By what least number 3600 must be multiplied to make it a perfect cube?

Options

A. 60
B. 50
C. 300
D. 450

Answer

To determine the least number by which 3600 must be multiplied to make it a perfect cube, we will follow these steps:

### Step 1: Prime Factorization of 3600
First, we need to find the prime factorization of 3600.

1. Divide 3600 by 2:
- \(3600 \div 2 = 1800\)
2. Divide 1800 by 2:
- \(1800 \div 2 = 900\)
3. Divide 900 by 2:
- \(900 \div 2 = 450\)
4. Divide 450 by 2:
- \(450 \div 2 = 225\)
5. Divide 225 by 3:
- \(225 \div 3 = 75\)
6. Divide 75 by 3:
- \(75 \div 3 = 25\)
7. Divide 25 by 5:
- \(25 \div 5 = 5\)
8. Divide 5 by 5:
- \(5 \div 5 = 1\)

Thus, the prime factorization of 3600 is:
\[ 3600 = 2^4 \times 3^2 \times 5^2 \]

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

- For \(2^4\), the exponent is 4. The nearest multiple of 3 is 3, so we need \(3 - 4 = 2\) more factors of 2.
- For \(3^2\), the exponent is 2. The nearest multiple of 3 is 3, so we need \(3 - 2 = 1\) more factor of 3.
- For \(5^2\), the exponent is 2. The nearest multiple of 3 is 3, so we need \(3 - 2 = 1\) more factor of 5.

### Step 3: Calculate the Least Number to Multiply
Now, we need to multiply 3600 by the factors we found:

- We need \(2^2\) (which is 4),
- We need \(3^1\) (which is 3),
- We need \(5^1\) (which is 5).

Thus, the least number we need to multiply by is:
\[ 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 \]

Calculating this:
1. \(4 \times 3 = 12\)
2. \(12 \times 5 = 60\)

### Final Answer
Therefore, the least number by which 3600 must be multiplied to make it a perfect cube is **60**.

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