( a, b, c \):
\[ a = x + 7 = 10 + 7 = 17 \]
\[ b = y + 10 = 10 + 10 = 20 \]
\[ c = z + 12 = 10 + 12 = 22 \]

### Step 7: Calculate \( a + b + c \)
Finally, we find:
\[ a + b + c = 17 + 20 + 22 = 59 \]

Thus, the least possible value of \( a + b + c \) is:

\[ \boxed{59} \]