The sum of the digits of the smallest number which, when multiplied by 1800, gives a perfect cube, is :

Options

A. a) 2
B. b) 3
C. c) 6
D. d) 8

Answer

To solve the problem of finding the sum of the digits of the smallest number which, when multiplied by 1800, gives a perfect cube, we can follow these steps:

### Step 1: Prime Factorization of 1800
First, we need to find the prime factorization of 1800.

- 1800 can be divided by 2:
- 1800 ÷ 2 = 900
- 900 ÷ 2 = 450
- 450 ÷ 2 = 225
- Now, we divide 225 by 3:
- 225 ÷ 3 = 75
- 75 ÷ 3 = 25
- Finally, we divide 25 by 5:
- 25 ÷ 5 = 5
- 5 ÷ 5 = 1

Thus, the prime factorization of 1800 is:
\[ 1800 = 2^3 \times 3^2 \times 5^2 \]

### Step 2: Determine the Exponents for a Perfect Cube
For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3.

- In \( 1800 = 2^3 \times 3^2 \times 5^2 \):
- The exponent of 2 is already 3 (which is a multiple of 3).
- The exponent of 3 is 2 (we need one more 3 to make it \( 3^3 \)).
- The exponent of 5 is 2 (we need one more 5 to make it \( 5^3 \)).

### Step 3: Find the Smallest Number to Multiply
To make the product a perfect cube, we need to multiply by \( 3^1 \) and \( 5^1 \):
\[ \text{Smallest number} = 3^1 \times 5^1 = 3 \times 5 = 15 \]

### Step 4: Calculate the Sum of the Digits
Now, we need to find the sum of the digits of the smallest number, which is 15:
- The digits of 15 are 1 and 5.
- Sum of the digits = \( 1 + 5 = 6 \)

### Final Answer
The sum of the digits of the smallest number which, when multiplied by 1800, gives a perfect cube is **6**.

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