What is the least number by which 6750 may be divided so that the quotient is a perfect cube?

Answer

To find the least number by which 6750 may be divided so that the quotient is a perfect cube, we can follow these steps:

### Step 1: Prime Factorization of 6750
First, we need to factorize the number 6750 into its prime factors.

- Start by dividing by the smallest prime number, which is 2:
\[ 6750 \div 2 = 3375 \]
- Next, we factor 3375. The sum of the digits (3 + 3 + 7 + 5 = 18) is divisible by 3, so we divide by 3:
\[ 3375 \div 3 = 1125 \]
- We can continue factoring 1125. Again, the sum of the digits (1 + 1 + 2 + 5 = 9) is divisible by 3:
\[ 1125 \div 3 = 375 \]
- Next, we factor 375. The sum of the digits (3 + 7 + 5 = 15) is also divisible by 3:
\[ 375 \div 3 = 125 \]
- Finally, we factor 125, which is \(5^3\):
\[ 125 \div 5 = 25 \quad (25 = 5^2) \quad \text{and} \quad 25 \div 5 = 5 \quad (5 = 5^1) \]

Thus, the complete prime factorization of 6750 is:
\[ 6750 = 2^1 \times 3^3 \times 5^3 \]

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

- In our factorization:
- The exponent of 2 is 1 (not a multiple of 3).
- The exponent of 3 is 3 (is a multiple of 3).
- The exponent of 5 is 3 (is a multiple of 3).

### Step 3: Adjust the Exponents
To make the exponent of 2 a multiple of 3, we need to increase it from 1 to 3. This means we need to divide by \(2^1\) (which is 2).

### Step 4: Conclusion
Therefore, the least number by which 6750 must be divided to make the quotient a perfect cube is:
\[ \text{Least number} = 2 \]

### Final Answer
The least number by which 6750 may be divided so that the quotient is a perfect cube is **2**.

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