Find the smallest number by which `2560` must be multiplied so that the product is a perfect cube.

Answer

To find the smallest number by which 2560 must be multiplied so that the product is a perfect cube, we will follow these steps:

### Step 1: Prime Factorization of 2560
We start by finding the prime factorization of 2560.

1. **Divide by 2** (since 2560 is even):
- 2560 ÷ 2 = 1280
2. **Divide by 2**:
- 1280 ÷ 2 = 640
3. **Divide by 2**:
- 640 ÷ 2 = 320
4. **Divide by 2**:
- 320 ÷ 2 = 160
5. **Divide by 2**:
- 160 ÷ 2 = 80
6. **Divide by 2**:
- 80 ÷ 2 = 40
7. **Divide by 2**:
- 40 ÷ 2 = 20
8. **Divide by 2**:
- 20 ÷ 2 = 10
9. **Divide by 2**:
- 10 ÷ 2 = 5
10. **Divide by 5**:
- 5 ÷ 5 = 1

Thus, the prime factorization of 2560 is:
\[ 2560 = 2^8 \times 5^1 \]

### Step 2: Analyze the Exponents
To form a perfect cube, all the exponents in the prime factorization must be multiples of 3.

- For \(2^8\): The exponent 8 is not a multiple of 3. The nearest multiple of 3 is 9. Thus, we need \(9 - 8 = 1\) additional factor of 2.
- For \(5^1\): The exponent 1 is not a multiple of 3. The nearest multiple of 3 is 3. Thus, we need \(3 - 1 = 2\) additional factors of 5.

### Step 3: Calculate the Smallest Number to Multiply
To make the product a perfect cube, we need to multiply by:
\[ 2^1 \times 5^2 \]

Calculating this gives:
\[ 2^1 = 2 \]
\[ 5^2 = 25 \]
Thus:
\[ 2 \times 25 = 50 \]

### Conclusion
The smallest number by which 2560 must be multiplied to make it a perfect cube is **50**.

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